Expert(Maths):__ Kindly provied me the solution of the following :- __

Q.No.1 How many five-digit multiples of 11 are there, if the five digits are 3, 4, 5, 6 and 7 same order ?

Q.No.2 The number of positive integers less than or equal to 100, which are not divisible by 2, 3 or 5, is :

(A) 24 (B) 26 (C) 29 (D) 32

Q.No.3 4^{16} + 4^{62} + 4^{63 } + 4^{64} is divisible by which number ?

__ MOST URGENT __

So, numbers 5,3,6,4,7 is a multiple of 11.

Here the numbers at odd places are 5, 6 and 7.

And it can be arranged in 3! ways.

Also the numbers at even places are 3 and 4.

And it can be arranged in 2! ways.

Therefore, total number of ways of such numbers = $3!\times 2!=3\times 2\times 1\times 2\times 1=12$

2) The positive integers divisible by 2 = $\frac{100}{2}$ = 50

The positive integers divisible by 3 = $\frac{99}{3}$ = 33

From these 33, 16 are even and divisible by 2 and 17 are odd and not divisible by 2. So we will take only 17, because the even numbers were already included in the 50 integers divisible by 2.

Now, the positive integers divisible by 5 = $\frac{100}{5}$ = 20

Here, 10 are odd and already included. 10 are even, but 3 of them are divisible by 3. So we take only 7.

Therefore, numbers which are not divisible by 2, 3 or 5 = 100 - 50 - 17 - 7 = 26

3) Please check your question once and do get back to us. As per my interpretation the first number should be ${4}^{61}$ instead of ${4}^{16}$.

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