Given
a, b, and c are distinct digits
Find
numbers abc and cba both be divisible by 7? If yes , then mention the number.
Step-by-step explanation
For divisibility of 7 , the last digit must be from 1 , 3 , 7 and 9
Let the possible number is abc , then cba will also be divisible by 7.
so , we can write it as ,
100a + 10b + c = 7x
100c + 10b + a = 7y
subtract both equations ,
100a+ 10b + c - 100c - 10b - a = 7x - 7y
99a - 99c = 7(x - y)
99(a - c) = 7(x - y)
hence , 99(a - c) must be divisible by 7 .
since 99 is not divisible 7 , (a - c) will be divisible by 7.
hence , only possible values of (a - c) are 0 and 7.
if a - c = 0 then a = c which is not possible as all digits are different.
if a - c = 7
possible values are a = 9 , 8 and c = 2 , 1
hence , multiples of 7 in the form is abc with a = 9 , 8 and c = 2 , 1 are
168 , 259 and their reverse are 861 and 952.
Final Answer
Hence ,
yes , there are numbers abc and cba which can be divisible by 7.
numbers are 168 , 259 and reverse is 861 and 952