IAS 2026 — CSAT Maths

The correct option is (b) - 111.

Solution

We are given the equation: $10^{m} x 1000 x n = 75^{25} x 25^{32} x 32^{75}$10mx1000xn=7525x2532x3275.
Since $1000 = 10^{3}$1000=103, the left side is: $10^{(m+3)} x n$10(m+3)xn.
We are told that n is not divisible by 10. This means that in the prime factorization of the right-hand side, all possible pairs of $(2 x 5)$(2x5) must be contained within the 10^(m+3) term. Therefore, (m+3) must be equal to the total number of trailing zeros, or the exponent of 10, in the product on the right.

Step 1: Prime Factorization of the right-hand side
$75^{25} = (3 x 5^{2})^{25} = 3^{25} x 5^{50}$7525=(3x52)25=325x550
$25^{32} = (5^{2})^{32} = 5^{64}$2532=(52)32=564
$32^{75} = (2^{5})^{75} = 2^{375}$3275=(25)75=2375

Step 2: Combine the factors
Right-hand side = $3^{25} x 5^{50} x 5^{64} x 2^{375}$325x550x564x2375
Right-hand side = $3^{25} x 5^{(50+64)} x 2^{375}$325x5(50+64)x2375
Right-hand side = $3^{25} x 5^{114} x 2^{375}$325x5114x2375

Step 3: Determine the power of 10
A power of 10 is formed by the pair $(2 x 5)$(2x5). The number of such pairs is determined by the lower exponent of 2 or 5.
Number of 5s = 114
Number of 2s = 375
Max pairs of $(2 x 5)$(2x5) = 114.
So, the expression can be written as: $(2 x 5)^{114} x 2^{(375-114)} x 3^{25}$(2x5)114x2(375−114)x325
= $10^{114} x (2^{261} x 3^{25})$10114x(2261x325)

Step 4: Solve for m
Comparing $10^{(m+3)} x n = 10^{114} x (2^{261} x 3^{25})$10(m+3)xn=10114x(2261x325):
Since $n = 2^{261} x 3^{25}$n=2261x325 (which is not divisible by 10 because it contains no factors of 5), we have:
$m + 3 = 114$m+3=114
$m = 114 - 3$m=114−3
$m = 111$m=111

Why the other options are incorrect

• Option (a) - 101: This result would occur if the total exponent of 10 was calculated as 104 instead of 114, likely by miscalculating the power of 5 in 75^25 or 25^32.
• Option (c) - 121: This value might be reached if one fails to subtract the 3 from the 1000 (10^3) term or adds it instead of subtracting it from the total power of 114.
• Option (d) - 131: This value is significantly higher and would require a much larger count of factor 5 in the prime factorization, which is not supported by the given numbers.

Key Concept

The exponent of 10 in a product is determined by the minimum of the exponents of its prime factors 2 and 5 (the "limiting factor" principle in prime factorization).