Given Number = 342
Remainder = 47

so, required number = 342 + 47 = 389

When the number 389 is divided by 19, the remainder is 9.

Required sum

= 100/2 [1 +100]
= 50 x 101
= 5050

The pages of the book may be divided into 10 groups:

(1 – 100) : contains ’11’ zeroes
(101 – 200) / (201-300) / (301-400) / (401-500) / (501-600) / (601-700) / (701-800) / (801-900) : contains ’20’ zeroes
(901- 1000) : contains ’21’ zeroes

so, total number of zeros required = 11 + 8 x 20 + 21 = 192

List of prime numbers from 100 to 200: 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199.

Total 21 prime numbers are there between 100 and 200.

Let P = n (n + 1) (n + 2) ( n + 3).
Then, n = 1 gives
P = (1 x 2 x 3 x 4) = 24

Let the three consecutive odd integers be X, X+2 and X+4
Then, according to the question
3X = 2(X+4) + 3
X = 11
Therefore, third integer = X+4 = 15

As the number comprising of all 1’s is obtained on multiplication by 7, so it means that 7 is the factor of that number.
Our answer will the smallest number comprising of all 1’s that will be divisible by 7.
So, let’s check.
Is 1 divisible by 7? – No
Is 11 divisible by 7? – No
Is 111 divisible by 7? – No
Is 1111 divisible by 7? – No
Is 11111 divisible by 7? – No
Is 1111111 divisible by 7? – Yes
So, 1111111/7 = 15,873

Here we have to find out all the integers between 700 to 1000, in which sum of the digits is 10, e.g. 703 ⟶ 7 + 0 + 3 = 10

These numbers have been listed below: 703, 712, 721, 730, 802, 811, 820, 901 and 910

Hence, there are 9 such integers in which the sum of the digits is 10.

The given numbers are: 2^40, 3^21, 4^18, and 8^12.
We can also write them as: 2^40, 3^21, 2^36, and 2^36.
So, we basically need to find the smallest one from among 2^36, and 3^21

We can rewrite 2^36 and 3^21 as:
2^12 and 3^7
4096 > 2187
Hence, 3^21 is the smallest number.