SOLUTION: Letter A substitutes a nonzero digit and AA is a two-digit number. The product of AA and A is always. F) even G)divisible by 3 H)divisible by 11 J) a perfect square

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Question 1144827: Letter A substitutes a nonzero digit and AA is a two-digit number. The product of AA and A is always.
F) even
G)divisible by 3
H)divisible by 11
J) a perfect square
Found 3 solutions by Edwin McCravy, ikleyn, richwmiller:
Answer by Edwin McCravy(20059) About Me  (Show Source):
You can put this solution on YOUR website!
"AA" where A is both the tens and units digit.
t=A, u=A
The value 10t+u becomes 10A+A which is 11A,
which is always divisible by 11.

Edwin

Answer by ikleyn(52797) About Me  (Show Source):
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
The problem ask about the product of AA*A?
1*11=11 11/11=1
2*22=44 44/11=4
3*33=99 99/11=9
4*44=176 176/11-16
5*55=275 275/11=25
6*66=396 396/11=36
7*77=539 539/11=49
8*88=704 704/11=64
9*99=891 891/11=81
F) even
G)divisible by 3
H)divisible by 11
J) a perfect square
F is not a solution since 11 is not even
G is not a solution since 11 is prime
H IS A SOLUTION since all can be divided by 11
I is not a solution since 11 is prime
This should be easily understood from Edwin's solution. He showed that all AA are multiples of 11
Since all AA are divisible by 11 then AA multiplied by anything else is also divisible by 11.