When 827 is added to 9A6, the result is 17B3. If 17B3 is divisible by 3, what is the largest possibl

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09 Mar 2022 Number System

When 827 is added to 9A6, the result is 17B3. If 17B3 is divisible by 3, what is the largest possible value of A?

OPtion

1) 3
2) 1
3) 6
4) 5
5) 7
6) 4
7) 2
8) 0
9) 8
10) None of these

Solution

8 2 7
+ 9 A 6
---------
1 7 B 3
There is 1 carry from sum of unit place and no carry from sum of hundred place.
For 17B3 to be divisible by 3, sum (1+7+B+3)=11+b should be divisible by 3.
Hence, B can be 1, 4, 7
As B cann't be less than 3. For B=4, A=1 and for B=7, A=4
.'. Largest value of A=4

Correct Option: 6 Best Solution (1)

Mayank Kandoi   2 years AGO

....8 2 7
+ 9 A 6
-----------------------
17 (2+A) 13
-----------------------
17 (3+A) 3

This means 3+A = B
By divisibility rule of 3, 17B3 is divisible by 3 only if B = 1, 4, 7
If B = 7, A = 4 (Highest possible)

Option 6

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