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Professor Halfbrain has recently made a fascinating discovery on the two integers $x=101$ and $y=110$. The professor first observed that the decimal representation of $x$ is a permutation of the decimal representation of $y$. While this observation is quite straightforward (and some might even say: quite boring), the really fascinating part of Halfbrain's discovery was

  • that the decimal digit sums of the numbers $2x$ and $2y$ coincide!
  • and that the same holds true for the digit sums of $3x$ and $3y$!!
  • and also for the digit sums of $4x$ and $4y$!!!
  • and for the digit sums of $5x$ and $5y$!!!!
  • and for the digit sums of $6x$ and $6y$!!!!!
  • and for the digit sums of $7x$ and $7y$!!!!!!
  • and for the digit sums of $8x$ and $8y$!!!!!!!
  • and for the digit sums of $9x$ and $9y$!!!!!!!!

I am aware of the fact that this sounds almost unbelievable, but please trust me: it really is true, just check it yourself. Based on these experimental results, Professor Halfbrain has now formulated a theorem with eight statements. At the current moment, he is busily working on the proofs.

Halfbrain's big theorem: For any two positive integers $x$ and $y$ for which the decimal representation of $x$ is a permutation of the decimal representation of $y$, the following eight statements hold true:

(a). The digit sum of $2x$ equals the digit sum of $2y$.
(b). The digit sum of $3x$ equals the digit sum of $3y$.
(c). The digit sum of $4x$ equals the digit sum of $4y$.
(d). The digit sum of $5x$ equals the digit sum of $5y$.
(e). The digit sum of $6x$ equals the digit sum of $6y$.
(f). The digit sum of $7x$ equals the digit sum of $7y$.
(g). The digit sum of $8x$ equals the digit sum of $8y$.
(h). The digit sum of $9x$ equals the digit sum of $9y$.

Perhaps the professor was a little bit too excited about his discovery, and perhaps not each and every one of his eight statements will stand the test of time.
Can you help the professor in finding out which statements are true and which ones are false?

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1 Answer 1

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(a) and (d) are correct, while the rest are false.

Reason

This is because the digit sum of 2N (or 5N) is just the sum of digit sums of 2*d_i (or 5*d_i) where d_i are the digits of N. This is not true for multiplication with any digit except 2 or 5.

Proof

The carry at each step on multiplication by 2 is going to be 0 or 1. As a result, adding the carry will not cause the result at the current step to go above a multiple of 10 since multiplying a digit with 2 never results in a 9.

Similarly, the carry at each step on multiplication by 5 is going to be between 0 and 4. Since multiplying a digit with 5 never results in anything but 0 or 5, adding the carry never makes the result go above the next multiple of 10.

This is not true for multiplication by other digits, and counterexamples can be found easily. For example, 34*3=102. The carry on multiplying 4 by 3 is 1. 3*3+1=10, taking it to the next multiple of 10. As a result, the digit sum is not the same as that of 43*3=129. The smallest counterexamples are

  • 3: 34,43
  • 4: 25,52
  • 6: 17,71
  • 7: 15,51
  • 8: 13,31
  • 9: 12,21
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  • $\begingroup$ And yet, 1+0+2 -> 3 and 1+2+9 -> 1+2 -> 3 holds; similarly 1+0+0 -> 1 and 2+0+8 -> 1+0 -> 1 holds! (i.e. adding up the digit sum until there is only one digit left). $\endgroup$ Feb 17, 2015 at 12:31
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    $\begingroup$ @No.7892142 That is not a digit sum :) $\endgroup$
    – dmg
    Feb 17, 2015 at 12:31
  • $\begingroup$ @dmg Obviously, it's the digit sum's digit sum! Just felt like noting that. $\endgroup$ Feb 17, 2015 at 12:32
  • $\begingroup$ @No. 7892142 : If you compute digit sums till only one digit remains, then all statements are true - since the recursive digit sum is just the remainder on dividing the number by 9. $\endgroup$ Feb 17, 2015 at 12:33
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    $\begingroup$ Maybe nice to know: the repeated digit sum is also called the digital root $\endgroup$
    – Ivo
    Feb 17, 2015 at 14:48

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