Easy math puzzles

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February 23rd, 2021 at 6:43:34 PM permalink

I think I killed the thread.

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

chevy

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February 23rd, 2021 at 6:51:50 PM permalink

Quote: gordonm888

How many distinct arrangements of the following three letter words can be formed in a 3x3 grid of 9 squares?
SEX
EEL
OFF
WOW
ZZZ

I admit to not understanding what you are asking. I count at least 13 letters needed (SEXELOFFWWZZZ) to form the 5 words, but only a 3x3=9 grid.

Obviously I am missing something (After I get past that hurdle, then I can fall back on just not being able to solve)

chevy

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February 23rd, 2021 at 7:23:49 PM permalink

Unless??????????
Are we supposed to fit one word into the 3x3 grid in as many ways as possible

For example SEX

S corner : 15 ways * 4 corners = 60
S middle of side : 19 ways * 4 middles = 76
S center : 16 ways

Total = 152

Or if orientation of the board does not matter
SEX on top

same as

X
E
S on side

Then 15 + 19 + 4 = 38

?????????????

Then repeat for the other words?

ThatDonGuy

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February 23rd, 2021 at 7:26:19 PM permalink

Quote: chevy
Quote: gordonm888

How many distinct arrangements of the following three letter words can be formed in a 3x3 grid of 9 squares?
SEX
EEL
OFF
WOW
ZZZ

I admit to not understanding what you are asking. I count at least 13 letters needed (SEXELOFFWWZZZ) to form the 5 words, but only a 3x3=9 grid.

Obviously I am missing something (After I get past that hurdle, then I can fall back on just not being able to solve)

What I think he means is this:
How many ways can each of the three sets of letters be placed in a 3x3 grid so the second letter is adjacent to the first and the third is adjacent to the second? Each one is meant to be treated separately.

I assume that EEL and OFF have the same number, since each spelling of EEL can be changed to one of OFF by replacing the L with an O, the Es with Fs, and going in the reverse direction.

chevy

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February 23rd, 2021 at 7:41:32 PM permalink

Quote: ThatDonGuy

What I think he means is this:
How many ways can each of the three sets of letters be placed in a 3x3 grid so the second letter is adjacent to the first and the third is adjacent to the second? Each one is meant to be treated separately.

I assume that EEL and OFF have the same number, since each spelling of EEL can be changed to one of OFF by replacing the L with an O, the Es with Fs, and going in the reverse direction.

Okay, that makes sense. Two more questions

1) Does orientation matter?
SEX. (top)
vs
X
E
S
on left side
2) Palindromes ruled out....but what about same 3 squares in different order?
Squares 1,2,4

E E
L

Form EEL by 124 and 214

chevy

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February 23rd, 2021 at 8:02:19 PM permalink

If orientation and sequence matter I get the same for EEL and SEX, so I will assume they don't as it makes the problem more interesting. Listing ways with first letter in corner, side middle, and center...not counting duplicate triples.

SEX = 15 + 19 + 4 = 38 ways???
EEL = 15 + 13 + 0 = 28 ways???

OFF = 28 per ThatDonGuy's comment.

ZZZ= 10 + 2 + 0 = 12 ways ???

charliepatrick

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February 24th, 2021 at 8:18:59 AM permalink

Quote: gordonm888
...
How many distinct arrangements of the following three letter words can be formed in a 3x3 grid of 9 squares?
SEX
EEL
OFF
WOW
ZZZ

For simplicity assume the squares are:-

A B C
D E F
G H J

Then there are 84 sets of three squares. They fall into three types
(i) Linear - e.g. A-B-C - you can only go in a forwards or backwards direction as you cannot reach A from C - so ABC or CBA.
(ii) Circular - e.g. A-B-D - you can reach A from D so they form a circle (or triangle) - so ABD BDA DAB DBA BAD ADB.
(iii) Impossible - e.g. A-B G - you cannot form a joined triangle. You only need these to check the total combinations is correct. (I couldn't see a mathematical way of working out the combinations, so just listed them all. With larger numbers one might write a program to go through them, but then there would be more "shapes" than linear and circular.)

Linear
SEX/EEL/OFF - you can go forwards or backwards (2)
WOW/ZZZ - you can only go forwards (as the backwards is an identical placement of letters) (1)
Circular
SEX - where all the letters are different you can use any of the six ways (6)
EEL/OFF/WOW - where one of the letters is different it can be in any of three places (3)
ZZZ - where all the letters are different only one comination counts (1)

Number	Triple type	SEX	EEL	OFF	WOW	ZZZ
32	Linear	2	2	2	1	1
16	Circular	6	3	3	3	1
		160	112	112	80	48

chevy

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February 24th, 2021 at 9:55:04 AM permalink

Your solution is more elegant than me. I did the problem by a more brute force approach. I think I get similar answers for EEL/OFF and ZZZ.

But if we consider SEX....I considered possibilities with S in corner, S in Middle of a side, S in center....I can't find where I come up short of your number.

using you grid labels

corner: (15 ways)
S at A, E can be B.....4 choices for X....(D,E,F,C)
S at A, E can be D.....4 choices for X....(B,E,H,G)
S at A, E can be E.....7 choices for X....(B,C,F,J,H,G,D)

middle of side : (19 ways)
S at B, E can be at A......2 choices for X.....(D,E)
S at B, E can be at C......2 choices for X.....(E,F)
S at B, E can be at D......4 choices for X.....(A,E,H,G)
S at B, E can be at F......4 choices for X.....(C,E,H,J)
S at B, E can be at E.....7 choices for X.....(A,D,G,H,J,F,C)

center : (16 ways)
S at E, E can be at the other 8 locations, and X can be either side of E (around the perimeter)

Total ways to have SEX
=4 corners + 4 middles sides + 1 center
=4*15+4*19+16
=152 ways

Anybody see where I am missing some SEX.....8 of them apparently???

charliepatrick

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February 24th, 2021 at 11:01:41 AM permalink

Quote: chevy
Anybody see where I am missing some SEX.....8 of them apparently???

If you go from E to A, then the next place can only be D or B. However if you go from E to B, then you can go to A C D or F. This adds two more for EB, and similarly for ED EF and EH.

gordonm888
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February 24th, 2021 at 3:06:58 PM permalink

Quote: chevy
Okay, that makes sense. Two more questions

1) Does orientation matter?
SEX. (top)
vs
X
E
S
on left side
2) Palindromes ruled out....but what about same 3 squares in different order?
Squares 1,2,4

E E
L

Form EEL by 124 and 214

Orientation does not matter. Words can be read from the bottom up or backwards or forwards, etc. They may be twisted into what I call a triangle, such as

SX
E

or

XE
S

or whatever.

Palindromes simply lack the ability to be sequenced backwards in a particular set of squares because that would not count as a distinct arrangement with respect to being sequenced forwards.

You are correct about off and eel having the same number of arrangements. I listed both of them just to have the complete set of symmetries for a 3-character string.

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

gordonm888
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February 24th, 2021 at 3:07:01 PM permalink

Quote: chevy
Okay, that makes sense. Two more questions

1) Does orientation matter?
SEX. (top)
vs
X
E
S
on left side
2) Palindromes ruled out....but what about same 3 squares in different order?
Squares 1,2,4

E E
L

Form EEL by 124 and 214

Orientation does not matter. Words can be read from the bottom up or backwards or forwards, etc. They may be twisted into what I call a triangle, such as

SX
E

or

XE
S

or whatever.

Palindromes simply lack the ability to be sequenced backwards in a particular set of squares because that would not count as a distinct arrangement with respect to being sequenced forwards.

You are correct about off and eel having the same number of arrangements. I listed both of them just to have the complete set of symmetries for a 3-character string.

