A seven digit number 62a57b0 is divisible by 132. how many pairs of(a,b) are possible
Answers
Given : A seven digit number 62a57b0 is divisible by 132
To find : possible pairs of(a,b)
Solution:
A seven digit number 62a57b0 is divisible by 13
132 = 11 * 4 * 3
number is divisible by 4 if last two digits are divisible by 4
b0 can be 00 , 20 , 40 , 60 , 80
hence b possible 0 , 2 , 4 , 6 , 8
number is divisible by 3 if Sum of digits divisible by 3
Hence 6 + 2 + a + 5 + 7 + b + 0
= 20 + a + b
=> a + b = 1 , 4 , 7 , 10 , 13 , 16
Number is Divisible by 11 if
subtract and then add the digits in an alternating pattern from left to right
and number got should be divisible by 11
6 - 2 + a - 5 + 7 - b + 0
= 6 + a - b
=> a - b = - 6 or a - b = 5
=> b - a = 6 or a - b = 5
Taking b - a = 6 possible b 6 , 8 possible a = 0 , 2
b = 6 a = 0 a + b = 6 does not satisfy a + b 1 , 4 , 7 , 10 , 13 , 16
b = 8 a = 2 ( 8 + 2 = 10) Satisfy a + b
Taking a - b = 5
possible b 0 , 2 , 4 possible a = 5 , 7 , 9
b = 4 , a = 9 a + b = 13 Satisfy
b = 2 , a = 7 a + b = 9 does not satisfy a + b 1 , 4 , 7 , 10 , 13 , 16
b = 0 , a = 5 a + b = 5 does not satisfy a + b 1 , 4 , 7 , 10 , 13 , 16
Two pairs of ab ( 2 , 8) & (9 , 4) are possible
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Answer:
2
Step-by-step explanation:
132 = 2^2 * 3 * 11
If the given number is divisible by 4, then its last 2 digits should be divisible by 4.
Hence, b belongs to (0,2,4,6,8).
If the given number is divisible by 3, then its sum of digits should be divisible by 3.
Hence, a+b belongs to (1,4,7,10,13,16)
If the given number is divisible by 11, then the difference of the sum of alternating digits should be divisible by 11.
Hence, | (6+a+7+0) - (2+5+b) | is divisible by 11.
Therefore, we get b=6+a or b+5=a.
Therefore, we get (a,b) as:
(0,6) (1,2) (2,8) (3,9) (5,0) (6,1) (7,2) (8,3) (9,4)
But since b belongs to (0,2,4,6,8) and a+b belongs to (1,4,7,10,13,16), we get (a,b) as:
(2,8) or (9,4).
Hence, there are 2 possible pairs of (a,b).
Hope it helps xD