A five-digit number AABAA is divisible by 33. Write all the numbers of this form.
Solution:
If a number is divisible by 33 then it has to be divisible by 11 and 3.
If a number is divisible by 11 then the difference of the sum of digits in the odd places and sum of digits in the even places is either multiple of 11 or zero
Sum of digits in odd places = A + B + A
Sum of digits in even places = A + A
(A + B + A) - (A + A) = 11
B = 11
But B cannot be two digit number hence we take,
(A + B + A) - (A + A) = 0
B = 0
Since the number AABAA is also divisible by 3 then the sum of its digits will be divisible by 3
A + A + B + A + A is divisible by 3. Since we have B = 0 we need 4A to be divisible by 3
For 4A to be divisible by 3
A = 3 or 6 or 9
The numbers which are possible are: 33033, 66066 and 99099.
✦ Try This: If a 7-digit number AAABAAA is divisible by 33 then what are all the number(s) of this form?
Since the number is divisible by 33, we can say that the number is divisible by 11 and 3. Therefore
(A + A + A + A) - (A + B + A) = 11
2A - B = 11 ----- (1)
Or
(A + A + A + A) - (A + B + A) = 0
4A - 2A - B = 0
2A = B ----- (2)
Since the number is divisible by 3 we have
6A + B is divisible by 3 ----- (3)
Since B = 2A we can say:
6A + 2A is divisible by 3
8A is divisible by 3
So A can be 3, 6, 9
If A = 3 B = 6
A = 6 B = 12
But B cannot be a two digit number, hence
A = 3
B = 6
The number is 3336333
☛ Also Check: NCERT Solutions for Class 8 Maths Chapter 16
NCERT Exemplar Class 8 Maths Chapter 13 Problem 48
A five-digit number AABAA is divisible by 33. Write all the numbers of this form.
Summary:
A five-digit number AABAA is divisible by 33. The possible numbers of this form 33033, 66066 and 99099.
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