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   Prodigitious digits

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Topic: Prodigitious digits (Read 1096 times)

pcbouhid
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Prodigitious digits
« on: Jan 10^th, 2006, 8:05am »

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For a positive integer n, let P(n) be the product of the nonzero base 10 digits of n.

Call n "prodigitious" if P(n) divides n.

What is the maximum number of consecutive prodigitious positive integers n?

« Last Edit: Jan 10^th, 2006, 8:06am by pcbouhid »

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JohanC
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Re: Prodigitious digits
« Reply #1 on: Jan 10^th, 2006, 2:27pm »

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towr
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Re: Prodigitious digits
« Reply #2 on: Jan 11^th, 2006, 8:00am »

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I can get up to twelve

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Grimbal
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Re: Prodigitious digits
« Reply #3 on: Jan 11^th, 2006, 9:06am »

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Depends how you feel about zero, I guess.

« Last Edit: Jan 11^th, 2006, 9:07am by Grimbal »

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JohanC
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Re: Prodigitious digits
« Reply #4 on: Jan 11^th, 2006, 11:15am »

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on Jan 11^th, 2006, 9:06am, Grimbal wrote:

Depends how you feel about zero, I guess.

Not really.
I'll give some very big hints:
- There never can be more than 13 consecutive numbers. The numbers ending in 3, 6 or 9 should be divisible by 3. So if N+3 ends with 3 and is divisible by 3, N+6 will end in 6 and be divisible by 3 as well as N+9 will end in 9 and will also be divisible by 3. But N+13 nor N-1cann't be.
- For two "prodigitious" numbers N and N+1 to be consecutive, all digits except the last one should be either 1 or 0. If there would be any other digit D it is impossible for both N and N+1 to be divisible by D.
- In the following, let N be such a number consisting of only 1's and 0's, ending in 0.
- Such numbers N+D ending with D=0,1,2 or 5 are quite trivial. This extends easily to N+10, N+11 and N+12.
- Numbers ending in D=4 will be divisible by 4 if there is at least one zero just before. Similarly for numbers ending in 8, two preceding zeros are needed. So our initial N should end with at least 3 zeros.
- Numbers N+D ending in D=9 will need their sum of digits be a multiple of 9. So the number of 1's in N needs to be a multiple of 9. Automatically will N+3 be divisible by 3 an N+6 be divisible by 6.
- The only more elaborate case to be considered is D=7. As N=0 is not allowed due to the positiveness constraint, we basically need a number N consisting of 9 (or 18 or 27 or ...) 1's with possibly some 0's in between, and 3 0's at the end. And that is divisible by 7 (N mod 7 = 1).
- The smallest number to consider is 111111111000, but that is not be divisible by 7.
- One digit more puts one 0 somewhere in between. Basically this can be written as N = 1111111111000 - 10^k where there are ten 1's in the first number and k > 2 and k < 12. Then N mod 7 = 1111111111000 mod 7 - 10^k mod 7 = 2 - 3^k mod 7 to be divisible by 7, which is true for k = 8. So N=1111011111000 seems to be the first number leading to 13 subsequent "prodigitious" numbers.

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