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Question

Find the maximum value of n such that 42×57×92×91×52×62×63×64×65×66×67 is perfectly divisible by 42n.

A
4
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B
3
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C
5
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D
6
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Solution

The correct option is B 3
42 = 7×3×2
Also, 57 = 3×19
92 = 2×2×23
91 = 7×13
52 = 2×2×13
62 = 2×31
63 = 3×3×7
64 = 2×2×2×2×2×2
65 = 5×13
66 = 2×3×11
67 = 1×67
In the expression given, there are three 7's and more than three 2's and 3's. Thus, Option (b) is correct.

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