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# Special numbers Watch

1. The repeat of a positive integer is obtained by writing it twice in a row (so for example the repeat of 213 is 213213). Is there a positive integer whose repeat is a square number?

i said call the repeat number abcabc and assume the result is true such that abcabc=n^2 for some n of the set of natural numbers.

Then abcabc=10*101[100a+10b+c]=n^2 (writing out the sum of each digit and factorising)

Then for LHS to be perfect square:P 100a+10b+c=10*101 as 1010 is not a square and is not decomposed of squares since 10=5*2.

But we also must restrict 0<a<9 and same for b and c. So the max value of 100a+10b+c is 100(9)+10(9)+9=999 (in other words it is at most a 3 digit number)

So as 1010>999 none such repeat numbers exist as perfect squares

Is this argument ok? infact i have no idea if the conclusion is right as no answer was published alongside the question.

(after typing this up and rereading the question i have realised that although the example was a 3 digit number, the number could be as long as you want: i think if this proof is right i could extend it and you would always have the maximum 99...999< SOMENUMBER needed. )

What do you think?
2. (Original post by newblood)
The repeat of a positive integer is obtained by writing it twice in a row (so for example the repeat of 213 is 213213). Is there a positive integer whose repeat is a square number?

i said call the repeat number abcabc and assume the result is true such that abcabc=n^2 for some n of the set of natural numbers.

Then abcabc=10*101[100a+10b+c]=n^2 (writing out the sum of each digit and factorising)

abcabc = 1001 * [100a+10b+c]=n^2
And 1001 factorises as 7,11,13.

Then for LHS to be perfect square:P 100a+10b+c=10*101 as 1010 is not a square and is not decomposed of squares since 10=5*2.
At this stage your requirement is that 1001 (corrected) divides 100a+10b+c and that the quotient is a square: Since all the prime factors of 1001 have a multiplicity of 1.

The rest of your argument follows, and works for a 3-digit number.

I don't have a full answer for the question, so can't help further. You may need a different line of attack for a general number.
3. (Original post by ghostwalker)
abcabc = 1001 * [100a+10b+c]=n^2
And 1001 factorises as 7,11,13.

At this stage your requirement is that 1001 (corrected) divides 100a+10b+c and that the quotient is a square: Since all the prime factors of 1001 have a multiplicity of 1.

The rest of your argument follows, and works for a 3-digit number.

I don't have a full answer for the question, so can't help further. You may need a different line of attack for a general number.
ah i must have made a slight arithmetic error there,

4. This is a Cambridge example sheet question, the tag is hardly appropriate...

(Original post by newblood)
...
I don't have time to read your working, but the conclusion is incorrect. Very strong hint: if is odd.

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Updated: April 15, 2014
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