A four-digit number aabb is divisible by 55. Then possible value(s) of b is/are
(a) 0 and 2
(b) 2 and 5
(c) 0 and 5
(d) 7
Solution:
The four digit number aabb is divisible by 11 because the the difference of the sum of numbers in the alternate positions is zero:
a a b b
1 2 3 4
Sum of numbers in position (1) and (3) = a + b
Sum of numbers in position (2) and (4) = a + b
The difference of their sums is given as
(a + b) - (a + b) = 0 ------ (1)
As the number rule, a number is divisible by 11 if the difference calculated as above is either zero or 11.
Since equation (1) shows the difference as zero we can state that aabb is divisible by11
Since aabb is divisible by 55 it means that it is divisible by 5 also because 55 = 11 × 5
Therefore aabb is also divisible by 5.
Now aabb will be divisible by 5 if b = 0 or b = 5.
The correct answer is alternative (c)
✦ Try This: A four-digit number 9b81 is divisible by 11. Then possible value(s) of b is (a) 3, (b) 1, (c) 6, (d) 5
A number is divisble by 11 when the difference of the sum of digits in alternate positions is either zero or eleven.
If the given rule is applied to the given number 9b81 then we have:
9 + 8 - (b + 1) = 0 or 11
Since b is a single digit number we can write:
9 + 8 - (b + 1) = 11
b = 17 - 1 - 11 = 5
Therefore the number 9b81 is 9581.
The correct choice is therefore (d).
☛ Also Check: NCERT Solutions for Class 8 Maths Chapter 16
NCERT Exemplar Class 8 Maths Chapter 13 Problem 6
A four-digit number aabb is divisible by 55. Then possible value(s) of b is/are (a) 0 and 2, (b) 2 and 5, (c) 0 and 5, (d) 7
Summary:
A four-digit number aabb is divisible by 55. Then possible values of b are 0 and 5.
☛ Related Questions:
- Let abc be a three digit number. Then abc + bca + cab is not divisible by (a) a + b + c, (b) 3, (c) . . . .
- A four-digit number 4ab5 is divisible by 55. Then the value of b - a is (a) 0, (b) 1, (c) 4, (d) 5
- If abc is a three digit number, then the number abc - a - b - c is divisible by (a) 9, (b) 90, (c) 1 . . . .
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