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Quote:teliotOkay, 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.

This is definitely true for every number through 9,999, which I just did in Excel. And late to party but also agree with Charlie results.

Hmmm .... not sure I like the "similar logic will apply" part. Can you formalize that?Quote:charliepatrickWe 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 5^{3}, 40 and 4^{3}, 3 and 3^{3}. 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. N^{3}-N = N * (N^{2}-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 M^{3}-10M to be a multiple of 3. This is the same as (M^{3}-M)-9*M. The former (M^{3}-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 (L^{3}-L)-999....999*L.

Hence it applies to all integers.

Quote:unJonDid it quick in my head but looks like:

153

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)?

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

Correct!

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

Quote:unJonDid it quick in my head but looks like:

153

P.S. This 89 second solve shatters the old speed record!

Quote:teliot...Hmmm .... not sure I like the "similar logic will apply" part. Can you formalize that?

^{3}) would be a multiple of 3. Now to show it applies to all larger integers.

Every integer (ABC....Z) can be considered A*10

^{n}+B*10

^{(n-1)}...Z. The sum of the integer cubes is A

^{3}+B

^{3}+...+Z

^{3}.

Consider the A part. The difference = A*10

^{n}-A

^{3}= A*(10

^{n}-1)+A-A

^{3}. The first part, being 999....999 is divisible by 3. The second part, by previous proof, is divisible by 3.

Hence A*10

^{n}-A

^{3}is divisible by 3.

Since this would apply to all the other digits in any number (B,C, .... Z); the condition is true for all integers.

Four congruent semicircles are arranged in a square. The radius of each circle is 1, what is the length of a side of the square?

Quote:Wizard

Four congruent semicircles are arranged in a square. The radius of each circle is 1, what is the length of a side of the square?

Call the centers of the topmost and rightmost circles O and P, respectively. And let point A be the upper right corner of the square.

Line segment OP is the hypotenuse of triangle OAP. OP = 2 and AP = 1. Therefore, OA = (2

^{2}-1

^{2})

^{1/2}= 3

^{1/2}.

Add 1 to the above length to get the side of the square: 3

^{1/2}+ 1

Quote:ChesterDog

Call the centers of the topmost and rightmost circles O and P, respectively. And let point A be the upper right corner of the square.

Line segment OP is the hypotenuse of triangle OAP. OP = 2 and AP = 1. Therefore, OA = (2^{2}-1^{2})^{1/2}= 3^{1/2}.

Add 1 to the above length to get the side of the square: 3^{1/2}+ 1

Winner, winner, chicken dinner!

Quote:charliepatrickEarlier the proof was given for (X-X^{3}) would be a multiple of 3. Now to show it applies to all larger integers.

Every integer (ABC....Z) can be considered A*10^{n}+B*10^{(n-1)}...Z. The sum of the integer cubes is A^{3}+B^{3}+...+Z^{3}.

Consider the A part. The difference = A*10^{n}-A^{3}= A*(10^{n}-1)+A-A^{3}. The first part, being 999....999 is divisible by 3. The second part, by previous proof, is divisible by 3.

Hence A*10^{n}-A^{3}is divisible by 3.

Since this would apply to all the other digits in any number (B,C, .... Z); the condition is true for all integers.

This appears to be an induction proof on the number of digits in the number. I'm still shaky on its logic, maybe just getting too old to make sense of things.

Here is my proof:

Quote:teliotOkay, 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.

I have a feeling I am repeating at least one earlier answer...

Let {a

_{0}, a

_{1}, a

_{2}, ...} be a set of integers in {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} such that N = a

_{0}+ 10 a

_{1}+ 100 a

_{2}+ ... + 10

^{k}a

_{k}+ ...

Since 10

^{k}- 1 = 9 (1 + 10 + 100 + ... + 10

^{k-1}), N = 3 (3 + 30 + ... +3 x 10

^{k-1}) + a

_{0}+ a

_{1}+ a

_{2}+ ..., this means that the difference between N and the sum of the cubes of its digits = a multiple of 3 + the sum of (each digit and the cube of that digit).

As has been pointed out earlier in the thread, for any digit K, K

^{3}- K = K (K

^{2}- 1) = (K-1) K (K+1);

since one of those must be a multiple of 3, K

^{3}- K is always a multiple of 3, so the difference between N and the sum of the cubes of its digits is the sum of multiples of 3.