bv8562 wrote:
I tried to do it this way and got only one combination of values for A and B.
For a number to be divisible by 45 it must be divisible by both 5 and 9
1. Divisibility rule for 9: Sum of the digits must be a multiple of 9.
2. Divisibility rule for 5: Last digit must be 0 or 5.
1+A+8+3+B = 12+A+B
if B=0 then A=6 => A+B=6 => Sum of digits = 18 (a multiple of 9)
if B=5 then
maximum value A can have = 9 => A+B = 14 => Sum of digits = 26 (NOT a multiple of 9)
Therefore, the only combination of values for A and B is 1 i.e A=6 B=0
chetan2u Bunuel Could you please explain me in detail the best way to solve this question and what am I doing wrong here?
You are wrong in coloured portion.
The total is 12+A+B.
When B is 0, then 12+A+0= 18 or A=6. We cannot have A+12 as 27, because A will become 2-digit number.
When B is 5, then 12+A+5= 18 or A=1. We cannot have A+17 as 27, because A will become 10, a 2-digit number.
for rectifying the error. I think I made this silly mistake because I was thinking only in terms of getting a sum of 27 (a multiple of 9) which made me ignore the other possibility. Anyway, I will make the corrections in my post. Thanks a lot for your time.