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 Repdigit Rigor (Posted on 2014-05-11)
Find all positive integers x, y, z so that:
• x2 + y = z and:
• Base ten expansion of each of x and y has precisely n digits and that of z has precisely 2n digits, where n > 1.
• All the digits of x are the same, all the digits of y are the same and all the digits of z are the same.
Prove that there are no others.

 No Solution Yet Submitted by K Sengupta No Rating

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 Solution Comment 2 of 2 |
For n=1 see Charlie's solution.
This is a proof for n>1

Let a be the repeated digit of x so x=a(10^n - 1)/9
Let b be the repeated digit of y so y=b(10^n - 1)/9
Let c be the repeated digit of z so z=c(10^(2n) - 1)/9
so x^2 + y = z becomes

a^2[10^(2n) - 2*10^n + 1]/81 + b(10^n - 1)/9 = c(10^(2n) - 1)/9

For values of a that are not multiples of 3, there won't be an integer value of c that balances the 10^(2n) term

If a=3 we get
[10^(2n) - 2*10^n + 1] + b(10^n - 1) = c(10^(2n) - 1)
to balance the 10^(2n) term, c must be 1
and it then follows that b=2 balances the rest.

If a=6 we get
4*[10^(2n) - 2*10^n + 1] + b(10^n - 1) = c(10^(2n) - 1)
to balance the 10^(2n) term, c must be 4
and it then follows that b=8 balances the rest.

If a=9 the c would have to be 3 but the b that balances the rest would be 18.  (The value of y would have overlapping 18's and look more like 199...998.)

 Posted by Jer on 2014-05-12 12:47:31

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