Question

A three digit number abc is cyclically reversed . find the quotients when the sum of these numbers is divided by 111.​

Answer 1

Answer:

Given:

Sum of 3-digit number abc and the number

obtained by changing the order of digits i.e. bca and cab.

∴abc+bca+cab

=100a+10b+c+1006+10c+a+100c+10a+b

=111a+111b+111c=111(a+b+c)

(i) When divided by 111, we get

111(a+b+c)÷111=a+b+c

(ii) When divided by (a+b+c), we get

111(a+b+c)÷(a+b+c)=111

(iii) When divided by 37, we get

111(a+b+c)÷37=3(a+b+c)  

(iv) When divided by 3, we get

111(a+b+c)÷3=37(a+b+c)

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