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Question

If [x] denotes the greatest integer function of x, then [(66+14)2n+1] is

A
an integer divisible by 2
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B
an integer divisible by 3
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C
an integer divisible by5
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D
an odd prime number
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Solution

The correct option is A an integer divisible by 2
Let (66+14)2n+1=I+F
where I is a natural number and 0<F<1
Let G=(6614)2n+1
0<G<1

Now,
I+FG=(66+14)2n+1(6614)2n+1=2(2n+1C1(66)2n14+2n+1C3(66)2n2143+)

As 1<FG<1, so
FG=0
Therefore,
I is an even integer.

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