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

gordonm888
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February 24th, 2021 at 3:16:42 PM permalink

Quote: charliepatrick
For simplicity assume the squares are:-
A B C
D E F
G H J
Then there are 84 sets of three squares. They fall into three types
(i) Linear - e.g. A-B-C - you can only go in a forwards or backwards direction as you cannot reach A from C - so ABC or CBA.
(ii) Circular - e.g. A-B-D - you can reach A from D so they form a circle (or triangle) - so ABD BDA DAB DBA BAD ADB.
(iii) Impossible - e.g. A-B G - you cannot form a joined triangle. You only need these to check the total combinations is correct. (I couldn't see a mathematical way of working out the combinations, so just listed them all. With larger numbers one might write a program to go through them, but then there would be more "shapes" than linear and circular.)

Linear
SEX/EEL/OFF - you can go forwards or backwards (2)
WOW/ZZZ - you can only go forwards (as the backwards is an identical placement of letters) (1)
Circular
SEX - where all the letters are different you can use any of the six ways (6)
EEL/OFF/WOW - where one of the letters is different it can be in any of three places (3)
ZZZ - where all the letters are different only one comination counts (1)
Number Triple type SEX EEL OFF WOW ZZZ

32
Linear
2
2
2
1
1

16
Circular
6
3
3
3
1

160
112
112
80
48

This is absolutely CORRECT, and using the same methodology that I use.

The methodology used by Charlie allows you to straightforwardly address how many three character strings would be on a 4x4 grid or more generally, on an n x n grid.

This methodology is more complex when you are analyzing strings that are 4, 5 and 6 characters in length because there are far more spatial arrangements that the string can be twisted into, especially for 4x4 and larger grids, Also there are far more symmetrical arrangements for repeated characters in the strings that need to be considered. I have never attempted a 7 character string.

And I really wish I was smart enough to figure out an approach for calculating how many ways a 9 character string can be arranged in a 3x3 grid. I've never really seriously attempted that.

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

chevy

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February 24th, 2021 at 3:32:49 PM permalink

Quote: charliepatrick
If you go from E to A, then the next place can only be D or B. However if you go from E to B, then you can go to A C D or F. This adds two more for EB, and similarly for ED EF and EH.

Thank you very much!!! I guess I ended my counting of SEX from the center square prematurely.

charliepatrick

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February 25th, 2021 at 2:11:12 AM permalink

Quote: gordonm888
....I really wish I was smart enough to figure out an approach for calculating how many ways a 9 character string can be arranged in a 3x3 grid....

I didn't know how to do this so initially did it by hand. However when the alternative answer of 152 came in I couldn't see the 8 missing ones, so only then knocked up a program to see which answer was correct. This could easily be extended for larger grids but doesn't find nor define which ones have which patterns. It can only really be used to confirm the total number of valid strings.
(Some lines omitted - for larger grids I'd do it the other way round as there are fewer valid links than invalid ones)


   n=0;
   for (i=1; i<=9; i++)
      {
      for (j=1; j<=9; j++)
         {
         for (k=1; k<=9; k++)
            {
            invalid=0;
            if (i==j) {invalid=1;};
            if (i==k) {invalid=1;};
            if (j==k) {invalid=1;};
            /* Check first move is legal */
            if ((i==1)&&(j==3)) {invalid=1;};
            if ((i==1)&&(j==6)) {invalid=1;};
etc.
            if ((i==9)&&(j==4)) {invalid=1;};
            if ((i==9)&&(j==7)) {invalid=1;};

            /* Check second move is legal */
            if ((j==1)&&(k==3)) {invalid=1;};
            if ((j==1)&&(k==6)) {invalid=1;};
etc.
            if ((j==9)&&(k==4)) {invalid=1;};
            if ((j==9)&&(k==7)) {invalid=1;};

            if (invalid==0)
               {
               n++;
               outline="N("+n+") ";
               outline+="I: "+i+" ";
               outline+="J: "+j+" ";
               outline+="K: "+k+" ";
               outline+="<BR>";
               document.write(outline);
               };
            };
         };
      };

Gialmere

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February 25th, 2021 at 8:28:54 AM permalink

You don't play golf but decide to try a bizarre online system for the game that uses meditation and self hypnosis.

Before playing, you intensely concentrate on two strokes of different lengths to use for the entire round, one a drive, the other an approach, and play directly toward the hole so that various combinations of the two distances will get you there.

The ball must go the full length on each stroke, but you may go beyond the hole with either stroke, then play back toward the hole. All strokes are on a straight line toward the hole and will go in the cup if the stroke and hole distances are equal.

Using this odd technique of selecting two exact distances only, what is your lowest score possible on the following nine-hole course?

Hole	Length	Score
#1	150 yds	?
#2	300 yds	?
#3	250 yds	?
#4	325 yds	?
#5	275 yds	?
#6	350 yds	?
#7	225 yds	?
#8	450 yds	?
#9	425 yds	?

Have you tried 22 tonight? I said 22.

Joeman

Joeman

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February 25th, 2021 at 9:20:39 AM permalink

I can get down to 28 strokes.

Similar to my actual golf game, my Drive would be 150 yd, and my Approach would be 25 yd. Out in 28:

Hole	Length	Score	Shots
#1	150 yds	1	150
#2	300 yds	2	150 + 150
#3	250 yds	4	150 + 150 - 25 - 25
#4	325 yds	3	150 + 150 + 25
#5	275 yds	3	150 + 150 - 25
#6	350 yds	4	150 + 150 + 25 + 25
#7	225 yds	4	150 + 25 + 25 + 25
#8	450 yds	3	150 + 150 + 150
#9	425 yds	4	150 + 150 + 150 - 25

"Dealer has 'rock'... Pay 'paper!'"

charliepatrick

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February 25th, 2021 at 9:23:34 AM permalink

I'll have to think about an analytical approach to this but I did like my first guess....

Dividing all the lengths by 25 gets 6 12 10 13 11 14 9 17 18 = 6 17 18 (9 thru 14). 7 and 3 (i.e. 175 and 75 yards) gives 26 shots. (2 4 2 3 3 2 3 3 4).

Wizard
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Wizard

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February 25th, 2021 at 9:28:53 AM permalink

Chag Purim Sameach!, my fellow math enthusiasts.

Here is my guess,

30

75 and 125 yards. I can complete the course with a score of 30.

I am not saying this is optimal, but have a good feeling about it.

I first divided all the yardages by 25 to get:

6
12
10
13
11
14
9
18
17

Then I just know somewhere that primes are good for constructing other numbers. With a little trial and error 3 and 5 seem to work well.

For example,
hole 1: 3+3
hole 2: 5+5+5-3
hole 3: 5+5

Can anyone beat my 30?

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

Wizard
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Wizard

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February 25th, 2021 at 9:55:42 AM permalink

Looks like Charlie is winning the WoV golf match so far and I'm losing, which is usually the case when I golf.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

ThatDonGuy

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February 25th, 2021 at 11:20:03 AM permalink

Quote: Wizard
Looks like Charlie is winning the WoV golf match so far and I'm losing, which is usually the case when I golf.

I did a brute force search on every pair of driver and approach clubs in 25-yard intervals up to 500 yards, and get the same answer Charlie got.

