Divisibility Test III
Number Theory
Level
1
Take any positive three digit number having distinct digits. Reverse the order of its digits to obtain a new number. Out of the two numbers, subtract the smaller number from the larger number. The number which you get after subtraction will always be divisible by
11 only
Both 9 and 11
9 only
None of the others
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Let the digits of the number be a , b , c , in that order. Then, the value of that number is 1 0 0 a + 1 0 b + c . Notice that a has been multiplied by 100 because it is in the hundreds place, b has been multiplied by 10 because it is in the tens place and c has been left as it is ( or multiplied by 1) because it is in the ones place.
When we reverse the order of digits, a goes in ones place and c goes in hundreds place but b stays as it is. Now, the value of the number becomes 1 0 0 c + 1 0 b + a . So, on subtraction we get,
( 1 0 0 a + 1 0 b + c ) − ( 1 0 0 c + 1 0 b + a ) = 1 0 0 a − a + 1 0 b − 1 0 b + c − 1 0 0 c = 9 9 a − 9 9 c = 9 9 ( a − c )
The number which we get after subtraction is a product of 99 and some natural number. So, it will always be divisible by 99, which in turn is divisible by both 9 and 11.