Problem:
Let a>1 and x>1 satisfy loga(loga(loga2)+loga24−128)=128 and loga(logax)=256. Find the remainder when x is divided by 1000.
Solution:
Simplify the first equation as follows.
loga(loga2)+loga24−128loga(24loga2)loga(224)224(23)(23)=a128=128+a128=a128⋅aa128=aa128⋅aa128=(aa128)(aa128)
Thus aa128=23. Letting a=2128b shows that (2128b)2b=23, which reduces to 3⋅128=b⋅2b. This implies that b=6, so a=2643, and
x=aa256=(2643)2(643⋅256)=2192.
Then x≡0(mod8). Euler's Theorem shows that 2192≡2−8≡256−1(mod125). Because 3⋅42=126≡1(mod125), it follows that 128 and 42 are inverses mod125. Thus 256 and 21 are inverses mod125, so x≡21(mod125). Because x≡0 (mod8) and x≡21(mod125), the Chinese Remainder Theorem implies that x≡896(mod1000).
The problems on this page are the property of the MAA's American Mathematics Competitions