Question

find the sum of all 3 digits divisible by 3 and 4

Answer 1

The numbers which are divisible by 4 and 3, is a multiple of 4*3 i.e.

they are multiple of 12.

the lowest 3 digit number which is divisible by 12 is 108.

the largest 3 digit number which is divisible by 12 is 83*12=996 [since 1000/12=83.33]

let the total number of 3 digit numbers which is divisible by 4 and 3 is n.

therefore 996=108+(n-1)*12

906-108=(n-1)*12

888=(n-1)*12

n-1=888/12=74

n=74+1=75

therefore the sum of the numbers which are divisible by 4 and 3 (i.e. divisible by 12) is

Sn = N/2 [a+(n-1)d]

S75 = 75/2[108+(75-1)12]

= 75/2[108+(74×12)]

= 75/2 (108+888)

= 75/2×996

= 37350

Answer 2

Answer:

Step-by-step explanation:

The numbers which are divisible by 3 and 4 are multiples of 3*4 i. e. they are multiples of 12

So, the lowest 3 digit number divisible by 12 is 108

The largest 3 digit number divisible by 12 is 996.

Let no. of terms divisible by 3 and 4 be n

Therefore, 996=108+(n-1)*12

996-108=12n-12

So n=900/12

=75

HOPE THIS HELPS YOU ✌