Question

Let abc be a three-digit number. Then 'abc-cba' is NOT divisible by [show with
explanation]
(a) 9
(b) 11
(c) 33
(d) 18​

Answer 1

Step-by-step explanation:

given,abc is a three digit numbers.

Then abc=100a+10b+c

and cba=100c+10b+a

therefore,abc-cba=(100a+10b+c)(100c+10b+c)

=100a-a+10b-10b+c-100

=99a-99c=99(a-c)

=abc-cba is divisible by 99

=abc-cba is divisible by 9,11 and 33,but not divisible by 10

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Answer 2

option (d)

abc - cba is not divisible by 18

Step-by-step explanation:

Given that ; abc is a three digit number

then

abc = 100a +10b+c

cba = 100c +10b +a

abc - cba = (100a +10b+c )- ( 100c +10b +a)

abc - cba = 100a-a+10b-10b+c-100

abc - cba = 99a-99c

abc - cba = 99(a-c)

then abc - cba is divisible by 99

therefore ,

abc - cba is divisible by 9,11,33 but not 18

hence , abc - cba is not divisible by 18

#Learn more:

If abc is a three digit number then the number abc+acb+bac+bca+cab+cba is always divisible by

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