Note that X indicates that reaching that hole is impossible

Long	Short	1	2	3	4	5	6	7	8	9	Score
50	25	3	6	5	7	6	7	5	9	9	57
75	25	2	4	4	5	5	6	3	6	7	42
75	50	2	4	4	5	4	5	3	6	6	39
100	25	3	3	4	4	4	5	3	6	5	37
100	50	2	3	3	X	X	4	X	5	X	X
100	75	2	3	3	4	3	4	3	5	5	32
125	25	2	4	2	5	3	4	3	6	5	34
125	50	3	3	2	4	4	4	3	5	4	32
125	75	2	4	2	3	3	4	3	4	5	30
125	100	3	3	2	3	4	3	2	4	4	28
150	25	1	2	4	3	3	4	4	3	4	28
150	50	1	2	3	X	X	3	X	3	X	X
150	75	1	2	X	X	X	X	2	3	X	X
150	100	1	2	2	X	X	3	X	3	X	X
150	125	1	2	2	4	2	6	7	3	3	30
175	25	2	4	4	3	5	2	3	6	5	34
175	50	3	3	4	4	3	2	2	4	5	30
175	75	2	4	2	3	3	2	3	4	3	26
175	100	4	3	3	5	2	2	6	3	4	32
175	125	6	2	2	7	5	2	3	6	3	36
175	150	1	2	7	2	9	2	5	3	8	39
200	25	3	5	3	5	4	4	2	4	3	33
200	50	2	3	2	X	X	3	X	3	X	X
200	75	2	4	4	3	2	3	3	5	4	30
200	100	X	2	X	X	X	X	X	X	X	X
200	125	4	8	2	2	3	5	6	3	7	40
200	150	1	2	3	X	X	2	X	3	X	X
200	175	12	9	5	11	7	2	3	6	4	59
225	25	4	4	2	5	3	6	1	2	3	30
225	50	3	5	5	3	2	4	1	2	5	30
225	75	2	2	X	X	X	X	1	2	X	X
225	100	5	3	4	2	7	3	1	2	3	30
225	125	10	6	2	3	9	2	1	2	5	40
225	150	1	2	X	X	X	X	1	2	X	X
225	175	6	12	10	5	3	2	1	2	9	50
225	200	12	7	3	9	5	11	1	2	2	52
250	25	5	3	1	4	2	5	2	4	5	31
250	50	3	2	1	X	X	3	X	3	X	X
250	75	2	4	1	2	5	4	3	6	3	30
250	100	2	3	1	X	X	2	X	3	X	X
250	125	X	X	1	X	X	X	X	X	X	X
250	150	1	2	1	X	X	3	X	3	X	X
250	175	4	8	1	3	13	2	6	12	2	51
250	200	6	3	1	X	X	5	X	2	X	X
250	225	12	5	1	7	3	9	1	2	15	55
275	25	6	2	2	3	1	4	3	6	7	34
275	50	3	6	5	2	1	6	2	4	4	33
275	75	2	4	6	5	1	2	3	6	3	32
275	100	6	3	5	8	1	4	9	3	7	46
275	125	2	4	2	7	1	10	11	6	3	46
275	150	1	2	4	12	1	9	7	3	2	41
275	175	12	6	14	11	1	2	9	2	13	70
275	200	4	8	13	15	1	3	6	12	5	67
275	225	6	12	10	3	1	14	1	2	7	56
275	250	12	3	1	5	1	7	18	15	13	75
300	25	6	1	3	2	2	3	4	7	6	34
300	50	3	1	2	X	X	2	X	4	X	X
300	75	2	1	X	X	X	X	2	3	X	X
300	100	X	1	X	X	X	X	X	X	X	X
300	125	9	1	2	11	8	4	5	10	2	52
300	150	1	1	X	X	X	X	X	2	X	X
300	175	10	1	4	9	12	2	15	11	3	67
300	200	X	1	X	X	X	X	X	X	X	X
300	225	4	1	X	X	X	X	1	2	X	X
300	250	6	1	1	X	X	3	X	7	X	X
300	275	12	1	20	3	1	5	18	13	11	84
325	25	6	2	4	1	3	2	5	6	5	34
325	50	3	6	5	1	2	7	3	6	3	36
325	75	2	4	2	1	7	6	3	6	5	36
325	100	7	3	6	1	10	5	2	4	2	40
325	125	6	12	2	1	5	8	9	2	11	56
325	150	1	2	11	1	14	4	8	3	13	57
325	175	2	4	10	1	7	2	13	6	9	54
325	200	15	9	4	1	17	14	12	3	11	86
325	225	14	6	16	1	11	18	1	2	3	72
325	250	4	8	1	1	15	17	6	12	19	83
325	275	6	12	10	1	1	14	21	18	5	88
325	300	12	1	20	1	22	3	18	11	9	97
350	25	6	3	5	2	4	1	6	5	4	36
350	50	3	2	3	X	X	1	X	3	X	X
350	75	2	4	8	7	2	1	3	6	2	35
350	100	3	3	2	X	X	1	X	2	X	X
350	125	14	9	2	5	13	1	2	4	8	58
350	150	1	2	5	X	X	1	X	3	X	X
350	175	X	X	X	X	X	1	X	X	X	X
350	200	2	4	7	X	X	1	X	6	X	X
350	225	7	14	4	19	9	1	1	2	16	73
350	250	9	6	1	X	X	1	X	3	X	X
350	275	4	8	15	17	1	1	6	12	3	67
350	300	6	1	10	X	X	1	X	5	X	X
350	325	12	24	20	1	22	1	18	9	7	114
375	25	6	4	6	3	5	2	7	4	3	40
375	50	3	6	5	2	3	7	4	8	2	40
375	75	2	2	X	X	X	X	3	2	X	X
375	100	8	3	7	11	2	6	12	5	10	64
375	125	X	X	2	X	X	X	X	X	X	X
375	150	1	2	X	X	X	X	2	3	X	X
375	175	18	14	8	17	11	2	5	10	7	92
375	200	5	10	16	7	13	4	19	15	18	107
375	225	2	4	X	X	X	X	1	2	X	X
375	250	X	X	1	X	X	X	X	X	X	X
375	275	16	6	18	13	1	20	11	22	15	122
375	300	4	1	X	X	X	X	6	3	X	X
375	325	6	12	10	1	25	14	23	18	3	112
375	350	12	24	20	26	22	1	18	7	5	135
400	25	6	5	7	4	6	3	8	3	2	44
400	50	3	3	4	X	X	2	X	2	X	X
400	75	2	4	3	2	9	8	3	6	7	44
400	100	X	2	X	X	X	X	X	X	X	X
400	125	3	6	2	10	2	14	15	9	5	66
400	150	1	2	2	X	X	5	X	3	X	X
400	175	9	18	15	8	5	2	2	4	14	77
400	200	X	X	X	X	X	X	X	X	X	X
400	225	16	7	10	18	21	4	1	2	12	91
400	250	2	4	1	X	X	9	X	6	X	X
400	275	24	21	4	16	1	11	9	18	23	127
400	300	X	1	X	X	X	X	X	X	X	X
400	325	4	8	26	1	17	19	6	12	21	114
400	350	6	12	10	X	X	1	X	3	X	X
400	375	12	24	20	26	22	28	18	5	3	158
425	25	6	6	8	5	7	4	9	2	1	48
425	50	3	6	5	3	4	7	5	9	1	43
425	75	2	4	10	9	3	2	3	6	1	40
425	100	9	3	8	2	13	7	3	6	1	52
425	125	12	2	2	15	11	6	7	14	1	70
425	150	1	2	6	17	2	13	10	3	1	55
425	175	6	12	2	5	19	2	9	18	1	74
425	200	18	11	5	14	8	17	2	4	1	80
425	225	8	16	22	13	19	10	1	2	1	92
425	250	21	15	1	23	7	4	18	9	1	99
425	275	2	4	22	9	1	14	17	6	1	76
425	300	14	1	4	11	16	23	21	13	1	104
425	325	18	6	20	1	13	22	27	24	1	132
425	350	4	8	17	19	28	1	6	12	1	96
425	375	6	12	10	29	27	14	25	18	1	142
425	400	12	24	20	26	22	28	18	3	1	154
450	25	6	7	9	6	8	5	9	1	2	53
450	50	3	4	5	X	X	3	X	1	X	X
450	75	2	3	X	X	X	X	3	1	X	X
450	100	4	3	3	X	X	2	X	1	X	X
450	125	8	16	2	2	7	11	12	1	15	74
450	150	1	2	X	X	X	X	X	1	X	X
450	175	17	9	20	16	2	2	13	1	19	99
450	200	9	5	2	X	X	8	X	1	X	X
450	225	X	X	X	X	X	X	1	1	X	X
450	250	5	10	1	X	X	7	X	1	X	X
450	275	10	20	7	12	1	4	15	1	9	79
450	300	2	1	X	X	X	X	X	1	X	X
450	325	21	11	4	1	23	18	16	1	13	108
450	350	11	6	13	X	X	1	X	1	X	X
450	375	4	8	X	X	X	X	6	1	X	X
450	400	6	12	10	X	X	14	X	1	X	X
450	425	12	24	20	26	22	28	18	1	1	152
475	25	6	8	10	7	9	6	9	2	3	60
475	50	3	6	5	4	5	7	6	9	2	47
475	75	2	4	4	3	11	10	3	6	9	52
475	100	10	3	9	14	3	8	15	7	13	82
475	125	18	12	2	7	17	2	3	6	11	78
475	150	1	2	15	2	19	6	11	3	18	77
475	175	14	2	6	13	17	2	21	16	5	96
475	200	6	12	19	22	2	5	9	18	8	101
475	225	18	8	2	11	5	14	1	2	23	84
475	250	11	22	1	19	25	16	2	4	7	107
475	275	24	18	10	7	1	26	21	12	23	142
475	300	15	1	25	17	12	4	7	14	27	122
475	325	2	4	14	1	9	26	19	6	11	92
475	350	9	18	4	25	11	1	30	27	20	145
475	375	20	6	22	15	31	24	13	26	17	174
475	400	4	8	30	32	19	21	6	12	23	155
475	425	6	12	10	31	29	14	27	18	1	148
475	450	12	24	20	26	22	28	18	1	34	185
500	25	6	9	10	8	10	7	9	3	4	66
500	50	3	5	5	X	X	4	X	2	X	X
500	75	2	4	12	11	4	3	3	6	2	47
500	100	X	3	X	X	X	X	X	X	X	X
500	125	X	X	2	X	X	X	X	X	X	X
500	150	1	2	7	X	X	2	X	3	X	X
500	175	3	6	14	2	10	2	18	9	13	77
500	200	X	2	X	X	X	X	X	X	X	X
500	225	9	18	15	5	2	21	1	2	11	84
500	250	X	X	1	X	X	X	X	X	X	X
500	275	22	13	16	27	1	10	2	4	21	116
500	300	X	1	X	X	X	X	X	X	X	X
500	325	30	27	17	1	22	4	12	24	19	156
500	350	2	4	9	X	X	1	X	6	X	X
500	375	X	X	4	X	X	X	X	X	X	X
500	400	X	6	X	X	X	X	X	X	X	X
500	425	4	8	19	21	32	34	6	12	1	137
500	450	6	12	10	X	X	14	X	1	X	X
500	475	12	24	20	26	22	28	18	36	34	220

