CAT 2023 Slot 3 Solved Paper

Question : 46 of 66

Marks: +1, -0

Let

{\n}

and

{\m}

be two positive integers such that there are exactly 41 integers greater than

8^m

and less than

8^n

, which can be expressed as powers of 2 . Then, the smallest possible value of

n+m

Solution:

It is given that there are exactly 41 numbers, which can be expressed as the power of two, and exist between

8^m

and

8^n

, (where

m

, and

n

are positive integers, and

m< n

)
Hence,

2^{3 m}< 41

numbers

< 2^{3 n}

Since,

m

is a positive integer, the least value of

m

is 1 . Therefore,

2^{3 m}=2^3

, hence, the 41 numbers between them are

2^4, 2^5, 2^6, . . ., 2^{44}

.
Then the lowest possible value of

8^n

2^{45}

. Hence, the smallest value of

{\n}

2^{45}=8^n⇒2^{3 n}=2^{45}⇒ n=15

Hence, the smallest value of

m+n

(15+1)=16

The correct option is D

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