If the number 22 *54 x46 x 10s ic610 x 1512 x814 x2016 xlOie x2S20 is divisible by 10n, then which one of the following is the maximum value of n?

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Q: 33 (CAPF/2018)
If the number 22 *54 x46 x 10s ic610 x 1512 x814 x2016 xlOie x2S20 is divisible by 10n, then which one of the following is the maximum value of n?

question_subject: 

Logic/Reasoning

question_exam: 

CAPF

stats: 

0,2,4,1,2,1,2

keywords: 

{'maximum value': [0, 0, 0, 1], 'number': [0, 0, 0, 2], '10n': [0, 0, 0, 1], 'x814 x2016 xloie x2s20': [0, 0, 0, 1], 'x46': [0, 0, 0, 1]}

To determine the maximum value of n for which the given number is divisible by 10n, we need to find the number of trailing zeros in the given number.

The number of trailing zeros in a number is determined by the power of 10 in its prime factorization.

Looking at the given number, we can see that it contains multiple factors of 10, which is equal to 2 × 5.

To find the number of trailing zeros, we need to find the minimum power of 10 in the prime factorization of the given number.

In the given number, we have 10^1, 10^2, and 10^3 as factors.

Therefore, the maximum value of n is 3, which corresponds to 10^3 or 1000.

Thus, the correct answer is option 4: 98.

Alert - correct answer should be option 3: 89.

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