Q79·CSAT · Prelims 2026
Quantitative — Modular Exponent Prime Factorization
NumericalNumber SystemArithmetic Problem● Hard
Question
If 10^m × 1000 × n=75^25 × 25^32 × 32^75, where n is not divisible by 10, then the value of m is
Options
a
101
bAnswer
111
c
121
d
131
Explanation
Let us find the value of m by breaking down both sides of the equation into their prime factorization bases (2^x × 5^y × 3^z) .
Step 1: Factor the right side components:
75^25 = (3 × 5²)^25 = 3^25 × 5^50
25^32 = (5²)^32 = 5^64
32^75 = (2^5)^75 = 2^375
Combine them: Right Side = 2^375 × 5^(50 + 64) × 3^25 = 2^375 × 5^114 × 3^25
Step 2: Factor the left side components:
10^m = (2 × 5)^m = 2^m × 5^m
1000 = 10³ = 2³ × 5³
Left Side = 2^(m+3) × 5^(m+3) × n
Step 3: Equate both sides:
2^(m+3) × 5^(m+3) × n = 2^375 × 5^114 × 3^25
Step 4: Isolate the powers of 10: Since n is not divisible by 10, it cannot contain any combined 2 × 5 pairs, meaning either all available factors of 2 or all available factors of 5 must be completely used up by the 10^(m+3) component on the left.
Since the right side has fewer factors of 5 (114) than factors of 2 (375), the maximum number of 10-base pairs is limited by the power of 5. Set the exponents equal:
m + 3 = 114 \implies m = 114 - 3 = 111
The number of trailing zeros or powers of 10 in a prime factorization product is always limited by the lower exponent between its factors of 2 and 5.
Answer: (b).
Question details
Year
2026
Paper
CSAT
Question
Q79
Section
Quantitative Aptitude
Sub-topic
Number System
Type
Arithmetic Problem
Difficulty
Hard
Source hint
Prime factors — base prime factorization equations
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