Advertisements
Advertisements
Question
The difference of three-digit number and the number obtained by putting the digits in reverse order is always divisible by 9 and ______.
Advertisements
Solution
The difference of three-digit number and the number obtained by putting the digits in reverse order is always divisible by 9 and 11.
Explanation:
Let abc be a three-digit number, then we have
abc – cba = (100a + 10b + c) – (100c + 10b + a)
= (100a – a) + (c – 100c)
= 99a – 99c
= 99(a – c)
= 9 × 11 × (a – c)
Hence, abc – cba is always divisible by 9, 11 and (a – c).
APPEARS IN
RELATED QUESTIONS
If \[\overline{{3x2}}\] is a multiple of 11, where x is a digit, what is the value of x?
Which of the following statement is true?
If a number exactly divides the sum of two numbers, it must exactly divide the numbers separately.
Now you try and change these numbers into special numbers —
132
The sum of a two-digit number and the number obtained by reversing the digits is always divisible by ______.
If
then A = ______ and B = ______.
A number is divisible by 11 if the differences between the sum of digits at its odd places and that of digits at the even places is either 0 or divisible by ______.
A three-digit number abc is divisible by 6 if c is an even number and a + b + c is a multiple of 3.
If AB × 4 = 192, then A + B = 7.
If AB + 7C = 102, where B ≠ 0, C ≠ 0, then A + B + C = 14.
Fill in the blank space in the same way.

