One-to-Fifty puzzle
October 5th, 2010 at 7:55:43 AM
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Last night, I had a dream about a puzzle. The rules to the puzzle were vivid enough that I remembered it the next morning. I haven't tried to solve it, and frankly don't even know if it has a solution, or if it would be "fun" to do. I just wanted to post it before it left my brain.
Using all of the integers 1 through 100 inclusive, can you construct 50 mathematical equations to give answers 1 through 50?
The numbers 1-100 should each be used only once. You may not repeat a number within an equation, or in the entire puzzle, save for the answers of the 50 equations.
In the dream, there was an optional rule: Equations could be a single number if you'd like. (ie 53 = 53)
If this rule was not in effect, that would mean each equation would use exactly two numbers.
Basic math functions are allowed. Add, subtract, multiply, divide. Use square roots and exponents if necessary.
Can this be solved?
Can you solve it with just plus, minus, times and divide?
Can it be solved with pairs, ie no single number equations?
I'll see if I can come up with a solution as well, but wanted to give some time for the forum members to tackle it.
Using all of the integers 1 through 100 inclusive, can you construct 50 mathematical equations to give answers 1 through 50?
The numbers 1-100 should each be used only once. You may not repeat a number within an equation, or in the entire puzzle, save for the answers of the 50 equations.
In the dream, there was an optional rule: Equations could be a single number if you'd like. (ie 53 = 53)
If this rule was not in effect, that would mean each equation would use exactly two numbers.
Basic math functions are allowed. Add, subtract, multiply, divide. Use square roots and exponents if necessary.
Can this be solved?
Can you solve it with just plus, minus, times and divide?
Can it be solved with pairs, ie no single number equations?
I'll see if I can come up with a solution as well, but wanted to give some time for the forum members to tackle it.
-Dween!
October 5th, 2010 at 9:22:56 AM
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Just bedween you and me, it seems simple.
1=1
2=2
.
.
.
.
.
50=50
1=1
2=2
.
.
.
.
.
50=50
October 5th, 2010 at 9:28:56 AM
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Where have you used the numbers 51 through 100?Quote: dmJust bedween you and me, it seems simple.
1=1
2=2
.
.
.
.
.
50=50
October 5th, 2010 at 9:34:17 AM
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If you're going to get technical, I withdraw my answer.
October 5th, 2010 at 9:40:34 AM
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Quote: dmIf you're going to get technical, I withdraw my answer.
hehe
Simplicity is the ultimate sophistication - Leonardo da Vinci
October 5th, 2010 at 9:44:12 AM
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Quote: DocWhere have you used the numbers 51 through 100?
Okay then - use the answer above, except the last one:
50 + SUM[53..98]*((100+51)-(99+52)) = 50
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice."
-- Girolamo Cardano, 1563
October 5th, 2010 at 9:45:59 AM
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BUT, I resubmit a NEW one.
1=1
2=2
.
.
24=24
26=26
27=27
.
.
50=50
100-99+98-97+96............+52-51=25
1=1
2=2
.
.
24=24
26=26
27=27
.
.
50=50
100-99+98-97+96............+52-51=25
October 5th, 2010 at 10:06:29 AM
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Quote: MathExtremistOkay then - use the answer above, except the last one:
50 + SUM[53..98]*((100+51)-(99+52)) = 50
I don't think that is allowable-used 50 twice(X+Y=X) and not X=X
October 5th, 2010 at 10:07:53 AM
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dm - you did not use 25 in any of your equations.
October 5th, 2010 at 10:18:20 AM
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Quote: dmQuote: MathExtremistOkay then - use the answer above, except the last one:
50 + SUM[53..98]*((100+51)-(99+52)) = 50
I don't think that is allowable-used 50 twice(X+Y=X) and not X=X
You make the rules, so you get to say what's allowable. The equation reduces to 50 + 0, which does indeed equal 50.
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice."
-- Girolamo Cardano, 1563
October 5th, 2010 at 10:19:47 AM
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Ok I got it:
99 - 51 = 48
98 - 52 = 46
...
76 - 74 = 2
then,
49 - 2 = 47
48 - 3 = 45
...
39 - 12 = 27
then,
37 - 14 = 23
36 - 15 = 21
...
26 - 25 = 1
then,
49 = 49
100 - 50 = 50
75 - 38 - 13 +1 = 25
99 - 51 = 48
98 - 52 = 46
...
