Question

if a b c are three digits of a three digit number prove that abc + cab + bca.is a multiple of 37?

Answer 1

Answer:

abc+bca+cab=(100a+10b+c)+(100b+10c+a)+(100c+10a+b)

abc+bca+cab=(100a+10b+c)+(100b+10c+a)+(100c+10a+b)=111a+111b+111c

abc+bca+cab=(100a+10b+c)+(100b+10c+a)+(100c+10a+b)=111a+111b+111c=111(a+b+c)

abc+bca+cab=(100a+10b+c)+(100b+10c+a)+(100c+10a+b)=111a+111b+111c=111(a+b+c)=3×37(a+b+c)

abc+bca+cab=(100a+10b+c)+(100b+10c+a)+(100c+10a+b)=111a+111b+111c=111(a+b+c)=3×37(a+b+c)Therefore the number abc+bca+cab is divisible by3,37 and (a+b+c) but not by 9.

Step-by-step explanation:

Hope it's helps you

Answer 2

Given:

a, b, c are three digits of a number.

To Find:

abc+ cab+ bca is a multiple of 37.

Solution:

As given in the question, a, b, c are three digits of a number. Then, let the number be a, b, c.

       ⇒ abc+ bca+ cab

       ⇒ (100a+ 10b+ 1c)+ (100b+ 10c+ 1a)+ (100c+ 10a+ 1b)

       ⇒ 100(a+ b+ c)+ 10(a+ b+ c)+ 1(a+ b+ c)

       ⇒ abc(111)

       ⇒ abc(37)³

  So, abc+ cab+ bca is divisible by 37.

Hence proved abc+ cab+ bca is a multiple of 37.