Question

If a nine-digit number 260A4B596 is divisible by 33, then find the number of possible values of A.

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Answer 1

Question :- if a nine digit number 260A4B596 is divisible by 33, Then find the number of possible values of A. ?

Solution :-

Since , 260A4B596 is divisible by 33 = 3 * 11 . Then it must be divisible by both 3 and 11.

we know that,

• if sum of all digits is divisible by 3, then the number also divisible by 3.
• A number to be divisible by 11 , the difference between the sum of the odd numbered digits (1st, 3rd, 5th...) and the sum of the even numbered digits (2nd, 4th...) must be divisible by 11.

So,

By Divisibility Rule of 3 we get :-

→ (2+6+0+A+4+B+5+9+6) ÷ 3

→ (32 + A + B) ÷ 3

So, Possible values of (A + B) are :-

• 33 - 32 = 1
• 36 - 32 = 4
• 39 - 32 = 7
• 42 - 32 = 10
• 45 - 32 = 13
• 48 - 32 = 16 .

By Divisibility Rule of 11 now, we get :-

→ { (6 + A + B + 9 ) - (2 + 0 + 4 + 5 + 6) } = 0, 11, 22, 33 ____

→ (15 + A + B) - (17) = 0, 11, 22, 33 ____

→ (A + B) - 2 = 0, 11, 22, 33 __________

From Both Conclusions , we get ,

• A + B ≠ 1 , as 1 - 2 = (-1) .
• A + B ≠ 4 as 4 - 2 = 2 .
• A + B ≠ 7 as 7 - 2 = 5 .
• A + B ≠ 10 as 10 - 2 = 8 .
• A + B = 13 as 13 - 2 = 11 .
• A + B ≠ 16 as 16 - 2 = 14 .

therefore, we can conclude that, (A + B) is equal to 13.

Hence, the Possible values of A and B are :-

• A = 4, B = 9
• A = 5 , B = 8
• A = 6 , B = 7
• A = 7 , B = 6
• A = 8 , B = 5
• A = 9 , B = 4

∴ The Number of Possible values of A are = {4,5,6,7,8,9} = Total 6 . (Ans.)

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