Question

The probability that a randomly chosen positive divisor of $10^{29}$ is an integer multiple of $10^{23}$ is: $a^2/b^2$. Then $b-a$ would be:

• A. 8
• B. 15
• C. 21
• D. 23
• E. 45

Hint:

Write $10^n$ as $2^n5^n$ so every divisor corresponds to a pair of exponents $(i,j)$. “Multiple of $10^k$” just means $i,j\ge k$. Count lattice points in that rectangle and divide by the total $(n+1)^2$.

Answer 1

The Correct Option is D

Step 1: Count total divisors of 10^29.
10^29 = (2 · 5)^29 = 2^29 · 5^29. A generic divisor is 2^i 5^j with 0 ≤ i,j ≤ 29.
Hence, total number of positive divisors:
N_total = (29+1)(29+1) = 30² = 900.

Step 2: Count divisors that are multiples of 10^23.
A divisor 2^i 5^j is a multiple of 10^23 = 2^23 5^23 iff i ≥ 23 and j ≥ 23 (with i,j ≤ 29).
Possible exponents: i ∈ {23,24,...,29} (7 choices) and j ∈ {23,24,...,29} (7 choices).
Therefore, N_fav = 7 × 7 = 49.

Step 3: Probability and identification of a,b.
P = N_fav / N_total = 49/900 = 7²/30².
Thus a = 7, b = 30 ⇒ b − a = 30 − 7 = 23.