# RMO 2014 Delhi Region Q.3

Suppose for some positive integers $r$ and $s$, the digits of $2^r$ is obtained by permuting the digits of $2^s$ in decimal expansion. Prove that $r=s$

• You can find rest of the problems here

• You can find the solutions here Note by Aneesh Kundu
6 years, 6 months ago

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## Comments

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@Aneesh Kundu Hello!!! Here is my approach..................... We can consider each power of two in its Binary representation, and notice that if we rearrange the numbers, either the digits will all not be used up or we will reach the same number.......hence proven!!

- 2 years, 10 months ago

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if r < or = s 2^s - 2^r = 2^r(2^n - 1) where r+n=s 9 divides 2^n-1 (2^r is a permutation of the digits of 2^s) This is only possible when 2^n=1 Therefore, n=0 which implies that r=s

- 5 years, 6 months ago

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what marks would you get to show that r-s<4 in this question < i could only proceed till here>

- 6 years, 5 months ago

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Suppose s ≤ r. If s < r then 2s < 2 r . Since the number of digits in 2s and 2r are the same, we have 2r < 10 × 2 s < 2 s+4. Thus we have 2s < 2 r < 2 s+4 which gives r = s + 1 or s + 2 or s + 3. Since 2r is obtained from 2s by permuting its digits, 2r − 2 s is divisible by 9. If r = s + 1, we see that 2r − 2 s = 2s and it is clearly not divisible by 9. Similarly, 2s+2 − 2 s = 3 × 2 s and 2s+3 − 2 s = 7 × 2 s and none of these is divisible by 9. We conclude that s < r is not possible. Hence r = s

- 6 years, 5 months ago

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What exactly is the question? Are leading 0's allowed or not?

- 6 years, 6 months ago

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Since its not mentioned in the question whether both the power have the same no digits or not, we need to construct a general case.

Its strange that they still haven't uploaded the official solutions till now.

- 6 years, 5 months ago

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That is the precise problem.Nothing is mentioned about the leading zeroes. I think if we consider the leading zeroes the standard of the question goes well above RMO. But i was stuck at this very juncture and failed to provide the above mod 9 solution.

- 6 years, 5 months ago

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Thee official key says the number of digits are same.I was trying to solve the more general case.I couldnt get it.But i noticed a strange thing 2^{34}=134217728 and 2^{30}=1073741824 and they have almost same digits that is they differ just 1 digit (2 and 4) I don`t think this may help But is it true? Thanks

- 6 years, 5 months ago

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Actually i should have felt that the general case is too difficult! the sad part is i felt it after submitting the answer script. Actually this means nothing now but still it was too much of spoilt milk not to cry over!! Jokes apart your observation is really interesting but i heard recently that someone have proved it,( though i haven't seen it myself)

- 6 years, 5 months ago

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I don't get it

- 6 years, 6 months ago

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i did not enjoy this problem at all! :(

- 6 years, 6 months ago

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I have also not solved this. But if we assume the number of digits of 2^r and 2^s are same here is a simple solution,

- 6 years, 6 months ago

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Without loss of generality let us assume r>s. Let r=s+k. where k>0 As 2^r and 2^s have same number of digits k<4. [ Since k>=4 will imply multiplying 2^s by a number >10 and thus ending up with a greater number of digits.] now as they have same digits their sum of digits is same SO they are congruent modulo 9. So 2^s+k - 2^s = 2^s( 2^k -1) ==0 (mod 9). Since (9,2^s)=1(i.e they are co-prime) this implies 2^k -1==0(mod 9). k<4 implies 2^k -1 can take values 1 , 3 or 7 all of which leads to a contradiction downright. So k=0 giving r=s.

- 6 years, 6 months ago

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Can you explain why you chose to use modulo 9 rather than modulo 3? Using modulo 3 allows k = 2 to work.

- 6 years, 6 months ago

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What do you mean? Modulo 3 will fail to contradict the case k=2 so modulo 9 is the only option left.

- 6 years, 6 months ago

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So one would have to first see that modulo 3 fails before you know to use modulo 9?

- 6 years, 6 months ago

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Why $k\leq 4$

Aren't we just rearranging the digits?

- 6 years, 6 months ago

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Of course!Permutation means rearrangement. k>=4 means we have to multiply 2^s by 16 or more which is >10.An that will lead us to more number of digits in 2^r

- 6 years, 6 months ago

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oh..sorry

Actually I misread $r=s+k$ as $r+s=k$, thats why I was wondering that $k<4$ could be a really useful result.

- 6 years, 6 months ago

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Apparantly this method fails when k=6n as 2^6n==1 (mod 9) which does not provide a contradiction. But this case can surely arise when THERE ARE A NUMBER OF ZEROES AS DIGITS IN 2^r.

- 6 years, 6 months ago

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Please anybody provide solution for this part

- 6 years, 6 months ago

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if $k=6n$ and $k<4$, then $\dfrac{2}{3}>n$ this leaves us with no positive integer values for $n$.

- 6 years, 6 months ago

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Noooo! We get k<4 only when we know that the number of digits in 2^r and 2^s are same. But if suppose there is a zero in 2^r. When we permute the digits the zero can bemade to come in front thus reducing the number of digits in 2^s. And then we cant say k<4.

- 6 years, 6 months ago

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This one went straight like a tangent over my head.

- 6 years, 6 months ago

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