Let \(\frac{m}{n}\) in lowest term, be the probability that a randomly chosen positive divisor of \(10^{99}\) is an integer multiple of \(10^{88}\). Find m+n.

\(10^{99}=2^{99}5^{99}\)

or, (99+1)(99+1)=10000 factors

those factors divisible by \(10^{88}\)

are bijection to number of factors \(10{11}=2^{11}5^{11}\) has, which is (11+1)(11+1)=144

one required probability =\(\frac{m}{n}=\frac{144}{10000}=\frac{9}{625}\)

m+n=634.

Subscribe to Cheenta at Youtube

---

Let S be the sum of the base 10 logarithms of all the proper divisors (all divisors of a number excluding itself) of 1000000. What is the integer nearest to S?

1000000=\(2^{6}5^{6}\) or, (6+1)(6+1)=49 divisors of which 48 are proper

\(log1+log2+log4+....+log1000000\)

\(=log(2^{0}5^{0})(2^{1}5^{0})(2^{2}5^{0})....(2^{6}5^{6})\)

power of 2 shows 7 times, power of 5 shows 7 times

total power of 2 and 5 shows=7(1+2+3+4+5+6)

=(7)(21)=147

for proper divisor taking out \(2^{6}5^{6}\)=147-6=141

or, \(S=log2^{141}5^{141}=log10^{141}=141\).

Subscribe to Cheenta at Youtube

---