76 - 74 = 2
then,
49 - 2 = 47
48 - 3 = 45
...
39 - 12 = 27
then,
37 - 14 = 23
36 - 15 = 21
...
26 - 25 = 1
then,
49 = 49
100 - 50 = 50
75 - 38 - 13 +1 = 25
October 5th, 2010 at 10:27:52 AM
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Quote: WestSiderdm - you did not use 25 in any of your equations.
Did you not see the last one?
October 5th, 2010 at 10:29:54 AM
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Quote: MathExtremistQuote: dmQuote: MathExtremistOkay then - use the answer above, except the last one:
50 + SUM[53..98]*((100+51)-(99+52)) = 50
I don't think that is allowable-used 50 twice(X+Y=X) and not X=X
You make the rules, so you get to say what's allowable. The equation reduces to 50 + 0, which does indeed equal 50.
DWEEN made the rules. I try to adhere to them.
October 5th, 2010 at 10:33:21 AM
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Quote: cardsharkOk I got it:
99 - 51 = 48
98 - 52 = 46
...
76 - 74 = 2
then,
49 - 2 = 47
48 - 3 = 45
...
39 - 12 = 27
then,
37 - 14 = 23
36 - 15 = 21
...
26 - 25 = 1
then,
49 = 49
100 - 50 = 50
75 - 38 - 13 +1 = 25
You used 25 in 2 different equations, and Y-X=X not allowed.
October 5th, 2010 at 10:37:12 AM
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Let me clarify:Quote: DweenUsing all of the integers 1 through 100 inclusive, can you construct 50 mathematical equations to give answers 1 through 50?
The numbers 1-100 should each be used only once. You may not repeat a number within an equation, or in the entire puzzle, save for the answers of the 50 equations.
You must use all 100 numbers for the "left side" portion of the equation. The 50 answers for the "right side" can be repeated.
For example, this is ok:
50 + 1 + 2 - 3 = 50
This is NOT ok:
25 + ( 5 * 5 ) = 50
Used 5 twice
Also NOT ok:
50 + 50 - ( 10 * 5 ) = 50
Used 50 twice
Cardshark used the following, and despite what dm says, this is ok:
Quote: Cardshark26 - 25 = 1
...
75 - 38 - 13 +1 = 25
I only see a 25 being used once. I don't believe I missed it any where else.
Has anyone come up with a solution that uses exactly two number per equation?
-Dween!
October 5th, 2010 at 10:40:41 AM
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Quote: dmBUT, I resubmit a NEW one.
1=1
2=2
.
.
24=24
26=26
27=27
.
.
50=50
100-99+98-97+96............+52-51=25
Yes...you never used 25 on the "left hand" side of any of these equations.
October 5th, 2010 at 11:08:14 AM
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Actually, that's one of the NEW rules. What a fuc-ing joke. Every time a good answer appears I guess the rules will be altered again.
I ain't wasting no more time on this dumb crap.
I ain't wasting no more time on this dumb crap.
October 5th, 2010 at 11:29:56 AM
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If you would please explain, quoting where you can, where you saw the rules you are talking about? Keep in mind this all originated from a dream, and my subconscious cannot be held responsible for any confusion you are experiencing.Quote: dmActually, that's one of the NEW rules. What a fuc-ing joke. Every time a good answer appears I guess the rules will be altered again.
I ain't wasting no more time on this dumb crap.
-Dween!
October 5th, 2010 at 11:42:16 AM
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Quote: cardsharkOk I got it:
49 - 2 = 47
...
then,
49 = 49
Almost - you've got 49 twice. I'm pretty sure my solution works, and so do any number of equivalents. I don't believe there is a solution such that each equation takes the form X _ Y = Z, where _ is some operator, though I can't rigorously prove this right now.
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice."
-- Girolamo Cardano, 1563
October 5th, 2010 at 11:53:18 AM
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Quote: DweenIf you would please explain, quoting where you can, where you saw the rules you are talking about? Keep in mind this all originated from a dream, and my subconscious cannot be held responsible for any confusion you are experiencing.
October 5th, 2010 at 11:59:53 AM
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Quote: WestSiderYes...you never used 25 on the "left hand" side of any of these equations.
October 5th, 2010 at 12:12:56 PM
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You didn't say the answers to the equations had to be CORRECT. Therefore:
43+9=87
14 x 23=99
etc.
What's so hard about that?
