Math, asked by saritathakurdp, 5 months ago

7. Let abc be a three digit number. Then abc+ bca + cab is not divisible
by
(a) a + b + c (b) 3
(c) 37
(d) 9​

Answers

Answered by SadiaSheikh00
3

Answer:

Answer is 9

Step-by-step explanation:

By simplifying the general form of abc, bca and cab, we will get = abc+bca+cab = 111(a+b+c) Hence, abc+bca+cab is divisible by 111 and also it is divisible by the factors of 111. Here, 3 and 7 are the factors of 111, and a+b+c is also a factor of 111(a+b+c). But 9 is not the factor of 111.

Answered by Tharun6741J
1

Answer:

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Step-by-step explanation:

By simplifying the general form of abc, bca and cab, we will get = abc+bca+cab = 111(a+b+c) Hence, abc+bca+cab is divisible by 111 and also it is divisible by the factors of 111. Here, 3 and 7 are the factors of 111, and a+b+c is also a factor of 111(a+b+c). But 9 is not the factor of 111.

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