Question. 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

D. 9

Step-by-step explanation:

Let

abc=100a+10b+c

bca=100b+10c+a

cab=100c+10a+b

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

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

Collect like terms

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

=111a + 111b + 111 c

Factorise

=111(a+b+c)

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

Factors of 111(a+b+c)= 1, 3, 37, 111, and (a+b+c)

abc + bca + cab is divisible by a+b+c because it is a factor of 111(a+b+c)

abc + bca + cab is divisible by 3 because 3 is a factor of 111(a+b+c)

abc + bca + cab is divisible by 37 because 37 is a factor of 111(a+b+c)

abc + bca + cab is not divisible by 9 because 9 is not a factor of 111(a+b+c)

answer: what if i dont want to show my work boi.

step-by-step explanation: im watching dbz and eating popcorn leave meh alone

people grow faster than other people.

step-by-step explanation:

should you really be posting this isn't math boi.