Last edited by: ThatDonGuy on Feb 25, 2021

charliepatrick

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February 25th, 2021 at 11:38:05 AM permalink

Thanks - the guess was only based on an observation...

the logic that you needed a long club to get near the higher numbers and with 7/3 I could see you could easily get 10, 14 and 17.

Gialmere

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Posts: 3044

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February 25th, 2021 at 5:20:56 PM permalink

Quote: charliepatrick
I'll have to think about an analytical approach to this but I did like my first guess....
Dividing all the lengths by 25 gets 6 12 10 13 11 14 9 17 18 = 6 17 18 (9 thru 14). 7 and 3 (i.e. 175 and 75 yards) gives 26 shots. (2 4 2 3 3 2 3 3 4).

Correct!!

Quote: ThatDonGuy
I did a brute force search on every pair of driver and approach clubs in 25-yard intervals up to 500 yards, and get the same answer Charlie got.

Note that X indicates that reaching that hole is impossible
Long Short 1 2 3 4 5 6 7 8 9 Score
50 25 3 6 5 7 6 7 5 9 9 57
75 25 2 4 4 5 5 6 3 6 7 42
75 50 2 4 4 5 4 5 3 6 6 39
100 25 3 3 4 4 4 5 3 6 5 37
100 50 2 3 3 X X 4 X 5 X X
100 75 2 3 3 4 3 4 3 5 5 32
125 25 2 4 2 5 3 4 3 6 5 34
125 50 3 3 2 4 4 4 3 5 4 32
125 75 2 4 2 3 3 4 3 4 5 30
125 100 3 3 2 3 4 3 2 4 4 28
150 25 1 2 4 3 3 4 4 3 4 28
150 50 1 2 3 X X 3 X 3 X X
150 75 1 2 X X X X 2 3 X X
150 100 1 2 2 X X 3 X 3 X X
150 125 1 2 2 4 2 6 7 3 3 30
175 25 2 4 4 3 5 2 3 6 5 34
175 50 3 3 4 4 3 2 2 4 5 30
175 75 2 4 2 3 3 2 3 4 3 26
175 100 4 3 3 5 2 2 6 3 4 32
175 125 6 2 2 7 5 2 3 6 3 36
175 150 1 2 7 2 9 2 5 3 8 39
200 25 3 5 3 5 4 4 2 4 3 33
200 50 2 3 2 X X 3 X 3 X X
200 75 2 4 4 3 2 3 3 5 4 30
200 100 X 2 X X X X X X X X
200 125 4 8 2 2 3 5 6 3 7 40
200 150 1 2 3 X X 2 X 3 X X
200 175 12 9 5 11 7 2 3 6 4 59
225 25 4 4 2 5 3 6 1 2 3 30
225 50 3 5 5 3 2 4 1 2 5 30
225 75 2 2 X X X X 1 2 X X
225 100 5 3 4 2 7 3 1 2 3 30
225 125 10 6 2 3 9 2 1 2 5 40
225 150 1 2 X X X X 1 2 X X
225 175 6 12 10 5 3 2 1 2 9 50
225 200 12 7 3 9 5 11 1 2 2 52
250 25 5 3 1 4 2 5 2 4 5 31
250 50 3 2 1 X X 3 X 3 X X
250 75 2 4 1 2 5 4 3 6 3 30
250 100 2 3 1 X X 2 X 3 X X
250 125 X X 1 X X X X X X X
250 150 1 2 1 X X 3 X 3 X X
250 175 4 8 1 3 13 2 6 12 2 51
250 200 6 3 1 X X 5 X 2 X X
250 225 12 5 1 7 3 9 1 2 15 55
275 25 6 2 2 3 1 4 3 6 7 34
275 50 3 6 5 2 1 6 2 4 4 33
275 75 2 4 6 5 1 2 3 6 3 32
275 100 6 3 5 8 1 4 9 3 7 46
275 125 2 4 2 7 1 10 11 6 3 46
275 150 1 2 4 12 1 9 7 3 2 41
275 175 12 6 14 11 1 2 9 2 13 70
275 200 4 8 13 15 1 3 6 12 5 67
275 225 6 12 10 3 1 14 1 2 7 56
275 250 12 3 1 5 1 7 18 15 13 75
300 25 6 1 3 2 2 3 4 7 6 34
300 50 3 1 2 X X 2 X 4 X X
300 75 2 1 X X X X 2 3 X X
300 100 X 1 X X X X X X X X
300 125 9 1 2 11 8 4 5 10 2 52
300 150 1 1 X X X X X 2 X X
300 175 10 1 4 9 12 2 15 11 3 67
300 200 X 1 X X X X X X X X
300 225 4 1 X X X X 1 2 X X
300 250 6 1 1 X X 3 X 7 X X
300 275 12 1 20 3 1 5 18 13 11 84
325 25 6 2 4 1 3 2 5 6 5 34
325 50 3 6 5 1 2 7 3 6 3 36
325 75 2 4 2 1 7 6 3 6 5 36
325 100 7 3 6 1 10 5 2 4 2 40
325 125 6 12 2 1 5 8 9 2 11 56
325 150 1 2 11 1 14 4 8 3 13 57
325 175 2 4 10 1 7 2 13 6 9 54
325 200 15 9 4 1 17 14 12 3 11 86
325 225 14 6 16 1 11 18 1 2 3 72
325 250 4 8 1 1 15 17 6 12 19 83
325 275 6 12 10 1 1 14 21 18 5 88
325 300 12 1 20 1 22 3 18 11 9 97
350 25 6 3 5 2 4 1 6 5 4 36
350 50 3 2 3 X X 1 X 3 X X
350 75 2 4 8 7 2 1 3 6 2 35
350 100 3 3 2 X X 1 X 2 X X
350 125 14 9 2 5 13 1 2 4 8 58
350 150 1 2 5 X X 1 X 3 X X
350 175 X X X X X 1 X X X X
350 200 2 4 7 X X 1 X 6 X X
350 225 7 14 4 19 9 1 1 2 16 73
350 250 9 6 1 X X 1 X 3 X X
350 275 4 8 15 17 1 1 6 12 3 67
350 300 6 1 10 X X 1 X 5 X X
350 325 12 24 20 1 22 1 18 9 7 114
375 25 6 4 6 3 5 2 7 4 3 40
375 50 3 6 5 2 3 7 4 8 2 40
375 75 2 2 X X X X 3 2 X X
375 100 8 3 7 11 2 6 12 5 10 64
375 125 X X 2 X X X X X X X
375 150 1 2 X X X X 2 3 X X
375 175 18 14 8 17 11 2 5 10 7 92
375 200 5 10 16 7 13 4 19 15 18 107
375 225 2 4 X X X X 1 2 X X
375 250 X X 1 X X X X X X X
375 275 16 6 18 13 1 20 11 22 15 122
375 300 4 1 X X X X 6 3 X X
375 325 6 12 10 1 25 14 23 18 3 112
375 350 12 24 20 26 22 1 18 7 5 135
400 25 6 5 7 4 6 3 8 3 2 44
400 50 3 3 4 X X 2 X 2 X X
400 75 2 4 3 2 9 8 3 6 7 44
400 100 X 2 X X X X X X X X
400 125 3 6 2 10 2 14 15 9 5 66
400 150 1 2 2 X X 5 X 3 X X
400 175 9 18 15 8 5 2 2 4 14 77
400 200 X X X X X X X X X X
400 225 16 7 10 18 21 4 1 2 12 91
400 250 2 4 1 X X 9 X 6 X X
400 275 24 21 4 16 1 11 9 18 23 127
400 300 X 1 X X X X X X X X
400 325 4 8 26 1 17 19 6 12 21 114
