If the number 2² x 5⁴ x 4⁶ x 10⁸ x 6¹⁰ x 15¹² x 8¹⁴ x 20¹⁶ x 10¹⁸ x 25

If the number 2² x 5⁴ x 4⁶ x 10⁸ x 6¹⁰ x 15¹² x 8¹⁴ x 20¹⁶ x 10¹⁸ x 25²⁰ is divisible by 10ⁿ, then which one of the following is the maximum value of n?

[amp_mcq option1=”78″ option2=”85″ option3=”89″ option4=”98″ correct=”option4″]

This question was previously asked in
UPSC CAPF – 2018
The correct answer is D) 98.
A number is divisible by 10ⁿ if its prime factorization contains at least n factors of 2 and n factors of 5. The maximum value of n is determined by the minimum exponent of 2 and 5 in the prime factorization of the given number.
The given number is N = 2² x 5⁴ x 4⁶ x 10⁸ x 6¹⁰ x 15¹² x 8¹⁴ x 20¹⁶ x 10¹⁸ x 25²⁰.
Let’s express all terms in terms of prime factors 2, 3, 5:
N = 2² * 5⁴ * (2²)⁶ * (2*5)⁸ * (2*3)¹⁰ * (3*5)¹² * (2³)¹⁴ * (2²*5)¹⁶ * (2*5)¹⁸ * (5²)²⁰
N = 2² * 5⁴ * 2¹² * 2⁸ * 5⁸ * 2¹⁰ * 3¹⁰ * 3¹² * 5¹² * 2⁴² * 2³² * 5¹⁶ * 2¹⁸ * 5¹⁸ * 5⁴⁰

Total power of 2: 2 + 12 + 8 + 10 + 42 + 32 + 18 = 124. So, 2¹²⁴.
Total power of 5: 4 + 8 + 12 + 16 + 18 + 40 = 98. So, 5⁹⁸.
Total power of 3: 10 + 12 = 22. So, 3²².

The number is 2¹²⁴ * 3²² * 5⁹⁸.
For this number to be divisible by 10ⁿ = 2ⁿ * 5ⁿ, the maximum possible value of n is min(124, 98) = 98.

Divisibility by 10ⁿ is equivalent to divisibility by (2*5)ⁿ, which means divisibility by 2ⁿ and 5ⁿ simultaneously. The number of trailing zeros in an integer is equal to the exponent of 10 in its prime factorization, which is the minimum of the exponents of 2 and 5.
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