Problem:
Positive real numbers a and b have the property that
loga+logb+loga+logb=100
and all four terms on the left are positive integers, where log denotes the base 10 logarithm. What is ab?
Answer Choices:
A. 1052
B. 10100
C. 10144
D. 10164
E. 10200 Solution:
Since loga is a positive integer, we get loga=x2 for some integer x; since loga=21loga is a positive integer, we get x=2m. Thus a=104m2; similarly b=104n2. Substituting, we get 2(m+n+m2+n2)=100, i.e. m(m+1)+n(n+1)=50. It follows that m,n≤6. The values of m(m+1) for m=1,…,6 are
m123456m(m+1)2612203042 Two of those values must add up to 50 and we see that 20+30=50, so m=4,n=5 and ab=104(m2+n2)=104(42+52), and our answer is (D)10164.
OR
Since all four terms on the left are positive integers, from loga, we know that both loga has to be a perfect square and a has to be a power of ten. The same applies to b for the same reason. Setting a and b to 10x and 10y, where x and y are the perfect squares, ab=10x+y. By listing all the perfect squares up to 142 (as 152 is larger than the largest possible sum of x and y of 200 from answer choice E ), two of those perfect squares must add up to one of the possible sums of x and y given from the answer choices (52,100,144,164, or 200).
Only a few possible sums are seen: 16+36=52,36+64=100,64+100=164,100+100=200, and 4+196=200. By testing each of these (seeing whether x+y+2x+2y=100 ), only the pair x=64 and y=100 work. Therefore, a and b are 1064 and 10100, and our answer is (D)10164.