A company security admin changes his password, every month where the password is divisible by 8 and 11. If password is prefixed with A and suffixed with B and the number provided to his branch is A818325B, then find the minimum sum of A + B?
Two numbers are in the ratio 8 : 11. If 15 is added to each number, the ratio becomes 11 : 14. Find the larger number before increment.
In the following matrix question, select the number which can be placed at the sign of question mark (?)
Recent Comments
Preeti
7 days ago
For divisibility by 8 last 3 digits 25B must be divisible by 8
256 is divisible by 8 so B =6
Then number will be A8183256.
For divisibility by 11
(A+1+3+5)-(8+8+2+6)
(A+9)-24
A-15
4-15=-11
A=4
Then sum of A+B=4+6
A+B=10 ans.
Shravya
8 days ago
Divisible by 8
Last three digits (25B) must be divisible by 8. 250+B by 8, B=6 (256÷8=32)
As Last three digits 256 is divisible by 8 so the number is A8183256.
Divisible by 11
Difference between sum of the digits in a number at odd and even places is 0 or divisible by 11
5+3+1+A=9+A
6+2+8+8=24
24-(9+A)/11=15-A/11
A=4
A=4,B=6 and A+B =4+6=10
Manswi Kadu
8 days ago
* Divisible by 8 → last 3 digits 25B must be divisible by 8.
Only 256 works.
So, B = 6
* Number becomes A8183256
* Divisible by 11 →
(A+1+3+5) - (8+8+2+6)=0 \text{ or multiple of 11}
A+9-24 = A-15
For a digit value, only:
A-15=-11
A=4
* Therefore:
A+B = 4+6 = 10