# Number Problem 2

Find the smallest number which should be added to the number 803642 so that it (i) may be divisible by 9; (ii) may be divisible by 11
Solution:
(i) In 803642, the sum of digits = 8 + 0 + 3 + 6 + 4 + 2 = 23. Since, the sum of the digits should be divisible by 9, the smallest number that is to be added is 4 since 23 + 4 = 27, which is divisible by 9
(ii) In 803642, the sum of the digits in the odd places = 2 + 6 + 0 = 8
Sum of the digits in the even places = 8+ 3 + 4 = 15
Therefore, in the units place, 15 – 8 = 7 should be added to make the sum of the digits in the odd places equal to the sum of the digits in the even places so that the number is divisible by 11.
Supply the missing digits marked by x in the following numbers if they are divisible by 11
(i) 7x563 (ii) 8276x57
Solution:
(i) If a number is divisible by 11, the difference of the sum of the digits in the odd places and sum of the digits in the even places should be either zero or a number divisible by 11
In 7x563, (7 + 5 + 3) – (x + 6) = 9 – x
9 – x should be equal to 0
i.e., 9 – x = 0; implies, x = 9
(ii) In 8276x57
(8 + 7 + x + 7) – (2 + 6 + 5) = (22 + x) – 13 = 9 + x
9 + x should be divisible by 11
Therefore, x should be 2
Therefore, x = 2.