SOLUTION: 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

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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
 ----------------
 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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