Srini wrote his class 10th board examination this year. When the resul...
Solution:
Given, the first 3 digits of the 6 digit number are 267.
Let the 6 digit number be ABC267.
It is given that the number is divisible by 11.
Divisibility rule of 11 states that the difference between the sum of the digits in the odd places and the sum of the digits in the even places of the number should be either 0 or a multiple of 11.
So, (C + A + B) - (6 + 2 + 7) = C + A + B - 15 should be either 0 or a multiple of 11.
C + A + B = 26, 37, 48, 59, 70, 81, 92
It is also given that the number is divisible by 13.
So, 13 divides ABC267.
13 divides (267000 + ABC)
13 divides (2A7 + BC0)
13 divides (2A7 - BC)
We can check the values of (2A7 - BC) for the possible values of C + A + B.
For C + A + B = 26, (2A7 - BC) can be 079, 192, 2A5, 3B8, 4C1, 5A4, 6B7, 7C0, 8A3, 9B6.
For C + A + B = 37, (2A7 - BC) can be 2B9, 3C2, 4A5, 5B8, 6C1, 7A4, 8B7, 9C0.
For C + A + B = 48, (2A7 - BC) can be 4C3, 5A6, 6B9, 7C2, 8A5, 9B8.
For C + A + B = 59, (2A7 - BC) can be 6B5, 7C8, 8A1, 9B4.
For C + A + B = 70, (2A7 - BC) can be 8C7, 9A0.
For C + A + B = 81, (2A7 - BC) can be 0B3, 1C6.
For C + A + B = 92, (2A7 - BC) can be 2C9.
Now, we know that the number is a multiple of 7.
Divisibility rule of 7 states that if we remove the last digit of the number, double it, and subtract it from the remaining part of the number, then the resulting number should be either 0 or a multiple of 7.
Let the unit digit of the number be x.
So, (ABC26 - 2x) should be either 0 or a multiple of 7.
When C + A + B = 26, (ABC26 - 2x) can be 737, 1057, 2377, 3327, 4657, 6627, 7287, 8617, 9287.
When C + A + B = 37, (ABC26 - 2x) can be 1187, 2517, 346