400 350 6 12 10 X X 1 X 3 X X
400 375 12 24 20 26 22 28 18 5 3 158
425 25 6 6 8 5 7 4 9 2 1 48
425 50 3 6 5 3 4 7 5 9 1 43
425 75 2 4 10 9 3 2 3 6 1 40
425 100 9 3 8 2 13 7 3 6 1 52
425 125 12 2 2 15 11 6 7 14 1 70
425 150 1 2 6 17 2 13 10 3 1 55
425 175 6 12 2 5 19 2 9 18 1 74
425 200 18 11 5 14 8 17 2 4 1 80
425 225 8 16 22 13 19 10 1 2 1 92
425 250 21 15 1 23 7 4 18 9 1 99
425 275 2 4 22 9 1 14 17 6 1 76
425 300 14 1 4 11 16 23 21 13 1 104
425 325 18 6 20 1 13 22 27 24 1 132
425 350 4 8 17 19 28 1 6 12 1 96
425 375 6 12 10 29 27 14 25 18 1 142
425 400 12 24 20 26 22 28 18 3 1 154
450 25 6 7 9 6 8 5 9 1 2 53
450 50 3 4 5 X X 3 X 1 X X
450 75 2 3 X X X X 3 1 X X
450 100 4 3 3 X X 2 X 1 X X
450 125 8 16 2 2 7 11 12 1 15 74
450 150 1 2 X X X X X 1 X X
450 175 17 9 20 16 2 2 13 1 19 99
450 200 9 5 2 X X 8 X 1 X X
450 225 X X X X X X 1 1 X X
450 250 5 10 1 X X 7 X 1 X X
450 275 10 20 7 12 1 4 15 1 9 79
450 300 2 1 X X X X X 1 X X
450 325 21 11 4 1 23 18 16 1 13 108
450 350 11 6 13 X X 1 X 1 X X
450 375 4 8 X X X X 6 1 X X
450 400 6 12 10 X X 14 X 1 X X
450 425 12 24 20 26 22 28 18 1 1 152
475 25 6 8 10 7 9 6 9 2 3 60
475 50 3 6 5 4 5 7 6 9 2 47
475 75 2 4 4 3 11 10 3 6 9 52
475 100 10 3 9 14 3 8 15 7 13 82
475 125 18 12 2 7 17 2 3 6 11 78
475 150 1 2 15 2 19 6 11 3 18 77
475 175 14 2 6 13 17 2 21 16 5 96
475 200 6 12 19 22 2 5 9 18 8 101
475 225 18 8 2 11 5 14 1 2 23 84
475 250 11 22 1 19 25 16 2 4 7 107
475 275 24 18 10 7 1 26 21 12 23 142
475 300 15 1 25 17 12 4 7 14 27 122
475 325 2 4 14 1 9 26 19 6 11 92
475 350 9 18 4 25 11 1 30 27 20 145
475 375 20 6 22 15 31 24 13 26 17 174
475 400 4 8 30 32 19 21 6 12 23 155
475 425 6 12 10 31 29 14 27 18 1 148
475 450 12 24 20 26 22 28 18 1 34 185
500 25 6 9 10 8 10 7 9 3 4 66
500 50 3 5 5 X X 4 X 2 X X
500 75 2 4 12 11 4 3 3 6 2 47
500 100 X 3 X X X X X X X X
500 125 X X 2 X X X X X X X
500 150 1 2 7 X X 2 X 3 X X
500 175 3 6 14 2 10 2 18 9 13 77
500 200 X 2 X X X X X X X X
500 225 9 18 15 5 2 21 1 2 11 84
500 250 X X 1 X X X X X X X
500 275 22 13 16 27 1 10 2 4 21 116
500 300 X 1 X X X X X X X X
500 325 30 27 17 1 22 4 12 24 19 156
500 350 2 4 9 X X 1 X 6 X X
500 375 X X 4 X X X X X X X
500 400 X 6 X X X X X X X X
500 425 4 8 19 21 32 34 6 12 1 137
500 450 6 12 10 X X 14 X 1 X X
500 475 12 24 20 26 22 28 18 36 34 220

If you're looking for another round of golf...

According to the programming website I found this puzzle at, in addition to the set {175, 75}, there is another set of numbers that will complete the course in the minimum of 26 strokes. They don't post the answer but, from ThatDonGuy's list, they're not a multiples of 25 (and possibly not even whole numbers).

--------------------------------

Have you tried 22 tonight? I said 22.

ThatDonGuy

Threads: 122
Posts: 6658

Joined: Jun 22, 2011

February 25th, 2021 at 6:23:01 PM permalink

Quote: Gialmere
If you're looking for another round of golf...

According to the programming website I found this puzzle at, in addition to the set {175, 75}, there is another set of numbers that will complete the course in the minimum of 26 strokes. They don't post the answer but, from ThatDonGuy's list, they're not a multiples of 25 (and possibly not even whole numbers).

I have checked every combination of clubs that are multiples of 0.1 yards up to 900, but can't find any other than the one posted solution.

Gialmere

Threads: 45
Posts: 3044

Joined: Nov 26, 2018

February 25th, 2021 at 8:32:56 PM permalink

Quote: ThatDonGuy
Quote: Gialmere
If you're looking for another round of golf...

According to the programming website I found this puzzle at, in addition to the set {175, 75}, there is another set of numbers that will complete the course in the minimum of 26 strokes. They don't post the answer but, from ThatDonGuy's list, they're not a multiples of 25 (and possibly not even whole numbers).

I have checked every combination of clubs that are multiples of 0.1 yards up to 900, but can't find any other than the one posted solution.

Oops! My bad. I messed up one of the hole lengths. Hole #8 is only 400 yds. Fortunately the 75/175 solve works for either length but, apologies to ThatDonGuy.

The correct length does, however, make the second solve much simpler.

Have you tried 22 tonight? I said 22.

charliepatrick

Threads: 39
Posts: 3011

Joined: Jun 17, 2011

February 25th, 2021 at 9:45:38 PM permalink

Quote: Gialmere
...
Oops! My bad. I messed up one of the hole lengths. Hole #8 is only 400 yds. Fortunately the 75/175 solve works for either length but, apologies to ThatDonGuy....

