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How many possible values are there for A and B such that 1A83B 12 Sep 2021, 22:23
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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
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B
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How many possible values are there for A and B such that 1A83B 13 Sep 2021, 01:05
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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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How many possible values are there for A and B such that 1A83B 13 Sep 2021, 01:27
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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.
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How many possible values are there for A and B such that 1A83B Updated on: 27 Jan 2022, 22:27
Originally posted by bv8562 on 27 Jan 2022, 08:36.
Last edited by bv8562 on 27 Jan 2022, 22:27, edited 1 time in total.
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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)
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How many possible values are there for A and B such that 1A83B 27 Jan 2022, 09:29
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bv8562
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.
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How many possible values are there for A and B such that 1A83B 27 Jan 2022, 22:19
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chetan2u
bv8562
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.

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 10 Jan 2024, 10:52
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