Question

find the sum of all 3 digit numbers which when divided by 16 leave reminder 7

Answer 1

Answer:

31407

Step-by-step explanation:

The first 3 digit number that leaves remainder 7 when divided by 16 is:

6×16 + 7 = 103

and the last one is

62×16 + 7 = 999.

The numbers to add up are the numbers of the AP

103, 119,..., 999

where a = 103 and d = 16.

The number of terms is:

n = 1 +  ( 999 - 103 ) / 16

  = 1 +  896 / 16

  = 1 + 56

  = 57

The sum of the first n terms of the AP is:

na + [ n(n-1)/2 ] × d

= 57×103 + [ 57×56/2 ] × 16

= 57×103 + 57×28×16

= 5871 + 25536

= 31407