Easy math puzzles
Poll
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35 members have voted
March 1st, 2021 at 8:23:53 AM
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Quote: GialmereHere's an easy Monday puzzle...
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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.
March 1st, 2021 at 8:46:15 AM
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Quote: GialmereHere'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)?
Bonus bonus question -- how many fish did Simon Peter catch (New Testament)?
Last edited by: teliot on Mar 1, 2021
Poetry website: www.totallydisconnected.com
March 1st, 2021 at 9:11:30 AM
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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
March 1st, 2021 at 9:23:12 AM
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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.
March 1st, 2021 at 9:49:18 AM
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Quote: teliot153 -- 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.
March 1st, 2021 at 11:34:07 AM
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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.
Poetry website: www.totallydisconnected.com
March 1st, 2021 at 12:55:26 PM
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I could only find 11 - so still looking!Quote: teliot...I found 12 solutions to the next smallest difference.
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
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
March 1st, 2021 at 1:03:40 PM
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I confess to miscounting! You got it. Here are the solutions with the difference <= 10:Quote: charliepatrickI 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
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
Poetry website: www.totallydisconnected.com
March 1st, 2021 at 1:05:44 PM
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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.
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.
Poetry website: www.totallydisconnected.com
March 1st, 2021 at 1:27:42 PM
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We can split any number into its constituent digits and consider each one in turn. So a number such as 543 look at the difference between 500 and 53, 40 and 43, 3 and 33. If each of these is divisible by 3, then the total will.
So the approach is to show it for one digit numbers, then two digit numbers, etc.
Consider 0<N<10. N3-N = N * (N2-1) = N * (N-1) * (N+1). One these must be a multiple of 3.
Now consider a number <100 which is 10M+N. Then we would like M3-10M to be a multiple of 3. This is the same as (M3-M)-9*M. The former (M3-M) is divisible by 3 (as per above) and 9*M is also divisible by 3. Hence the difference is divisible by 3. This proves the case for two digit numbers.
Similar logic will apply to any other digit in the number since it will split into (L3-L)-999....999*L.
Hence it applies to all integers.
So the approach is to show it for one digit numbers, then two digit numbers, etc.
Consider 0<N<10. N3-N = N * (N2-1) = N * (N-1) * (N+1). One these must be a multiple of 3.
Now consider a number <100 which is 10M+N. Then we would like M3-10M to be a multiple of 3. This is the same as (M3-M)-9*M. The former (M3-M) is divisible by 3 (as per above) and 9*M is also divisible by 3. Hence the difference is divisible by 3. This proves the case for two digit numbers.
Similar logic will apply to any other digit in the number since it will split into (L3-L)-999....999*L.
Hence it applies to all integers.
