What is the smallest three digit palindrome divisible by 18?
Dear Student,
Three digit palindrome numbers are 101,111,121,131,...So, for divisibility by 18 number must be divisible by 2 and 9 simultaneously.
So, for divisibility by 2, unit digit should be an even number.
So, 101,111,121,131,141,151....1,91 are not possible.
Then we check 202,212,222,232,242,252,...,292.
For divisibility by 9, sum of digits should be divisible by 9.
202=4, Not divisible
212=5, Not divisible
222=6, Not divisible
232=7, Not divisible
242=8, Not divisible
252=9, Divisible
So the required number is 252.
Regards!
Three digit palindrome numbers are 101,111,121,131,...So, for divisibility by 18 number must be divisible by 2 and 9 simultaneously.
So, for divisibility by 2, unit digit should be an even number.
So, 101,111,121,131,141,151....1,91 are not possible.
Then we check 202,212,222,232,242,252,...,292.
For divisibility by 9, sum of digits should be divisible by 9.
202=4, Not divisible
212=5, Not divisible
222=6, Not divisible
232=7, Not divisible
242=8, Not divisible
252=9, Divisible
So the required number is 252.
Regards!