What smallest number should be added to 8444 such that the sum is completely divisible by 7?

Solution :

(i) 39 M 2

The given number = 39 M 2

A number is divisible by 11, if the difference of sum of its digits in odd places from the right side and the sum of its digits in even places from the right side is divisible by 11.

Sum of its digits in odd places = 3 + M

Sum of its digits in even place = 9 + 2 = 11

Their difference :

\Rightarrow11-\left(3-M\right)=0

\Rightarrow11-3-M=0

\Rightarrow M=8

Therefore, value of M is 8.

(ii) 3 M 422

The given number = 3 M 422

A number is divisible by 11, if the difference of sum of its digits in odd places from the right side and the sum of its digits in even places from the right side is divisible by 11.

Sum of its digits in odd places = 3 + 4 + 2 = 9

Sum of its digits in even places = M + 2

Their difference :

\Rightarrow9-\left(2+M\right)=0

\Rightarrow9-2-M=0

\Rightarrow M=7

Therefore, value of M is 7.

(iii) 70975 M

THe given number = 70975 M

A number is divisible by 11, if the difference of sum of its digits in odd places from the right side and the sum of its digits in even places from the right side is divisible by 11.

Sum of its digits in odd places = 0 + 7 + M = 7 + M

Sum of its digit in even places = 5 + 9 + 7 = 21

Their difference :

\Rightarrow21-\left(7+M\right)=0

\Rightarrow21-7-M=0

\Rightarrow M=14

Therefore, value of M is 14.

(iv) 14 M 75

The given number = 14 M 75

A number is divisible by 11, if the difference of sum of its digits in odd places from the right side and the sum of its digits in even places from the right side is divisible by 11.

Sum of its digit in odd places = 1 + M + 5 = M + 6

Sum of its digit in even places = 4 + 7 = 11

Their difference :

\Rightarrow11-\left(6+M\right)=0

\Rightarrow11-6-M=0

\Rightarrow M=5

Therefore, value of M is 5.

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Answer : Option B

Explanation :

8444 ÷ 7 = 1206, remainder = 2 7 - 2 = 5 Hence, 5 should be added to 8444 such that the sum is completely divisible by 7.

Page 2

Answer : Option C

Explanation :

These numbers are 18, 24, 30, 36...........,78.

This is an  A.P. in which  a = 18,  d = (24 - 18) = 6   and  `l` = 78

Let the number of these terms be n. Then

`t_n = 78  rArr     a + (n -  1) d = 78`

`rArr   18 + (n - 1) xx 6 = 78`

`rArr     (n - 1) xx 6 = 60`

`rArr    (n - 1) = 10`

`rArr  n = 11`

Required number of numbers = 11 .

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