Math, asked by shashwatraj682, 4 months ago

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

Answered by av60464

The very first thing which is a must for you to keep in mind is : If the given 9 digit no. is divisble by 3 then it would also be divisible by 33 as 3 is the factor of 33.
The second thing is to follow the accurate divisibility process so as to get the answer
now, 2+6+0+A+4+B+5+9+6=32+A+B

The nearest whole no . after 32 which is divisible by 3 is 33 so A =0 or 1 and b=0or 1

Answered by RvChaudharY50

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 :-

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 ,

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

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

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

Learn more :-

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