Maths question :)

[Andiio]

" I take a two-digit positive number and add to it the number obtained by reversing the digits. For how many two-digit numbers is the result of this process a perfect square? "

The solutions say:
Let the two-digit number be 10x+y
Reversing the digits gives 10y+x
Thus 10x+y+10y+x = 11(x+y) where 1 <_ x + y <_ 18  (*<_ = the 'less than or equal to' equality sign)

For 11(x+y) to be a perf. square, x+y = 11, and the possible numbers are 29, 38, 47, 56, 65, 74, 83, 92, eight numbers in total.


I get the question; but what I don't get is how they get '10x+y'? Is it because there are only 10 two-digit numbers?

[pooshwaltzer]

99+99 = 198 which is largest value in the domain set. Closest square is 14^2 = 196. So 1^2, 2^2, 3^2 ... 14^2 = 14 distinct solutions.

[Andiio]

99+99 = 198 which is largest value in the domain set. Closest square is 14^2 = 196. So 1^2, 2^2, 3^2 ... 14^2 = 14 distinct solutions.

Solutions are up there; this is from the Westpac Maths Comp. The answer is 8 :\

[pooshwaltzer]

Ah Poosh, you dill. 1^2 thru to 6^2 are all void due to no DOUBLE digit additions arriving at those answers. Series starts at 7^2 ... thus 8 unique solns.

[Andiio]

Good point Poosh, but what about numbers in the series like 10^2?
10^2 = 100
        = 1 + 0 + 0
        = 1; a single digit number :S

[pooshwaltzer]

My bad. Ignore prev posts...I'm operating on a complete tangency here.

10x+y = some multiple of 10, ie. x, plus the single digit.

10y+x is the inverse function.

The ONLY possible sq is 11^2 = 121 so 8 pairs of conjugates giving one single soln.

SOZ for the confusion. I need more sleep.