Question

if 3a52895b7 is exactly divisible by 3 and 11 both, then find the value of a +b (where a and b are digits from 0 to 9) ​

Answer 1

For the given number to be divisible by 11, the condition is the sum of digits in odd places must be equal to sum of digits in even place.

i.e. 3+5+8+5+7=a+2+9+b

28-11=a+b

a+b=17

Answer 2

Step-by-step explanation:

I think it will help you in solve this please comment if answer is r8ght