How many possible values are there for A and B such that 1A83B

Question

How many possible values are there for A and B such that 1A83B is divisible by 45?
A. 1
B. 2
C. 0
D. 3
E. 5

yashikaaggarwal · Answer

The quickest Solution:
We know,
If the sum of all numbers of the given number add upto the multiple of 3 it is divisible by 3
45 is divisible by 3
But there are only 2 two digit numbers (45 & 90) which are divisible by both 3 & 45.
Hence, answer is B

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vrsthe1 · Answer

for 1A83B to be divisible by 45, it has to be divisible by both 3 & 5.
to be divisible by 5, B has to be 5 or 0. 
to be divisible by 3, A+B has to be divisible by 3 (as 1+8+3 = 12).
if we fix B=0 remaining options for A are either 3, 6 or 9. but only 16830 is divisible by 45.
if we fix B=5 remaining options for A are either 1 or 4. but 11835 is not divisible by 45.

so there are only 2 possible options for A & B for the number to be divisible by 45 those are A=6 & B=0 and A=4 & B=5.
Hence option 'B'.

hathunguyen254  thanks for pointing out my error and I really got lucky.

bv8562 · Answer

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  A=1  =>  A+B=6  =>  Sum of digits  =  18  (a multiple of 9)

Therefore, there are  TWO  combination of values for A and B i.e  (A,B) = (6,0) & (1,5) (B)

chetan2u · Answer

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.

bv8562 · Answer

Thanks chetan2u 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.

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How many possible values are there for A and B such that 1A83B

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How many possible values are there for A and B such that 1A83B

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