Question 1172751: The five-digit numbers 8a2bc and 3a8bc are both perfect squares. What is the product of a x c?
Previous attempts: Using square rules (If c = 5, then b MUST = 2). Narrowing down possible endings to square endings: 1, 4, 5, 6, 9
Difficulty: Unable to find letter a, as well as find substitutions that work for both five-digit numbers
Thanks in advance!
Answer by greenestamps(13198)   (Show Source):
You can put this solution on YOUR website!
The three missing digits are the same in both numbers.
So, given that the two numbers are m^2 and n^2, look at m^2-n^2 = (m+n)(m-n):
m^2 = 8a2bc
n^2 = 3a8bc
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m^2-n^2 = 49400
(m+n)(m-n) = 49400
Look for a factorization of 49400 into the product of two integers of the form m+n and m-n:
(m+n)(m-n) = 49400 = (494)(100)
m+n = 494; m-n = 100 --> m = 297, n = 197
297^2 = 88209
197^2 = 38809
a = 8; b = 0; c = 9
ANSWER: ac = 8*9 = 72
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NOTE....
m^2=8a2bc is greater than 80000; that means m is between sqrt(80000) = 282 and 300;
n^2=3a8bc is between 30000 and 40000; that means n is between sqrt(30000) = 173 and 200
In the above solution, there are many other factorizations of 49400 into the product of two integers of the form m+n and m-n; but none of the others produces values of m and n that meet those requirements.
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