Question

Maximum value of $n$ for which $40^n$ divides $60!$ is

Hint:

When checking divisibility of factorials, always compare prime powers separately and choose the minimum.

Answer 1

Correct Answer: 14

Step 1: Prime factorize the base

40 = 2³ × 5

Therefore,

40ⁿ = 2³ⁿ × 5ⁿ

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Step 2: Find the highest power of 2 in 60!

Using Legendre’s formula:

v₂(60!) = ⌊60/2⌋ + ⌊60/4⌋ + ⌊60/8⌋ + ⌊60/16⌋ + ⌊60/32⌋

= 30 + 15 + 7 + 3 + 1

= 56

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Step 3: Find the highest power of 5 in 60!

v₅(60!) = ⌊60/5⌋ + ⌊60/25⌋

= 12 + 2

= 14

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Step 4: Determine the maximum possible value of n

For 40ⁿ to divide 60!, both conditions must be satisfied:

2³ⁿ ≤ 2⁵⁶ ⇒ n ≤ 18

5ⁿ ≤ 5¹⁴ ⇒ n ≤ 14

Hence, the limiting factor is the power of 5.

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Final Answer:

The maximum value of n is
14