Show that a number with 160 digits has a prime power divisor that is at least 100. This is not true if we want a prime divisor that is at least 100.

Question

I really don't know how to approximate the problem

Thanks

Answer 1

This is really a pigeonhole problem. The largest number with no prime power divisor$>100$ is the product of all primes raised to the maximum powers below $100$, thus

$2^6×3^4×5^2×...×97^1$

The base 10 logarithm of this number is computed as between $40$ and $41$, but even without a calculator the logarithm cannot exceed $48$ as there are only $24$ primes in the product. So, with the more conservative estimate, even $49$ digits is enough to certify a prime power divisor $>100$ by contradiction.

Answer 2

Hint: the prime powers that don't exceed $100$ are $2^6,3^4,5^2,7^2,11,13,17,19,\dots,79,83,89,97$. What do you get when you multiply them all together?