Sign in
Please select an account to continue using cracku.in
↓ →
For a number to be divisible by 88, it has to be divisible by both 11 & 8 .
Divisibility rule of 8 : The number formed by last three digits of a given number has to be divisible by 8.
So, "65b" has to be a multiple of 8.
This is only possible at b = 6.
Divisibility rule of 11 : A number is divisible by 11 if the difference between the sum of the digits at the odd places and the sum of the digits at the even places is either 0 or a multiple of 11.
So, Sum of the digits at odd places = 5 + 3 + 4 + 5 = {17}.
Sum of the digits at even places = a + 2 + 6 + b = {a + b + 8}.
The difference is |a+b-9| .
As we deduced 'b' is 6 : |a-3| which has to zero only, because it can't be 11 as the maximum value 'a' can take is a single digit number.
Therefore, a = 3.
Hence the value of 2a + 5b = 2(3) + 5(6) = 36.
Click on the Email ☝️ to Watch the Video Solution
Crack IPMAT 2026 with Cracku
Educational materials for IPMAT and IIMB UG preparation
Ask our AI anything
AI can make mistakes. Please verify important information.
AI can make mistakes. Please verify important information.