Yes 150 125 works (1 2 2 4 2 5 4 3 3). 200/75 and 250/75 now get close with 27 shots.
btw I couldn't see how fractional ones could easily work as to get to 26 strokes you need at least one hole which is a 2. So if the clubs were X+fraction and Y-fraction (fraction=1/2 1/3 etc) then the only way to get whole numbers is X-Y 2X-2Y nX etc. Apart from 6 and 12 you have no others that are double.
Also you know that some combinations can't work, such as when both clubs are a multiple of 50 or where one club is a multiple of the other. However if using a brute force method that's not worth worrying about.

teliot

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Posts: 2871

Joined: Oct 19, 2009

February 25th, 2021 at 10:10:00 PM permalink

This very nice problem about golfing is also related to the postage stamp problem and minimizing total stamp usage across a variety of postages. Well, maybe not that related since you can't have negative stamps.

Postage Stamp Problem
Given stamps with postage x and y, relatively prime, you can always make every postage above (x - 1) * (y - 1).

I believe finding a closed form solution for the postage stamp problem for three postages is still unsolved.

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ThatDonGuy

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Posts: 6658

Joined: Jun 22, 2011

February 26th, 2021 at 7:15:42 AM permalink

Quote: Gialmere
Oops! My bad. I messed up one of the hole lengths. Hole #8 is only 400 yds. Fortunately the 75/175 solve works for either length but, apologies to ThatDonGuy.

The correct length does, however, make the second solve much simpler.

I checked every possible pair of different clubs that are multiples of 1/5, 1/6, 1/7, and 1/8 yard up to 1000 yards, and still get 175/75 as the only 26-stroke solution.

Update: now up through 1/14 of a yard - still only one solution for 26 (and two for 27).

Last edited by: ThatDonGuy on Feb 26, 2021

gordonm888
Administrator

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February 26th, 2021 at 6:18:21 PM permalink

EDIT:
If the 8th hole is only 400 yds, rather than 450:

250 and 75 seems to yield just 27 strokes

150 =75+75_____2
300=75 +75+ 75+75__4
250 = 250 ________1
325 = 250 + 75______2
275 = 250 -75 -75 -75 +250_________5
350 = 250 -75 -75 +250_______4
225 = 75 +75 +75 _______3
400 = 250 +75 +75 _________________3
425 = 250 -75 +250 ____3

EDit: I see that ThatDonGuy has already overwhelmed the problem.

Last edited by: gordonm888 on Feb 26, 2021

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

teliot

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February 26th, 2021 at 7:02:26 PM permalink

On the topic of Golfing, let me create a different problem that is based on efficient stamp totals.

Suppose the Post Office issues two values of stamps X cents and Y cents. What are the most efficient values for these two stamps X & Y so that we can make each postage from $1.20 to $2.00, using the least number of stamps overall?

Yeah, "Easy" Math Puzzles. :)

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gordonm888
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February 26th, 2021 at 7:18:25 PM permalink

Quote: teliot
On the topic of Golfing, let me create a different problem that is based on efficient stamp totals.

Suppose the Post Office issues two values of stamps X cents and Y cents. What are the most efficient values for these two stamps X & Y so that we can make each postage from $1.20 to $2.00, using the least number of stamps overall?

Yeah, "Easy" Math Puzzles. :)

My quick guess is 12 cents and 11 cents.

We need two numbers -one, call it n, that is even and the other that has multiples that are capable of making xmodn, x=1...n-1.

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

teliot

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February 26th, 2021 at 9:46:02 PM permalink

Quote: gordonm888

My quick guess is 12 cents and 11 cents.

We need two numbers -one, call it n, that is even and the other that has multiples that are capable of making xmodn, x=1...n-1.

Your answer requires a total of 1118 stamps. My answer requires 1072 stamps.

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charliepatrick

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February 27th, 2021 at 1:34:02 AM permalink

Quote: teliot
...My answer requires 1072 stamps.

I think you accidentally (or not) gave a hint earlier as 120 = (13-1)*(11-1) and both 13 and 11 are primes; and they seem to work. The early values can be achieved in either 10 or 11 stamps, e.g 120 = 5*each, 121=11*11c, 122 is 4*11c+6*13c, so to get 2 more you replace one 11 with a 13. At the end 199 needs 17 stamps, and 200 (which is 208-8 so 12*13+4*11 requires 16 stamps.) The total is 1072.

teliot

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February 27th, 2021 at 6:04:09 AM permalink

Quote: charliepatrick
I think you accidentally (or not) gave a hint earlier as 120 = (13-1)*(11-1) and both 13 and 11 are primes; and they seem to work. The early values can be achieved in either 10 or 11 stamps, e.g 120 = 5*each, 121=11*11c, 122 is 4*11c+6*13c, so to get 2 more you replace one 11 with a 13. At the end 199 needs 17 stamps, and 200 (which is 208-8 so 12*13+4*11 requires 16 stamps.) The total is 1072.

Yes!

Now, what's the answer if we also need to make a postage of $1.19? This is where it gets slightly evil.