43+9=87
14 x 23=99
etc.
What's so hard about that?
The fact that a believer is happier than a skeptic is no more to the point than the fact that a drunken man is happier than a sober one. The happiness of credulity is a cheap and dangerous quality.---George Bernard Shaw
October 5th, 2010 at 1:27:34 PM
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It is simply:
(100-75)/(99-74)*(98-73)/(97-72)*(96-71)/...............(76-51) = 25
The other 49 are 1 =1 , 2= 2,....24 = 24, then 26 = 26 to 50 =50.
Edit: Ahh! Modification. i didn't use 25, so:
(100-75)/(99-74)*(98-73)/(97-72)*(96-71)/...............(76-51)/25 = 1
The other 49 are 2= 2,....50 =50.
Edit #2: Now I've used 25 twice. This is becoming a headache!
Edit #3
(100-75)/(99-74)*(98-73)/(97-72)*(96-71)/...............(76-51) = 25
The other 49 are 1^25 =1 , then 2= 2,....24 = 24, then 26 = 26 to 50 =50.
(100-75)/(99-74)*(98-73)/(97-72)*(96-71)/...............(76-51) = 25
The other 49 are 1 =1 , 2= 2,....24 = 24, then 26 = 26 to 50 =50.
Edit: Ahh! Modification. i didn't use 25, so:
(100-75)/(99-74)*(98-73)/(97-72)*(96-71)/...............(76-51)/25 = 1
The other 49 are 2= 2,....50 =50.
Edit #2: Now I've used 25 twice. This is becoming a headache!
Edit #3
(100-75)/(99-74)*(98-73)/(97-72)*(96-71)/...............(76-51) = 25
The other 49 are 1^25 =1 , then 2= 2,....24 = 24, then 26 = 26 to 50 =50.
October 5th, 2010 at 4:05:47 PM
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In that case, it's far easier to do
2 = 2
3 = 3
4 = 4
...
50 = 50
and then
1 ^ (any expression involving the numbers 51..100) = 1
Or if exponentiation is forbidden, my earlier one
1 + ((100+51)-(99+52))*(any expression involving the numbers 53..98) = 1
2 = 2
3 = 3
4 = 4
...
50 = 50
and then
1 ^ (any expression involving the numbers 51..100) = 1
Or if exponentiation is forbidden, my earlier one
1 + ((100+51)-(99+52))*(any expression involving the numbers 53..98) = 1
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice."
-- Girolamo Cardano, 1563
October 5th, 2010 at 4:45:03 PM
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Quote: MathExtremistIn that case, it's far easier to do
2 = 2
3 = 3
4 = 4
...
50 = 50
and then
1 ^ (any expression involving the numbers 51..100) = 1
Or if exponentiation is forbidden, my earlier one
1 + ((100+51)-(99+52))*(any expression involving the numbers 53..98) = 1
I thought of that after. I think exponents s/b excluded.
October 5th, 2010 at 4:52:02 PM
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Here's an actual solution. All but one equation is of the form A - B = C; one uses multiplication. I do not believe it is possible to do with just subtraction.
100 - 50 = 50
98 - 49 = 49
96 - 48 = 48
94 - 47 = 47
92 - 46 = 46
90 - 45 = 45
88 - 44 = 44
86 - 43 = 43
84 - 42 = 42
82 - 41 = 41
80 - 40 = 40
78 - 39 = 39
76 - 38 = 38
74 - 37 = 37
72 - 36 = 36
70 - 35 = 35
51 - 17 = 34
52 - 19 = 33
53 - 21 = 32
54 - 23 = 31
99 - 69 = 30
97 - 68 = 29
95 - 67 = 28
93 - 66 = 27
91 - 65 = 26
89 - 64 = 25
87 - 63 = 24
85 - 62 = 23
83 - 61 = 22
81 - 60 = 21
79 - 59 = 20
77 - 58 = 19
75 - 57 = 18
73 - 56 = 17
71 - 55 = 16
31 - 16 = 15
32 - 18 = 14
33 - 20 = 13
34 - 22 = 12
25 - 14 = 11
15 - 5 = 10
13 - 4 = 9
10 - 2 = 8
7 * 1 = 7
9 - 3 = 6
11 - 6 = 5
12 - 8 = 4
27 - 24 = 3
28 - 26 = 2
30 - 29 = 1
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice."
-- Girolamo Cardano, 1563
