Question

Harsha received price money in a jackpot. He expressed the price money in the following way.
Consider a number 'N' that can be expressed in the form of prime factors 3, 5, 7 and 11. The total number of factors of N is 32. The difference between the largest and the smallest such number possible, given that each of these prime factors occur at least once, is the price money that Harsha won. What is the price money that Harsha won in the jackpot?

• A. Rs. 45,640
• B. Rs. 91,280
• C. Rs. 1,29,360
• D. Rs. 1,82,560
• E. Rs. 2,73,840

Hint:

The 4 exponents+1 must multiply to 32 = \(2^5\); only split into 4 positive parts is 2+1+1+1, so exponents are a permutation of (3,1,1,1). Put the highest power on the biggest prime to maximise, smallest prime to minimise.

Answer 1

The Correct Option is C

Step 1: Set up N.
Let \(N = 3^a \times 5^b \times 7^c \times 11^d\), with \(a, b, c, d \ge 1\) since each prime occurs at least once.

Step 2: Use the factor-count formula.
Number of factors of N \(= (a+1)(b+1)(c+1)(d+1) = 32 = 2^5\).
Since each of \(a+1, b+1, c+1, d+1\) is at least 2, and their product is a pure power of 2, each of them must itself be a power of 2. Writing \(a+1 = 2^{k_1}\) etc., we need \(k_1+k_2+k_3+k_4 = 5\) with each \(k_i \ge 1\). The only way to write 5 as a sum of four positive integers is \(2+1+1+1\).
So \((a+1, b+1, c+1, d+1)\) is a permutation of \((4, 2, 2, 2)\), which means \((a, b, c, d)\) is a permutation of \((3, 1, 1, 1)\).

Step 3: Maximise N.
To make N largest, put the biggest exponent (3) on the largest prime (11), and exponent 1 on the rest:
\(N_{max} = 3^1 \times 5^1 \times 7^1 \times 11^3 = 3 \times 5 \times 7 \times 1331 = 105 \times 1331 = 1,39,755\).

Step 4: Minimise N.
To make N smallest, put the biggest exponent (3) on the smallest prime (3):
\(N_{min} = 3^3 \times 5^1 \times 7^1 \times 11^1 = 27 \times 385 = 10,395\).

Step 5: Find the difference.
\(N_{max} - N_{min} = 1,39,755 - 10,395 = 1,29,360\).
So Harsha won Rs. 1,29,360.