1, 2, Total = 6560
1, 3, Total = 4415
1, 4, Total = 3362
1, 5, Total = 2747
1, 6, Total = 2349
1, 7, Total = 2075
1, 8, Total = 1886
1, 9, Total = 1743
1, 10, Total = 1640
1, 11, Total = 1559
1, 12, Total = 1496
1, 13, Total = 1451
1, 14, Total = 1431
1, 15, Total = 1389
1, 16, Total = 1394
1, 17, Total = 1367
1, 18, Total = 1400
1, 19, Total = 1379
1, 20, Total = 1394
1, 21, Total = 1399
1, 22, Total = 1445
1, 23, Total = 1419
1, 24, Total = 1395
1, 25, Total = 1511
1, 26, Total = 1504
1, 27, Total = 1509
1, 28, Total = 1550
1, 29, Total = 1571
1, 30, Total = 1508
1, 31, Total = 1559
1, 32, Total = 1640
1, 33, Total = 1751
1, 34, Total = 1793
1, 35, Total = 1757
1, 36, Total = 1739
1, 37, Total = 1739
1, 38, Total = 1757
1, 39, Total = 1793
1, 40, Total = 1886
1, 41, Total = 1919
1, 42, Total = 1927
1, 43, Total = 1949
1, 44, Total = 1985
1, 45, Total = 2035
1, 46, Total = 2099
1, 47, Total = 2177
1, 48, Total = 2269
1, 49, Total = 2375
1, 50, Total = 2495
1, 51, Total = 2479
1, 52, Total = 2420
1, 53, Total = 2367
1, 54, Total = 2320
1, 55, Total = 2279
1, 56, Total = 2244
1, 57, Total = 2215
1, 58, Total = 2192
1, 59, Total = 2175
1, 60, Total = 2223
1, 61, Total = 2339
1, 62, Total = 2465
1, 63, Total = 2601
1, 64, Total = 2747
1, 65, Total = 2903
1, 66, Total = 3069
1, 67, Total = 3245
1, 68, Total = 3230
1, 69, Total = 3219
1, 70, Total = 3212
1, 71, Total = 3209
1, 72, Total = 3210
1, 73, Total = 3215
1, 74, Total = 3224
1, 75, Total = 3237
1, 76, Total = 3254
1, 77, Total = 3275
1, 78, Total = 3300
1, 79, Total = 3329
1, 80, Total = 3362
1, 81, Total = 3399
1, 82, Total = 3440
1, 83, Total = 3485
1, 84, Total = 3534
1, 85, Total = 3587
1, 86, Total = 3644
1, 87, Total = 3705
1, 88, Total = 3770
1, 89, Total = 3839
1, 90, Total = 3912
1, 91, Total = 3989
1, 92, Total = 4070
1, 93, Total = 4155
1, 94, Total = 4244
1, 95, Total = 4337
1, 96, Total = 4434
1, 97, Total = 4535
1, 98, Total = 4640
1, 99, Total = 4749
1, 100, Total = 4862
1, 101, Total = 4879
1, 102, Total = 4797
1, 103, Total = 4715
1, 104, Total = 4633
1, 105, Total = 4551
1, 106, Total = 4469
1, 107, Total = 4387
1, 108, Total = 4305
1, 109, Total = 4223
1, 110, Total = 4141
1, 111, Total = 4059
1, 112, Total = 3977
1, 113, Total = 3895
1, 114, Total = 3813
1, 115, Total = 3731
1, 116, Total = 3649
1, 117, Total = 3567
1, 118, Total = 3485
1, 119, Total = 3403
2, 3, Total = 4387
2, 5, Total = 2713
2, 7, Total = 2042
2, 9, Total = 1706
2, 11, Total = 1522
2, 13, Total = 1419
2, 15, Total = 1359
2, 17, Total = 1342
2, 19, Total = 1346
2, 21, Total = 1362
2, 23, Total = 1384
2, 25, Total = 1445
2, 27, Total = 1477
2, 29, Total = 1531
2, 31, Total = 1537
2, 33, Total = 1657
2, 35, Total = 1705
2, 37, Total = 1727
2, 39, Total = 1785
2, 41, Total = 1879
2, 43, Total = 1927
2, 45, Total = 2003
2, 47, Total = 2107
2, 49, Total = 2239
2, 51, Total = 2350
2, 53, Total = 2332
2, 55, Total = 2326
2, 57, Total = 2332
2, 59, Total = 2350
2, 61, Total = 2439
2, 63, Total = 2605
2, 65, Total = 2791
2, 67, Total = 2997
2, 69, Total = 3022
2, 71, Total = 3055
2, 73, Total = 3096
2, 75, Total = 3145
2, 77, Total = 3202
2, 79, Total = 3267
2, 81, Total = 3340
2, 83, Total = 3421
2, 85, Total = 3510
2, 87, Total = 3607
2, 89, Total = 3712
2, 91, Total = 3825
2, 93, Total = 3946
2, 95, Total = 4075
2, 97, Total = 4212
2, 99, Total = 4357
2, 101, Total = 4510
2, 103, Total = 4469
2, 105, Total = 4428
2, 107, Total = 4387
2, 109, Total = 4346
2, 111, Total = 4305
2, 113, Total = 4264
2, 115, Total = 4223
2, 117, Total = 4182
2, 119, Total = 4141
3, 4, Total = 3300
3, 5, Total = 2681
3, 7, Total = 2009
3, 8, Total = 1813
3, 10, Total = 1562
3, 11, Total = 1485
3, 13, Total = 1383
3, 14, Total = 1353
3, 16, Total = 1322
3, 17, Total = 1303
3, 19, Total = 1309
3, 20, Total = 1311
3, 22, Total = 1345
3, 23, Total = 1353
3, 25, Total = 1397
3, 26, Total = 1408
3, 28, Total = 1468
3, 29, Total = 1491
3, 31, Total = 1513
3, 32, Total = 1566
3, 34, Total = 1642
3, 35, Total = 1661
3, 37, Total = 1685
3, 38, Total = 1723
3, 40, Total = 1782
3, 41, Total = 1839
3, 43, Total = 1893
3, 44, Total = 1927
3, 46, Total = 2009
3, 47, Total = 2057
3, 49, Total = 2167
3, 50, Total = 2229
3, 52, Total = 2269
3, 53, Total = 2293
3, 55, Total = 2297
3, 56, Total = 2328
3, 58, Total = 2343
3, 59, Total = 2381
4, 5, Total = 2648
4, 7, Total = 1973
4, 9, Total = 1636
4, 11, Total = 1448
4, 13, Total = 1346
4, 15, Total = 1287
4, 17, Total = 1271
4, 19, Total = 1271
4, 21, Total = 1285
4, 23, Total = 1308
4, 25, Total = 1343
4, 27, Total = 1401
4, 29, Total = 1451
4, 31, Total = 1481
4, 33, Total = 1566
4, 35, Total = 1619
4, 37, Total = 1661
4, 39, Total = 1721
5, 6, Total = 2214
5, 7, Total = 1939
5, 8, Total = 1741
5, 9, Total = 1599
5, 11, Total = 1411
5, 12, Total = 1353
5, 13, Total = 1307
5, 14, Total = 1282
5, 16, Total = 1243
5, 17, Total = 1231
5, 18, Total = 1230
5, 19, Total = 1227
5, 21, Total = 1243
5, 22, Total = 1266
5, 23, Total = 1273
5, 24, Total = 1282
5, 26, Total = 1339
5, 27, Total = 1353
5, 28, Total = 1383
5, 29, Total = 1411
6, 7, Total = 1904
6, 11, Total = 1374
6, 13, Total = 1271
6, 17, Total = 1188
6, 19, Total = 1194
6, 23, Total = 1242
7, 8, Total = 1670
7, 9, Total = 1527
7, 10, Total = 1418
7, 11, Total = 1337
7, 12, Total = 1277
7, 13, Total = 1235
7, 15, Total = 1177
7, 16, Total = 1160
7, 17, Total = 1157
7, 18, Total = 1155
7, 19, Total = 1157
7, 20, Total = 1159
8, 9, Total = 1490
8, 11, Total = 1300
8, 13, Total = 1198
8, 15, Total = 1137
8, 17, Total = 1123
9, 10, Total = 1344
9, 11, Total = 1263
9, 13, Total = 1159
9, 14, Total = 1121
10, 11, Total = 1226
10, 13, Total = 1121
11, 12, Total = 1128

Last edited by: teliot on Feb 27, 2021

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ThatDonGuy

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February 27th, 2021 at 7:12:35 AM permalink

Quote: teliot
On the topic of Golfing, let me create a different problem that is based on efficient stamp totals.

Suppose the Post Office issues two values of stamps X cents and Y cents. What are the most efficient values for these two stamps X & Y so that we can make each postage from $1.20 to $2.00, using the least number of stamps overall?

Yeah, "Easy" Math Puzzles. :)

Because I have nothing better to do at 7 AM on a Saturday morning...

(13, 11) uses 1072 stamps:

Total	13	11	Total	13	11	Total	13	11	Total	13	11
120	5	5	140	4	8	160	3	11	180	13	1
121	0	11	141	10	1	161	9	4	181	8	7
122	6	4	142	5	7	162	4	10	182	14	0
123	1	10	143	11	0	163	10	3	183	9	6
124	7	3	144	6	6	164	5	9	184	4	12
125	2	9	145	1	12	165	11	2	185	10	5
126	8	2	146	7	5	166	6	8	186	5	11
127	3	8	147	2	11	167	12	1	187	11	4
128	9	1	148	8	4	168	7	7	188	6	10
129	4	7	149	3	10	169	13	0	189	12	3
130	10	0	150	9	3	170	8	6	190	7	9
131	5	6	151	4	9	171	3	12	191	13	2
132	0	12	152	10	2	172	9	5	192	8	8
133	6	5	153	5	8	173	4	11	193	14	1
134	1	11	154	11	1	174	10	4	194	9	7
135	7	4	155	6	7	175	5	10	195	15	0
136	2	10	156	12	0	176	11	3	196	10	6
137	8	3	157	7	6	177	6	9	197	5	12
138	3	9	158	2	12	178	12	2	198	11	5
139	9	2	159	8	5	179	7	8	199	6	11
									200	12	4

teliot

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February 27th, 2021 at 7:30:55 AM permalink

Quote: ThatDonGuy
Because I have nothing better to do at 7 AM on a Saturday morning...

