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

Hint:

32 as a product of four factors, each at least 2, splits only as 2, 2, 2 and 4. So one prime gets exponent 3 and the rest get exponent 1; put exponent 3 on the largest prime for the biggest N, and on the smallest prime for the smallest N.

Answer 1

The Correct Option is C

Step 1: Set up the factor-count equation.
N is built from the primes 3, 5, 7 and 11, each occurring at least once, so we can write \( N = 3^a \times 5^b \times 7^c \times 11^d \) with a, b, c, d all at least 1.
The total number of factors of N is \( (a+1)(b+1)(c+1)(d+1) \), and this must equal 32.

Step 2: Split 32 into 4 factors, each at least 2.
Since each of a, b, c, d is at least 1, each of (a+1), (b+1), (c+1), (d+1) is at least 2. As 32 = \(2^5\), and we need four such factors multiplying to 32, the only way to split the five 2's among four slots, each getting at least one, is 2, 2, 2 and 4.
So one of (a+1), (b+1), (c+1), (d+1) equals 4, and the other three equal 2. In terms of the exponents, one of a, b, c, d equals 3, and the other three equal 1.

Step 3: Assign the exponent 3 to get the largest N.
To make N as large as possible, we should give the higher exponent, 3, to the biggest prime, 11, and leave exponent 1 on the smaller primes.
\[ N_{max} = 3^1 \times 5^1 \times 7^1 \times 11^3 = 3 \times 5 \times 7 \times 1331 = 105 \times 1331 = 1,39,755 \]

Step 4: Assign the exponent 3 to get the smallest N.
To make N as small as possible, we give the higher exponent, 3, to the smallest prime, 3, and leave exponent 1 on the rest.
\[ N_{min} = 3^3 \times 5^1 \times 7^1 \times 11^1 = 27 \times 5 \times 7 \times 11 = 27 \times 385 = 10,395 \]

Step 5: Find the difference.
\[ N_{max} - N_{min} = 1,39,755 - 10,395 = 1,29,360 \]

Final Answer:
The price money that Harsha won is Rs. 1,29,360. \[ \boxed{Rs.\ 1,29,360} \]