ATAR Notes: Forum
VCE Stuff => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematics => Topic started by: Andiio on August 04, 2010, 09:27:30 pm
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" 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?
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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.
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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 :\
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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.
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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
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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.