(13, 11) uses 1072 stamps:
Total 13 11 Total 13 11 Total 13 11 Total 13 11
120 5 5 140 4 8 160 3 11 180 13 1
121 0 11 141 10 1 161 9 4 181 8 7
122 6 4 142 5 7 162 4 10 182 14 0
123 1 10 143 11 0 163 10 3 183 9 6
124 7 3 144 6 6 164 5 9 184 4 12
125 2 9 145 1 12 165 11 2 185 10 5
126 8 2 146 7 5 166 6 8 186 5 11
127 3 8 147 2 11 167 12 1 187 11 4
128 9 1 148 8 4 168 7 7 188 6 10
129 4 7 149 3 10 169 13 0 189 12 3
130 10 0 150 9 3 170 8 6 190 7 9
131 5 6 151 4 9 171 3 12 191 13 2
132 0 12 152 10 2 172 9 5 192 8 8
133 6 5 153 5 8 173 4 11 193 14 1
134 1 11 154 11 1 174 10 4 194 9 7
135 7 4 155 6 7 175 5 10 195 15 0
136 2 10 156 12 0 176 11 3 196 10 6
137 8 3 157 7 6 177 6 9 197 5 12
138 3 9 158 2 12 178 12 2 198 11 5
139 9 2 159 8 5 179 7 8 199 6 11
200 12 4

Ha! I was writing c code at 5:30 this morning.

Correct, of course.

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Wizard
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Wizard

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February 27th, 2021 at 7:45:24 AM permalink

I'm sure we're all wondering if there is at least a short cut to the right answer to such problems or is brute force the only way? I suspect in general the coinages will be semi-prime to each other, but I can't put into words why.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

teliot

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February 27th, 2021 at 8:05:12 AM permalink

Quote: Wizard
I'm sure we're all wondering if there is at least a short cut to the right answer to such problems or is brute force the only way? I suspect in general the coinages will be semi-prime to each other, but I can't put into words why.

Yes, the coinages must be relatively prime. If d is a divisor of a & b, then d is also a divisor of ax + by for all x, y. Ergo, you can only make coinages that are divisible by gcd(a,b).

As for your interesting question of getting a short cut, that's why I gave the example of $1.19. The answer doesn't fit the most obvious short cut, namely to minimize the difference |a-b| while maximizing the product (a-1)*(b-1) <= N (where N is the smallest coinage that needs to be made).

Last edited by: teliot on Feb 27, 2021

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charliepatrick

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February 27th, 2021 at 8:52:25 AM permalink

This is only an observation but when the difference between two values was 2, one had to devise two sets of numbers.

For instance if you were using 5 and 7, 25 would be 5*5 and 0*7. Reducing the number of 5's by 1 and adding a 7 instead, would add 2 (27=4*5+1*7 etc.). So this creates 25 27 29 31 33 35. 28 is 4*7 so 26 is 3*7+1*5. 30 is 6*5 which starts the next series up to 42; 7*5 starts 35....

Presumably this process continues all the way up. So you might be able to determine how this goes based on the pattern and end conditions.

Gialmere

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March 1st, 2021 at 8:22:24 AM permalink

Here's an easy Monday puzzle...

Have you tried 22 tonight? I said 22.

unJon

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March 1st, 2021 at 8:23:53 AM permalink

Quote: Gialmere
Here's an easy Monday puzzle...

Did it quick in my head but looks like:

153

The race is not always to the swift, nor the battle to the strong; but that is the way to bet.

teliot

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March 1st, 2021 at 8:46:15 AM permalink

Quote: Gialmere
Here's an easy Monday puzzle...

153 -- which, coincidentally, is the smallest integer > 1 that is the sum of the cubes of its digits. 153 = 1^3 + 5^3 + 3^3. There are three more positive integers > 1 with this property. Bonus easy Monday question -- What are they?

Bonus bonus question -- how many fish did Simon Peter catch (New Testament)?

Last edited by: teliot on Mar 1, 2021

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ThatDonGuy

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March 1st, 2021 at 9:11:30 AM permalink

By 891, one of 1, 8, 9 is in the code, but by 849, neither 8 nor 9 are in the code, so 1 is in the code, but not the third digit
By 317, 1 is not the second digit; therefore, 1 is the first digit
By 793, either 9 is the second digit or 3 is the third digit, but by 849, 9 is not in the code, so 3 is the third digit
By 725, either 7 or 5 is the second digit, but by 793, since 3 is in the code, 7 is not, so 5 is the second digit
The code is 153

unJon

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March 1st, 2021 at 9:23:12 AM permalink

Quote: ThatDonGuy

By 891, one of 1, 8, 9 is in the code, but by 849, neither 8 nor 9 are in the code, so 1 is in the code, but not the third digit
By 317, 1 is not the second digit; therefore, 1 is the first digit
By 793, either 9 is the second digit or 3 is the third digit, but by 849, 9 is not in the code, so 3 is the third digit
By 725, either 7 or 5 is the second digit, but by 793, since 3 is in the code, 7 is not, so 5 is the second digit
The code is 153

You can solve also without reference to the 849 box.

ETA:

Comparing 725 and 793 box you can eliminate 7. From 317 you then know 3 and 1 are correct but not in those positions. Looking at 793 you can put 3 in last position. Then looking at 891 leaves the 1 in first position. Finally looking at 725 you see that 5 has to be the middle digit.

The race is not always to the swift, nor the battle to the strong; but that is the way to bet.

gordonm888
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March 1st, 2021 at 9:49:18 AM permalink

Quote: teliot
153 -- which, coincidentally, is the smallest integer > 1 that is the sum of the cubes of its digits. 153 = 1^3 + 5^3 + 3^3. There are three more positive integers > 1 with this property. Bonus easy Monday question -- What are they?

Bonus bonus question -- how many fish did Simon Peter catch (New Testament)?

370
371
407

At first, I thought the question was "prime numbers that are the sum of their digits," and I couldn't find any.

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

teliot

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March 1st, 2021 at 11:34:07 AM permalink

0
1
153
370
371
407

Ok, here is a really easy follow-up question. The list in the spoiler gives all six solutions to "sum of the cubes of the digits equals the number."

My follow-up question is to find all integers that minimize the difference when it is greater than 0, that is:

minimize |(sum of cube of digits of number) - number| > 0

There are six solutions when this difference is equal to 0. I found 12 solutions to the next smallest difference.

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charliepatrick

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March 1st, 2021 at 12:55:26 PM permalink

Quote: teliot
...I found 12 solutions to the next smallest difference.

I could only find 11 - so still looking!

N: 12 C: 9 diff: 3
N: 30 C: 27 diff: 3
N: 31 C: 28 diff: 3
N: 32 C: 35 diff: 3
N: 255 C: 258 diff: 3
N: 365 C: 368 diff: 3
N: 437 C: 434 diff: 3
N: 474 C: 471 diff: 3
N: 747 C: 750 diff: 3
N: 856 C: 853 diff: 3
N: 1799 C: 1802 diff: 3

teliot

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March 1st, 2021 at 1:03:40 PM permalink

Quote: charliepatrick
I could only find 11 - so still looking!
N: 12 C: 9 diff: 3
N: 30 C: 27 diff: 3
N: 31 C: 28 diff: 3
N: 32 C: 35 diff: 3
N: 255 C: 258 diff: 3
N: 365 C: 368 diff: 3
N: 437 C: 434 diff: 3
N: 474 C: 471 diff: 3
N: 747 C: 750 diff: 3
N: 856 C: 853 diff: 3
N: 1799 C: 1802 diff: 3

I confess to miscounting! You got it. Here are the solutions with the difference <= 10:

1, 0
153, 0
370, 0
371, 0
407, 0
12, 3
30, 3
31, 3
32, 3
255, 3
365, 3
437, 3
474, 3
747, 3
856, 3
1799, 3
22, 6
226, 6
372, 6
1079, 6
10, 9
11, 9
125, 9
216, 9
417, 9
566, 9
675, 9
766, 9
872, 9
873, 9
962, 9
963, 9

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teliot

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March 1st, 2021 at 1:05:44 PM permalink

Okay, which leads me to a cute little theorem-ette that is on the slightly non-trivial side (but not that non-trivial), related to the 153 question.

Let N > 0 be any integer. Let S be the sum of the cubes of its digits. Show that the difference (N-S) is always divisible by 3.

For example, N = 1263. S = 1^3 + 2^3 + 6^3 + 3^3 = 1 + 8 + 216 + 27 = 252. N - S = 1011= 3*337.

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charliepatrick