Question

Chứng tỏ rằng abc+bca+cab chia hết cho 111

Answer 1

Ta có:

abc = a100 + b10 + c

bca = b100 + c10 + a

cab = c100 + a10 + b

=> abc + bca + cab = (a100 + b100 + c100) + (b10 + c10 + a10) + (c + a + b) = (a + b + c)*100 + (a + b + c)*10 + (a + b + c)*1

= (a + b + c) * ( 100 + 10 + 1) = (a + b + c)*111 chia hết cho 111

=> abc + cab + bca chia hết cho 111

Answer 2

abc + bca + cab

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

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

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

Vậy abc + bca + cab chia hết cho 111

Answer 3

Ta có:  abc+bca+cab

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

=111a+111b+111c

=111(a+b+c) lun lun chia hết cho11

Vậy..............

tk mình nhaaaaaaaa!!!!!!!.........Hihi