[MOVED] AAB * B = CB5B: Find A, B, and C

[Mustang1910]

Can anyone help me with the answer to this problem:

AAB
x B
____
CB5B

Find A,B, and C. Explain your answer.

 

[soroban]

Re: Problem

Hello, Mustang1910!

. . \(\displaystyle \L\begin{array}{cccc} & A & A & B \\ & \times & & B \\ \hline
C & B & 5 & B \end{array}\)

Find \(\displaystyle A,\,B,\text{ and }C.\;\)Explain your answer.

In the first (rightmost) multiplication, \(\displaystyle B^2\) ends in \(\displaystyle B.\)
Hence: \(\displaystyle \:B \:=\:0,\,1,\,5,\,6\)


If \(\displaystyle B\,=\,0\), the product would be \(\displaystyle 0.\;\) Hence: \(\displaystyle B\,\neq\,0.\)


If \(\displaystyle B\,=\,1\), the product would be \(\displaystyle AAB.\;\)Hence: \(\displaystyle B\,\neq\,1.\)


Suppose \(\displaystyle B\,=\,5\), then we have:
. . \(\displaystyle \L\begin{array}{cccc} & A & A & 5 \\ & \times & & 5 \\ \hline
C & 5 & 5 & 5 \end{array}\)

The first multiplication gives us: \(\displaystyle 5\,\times\,5\:=\:25.\;\)There is 2 to "carry."

The second mutliplication is: \(\displaystyle \:5\,\times\,A\,+\,2\)
But \(\displaystyle 5\times A\) will end in 0 or 5.
. . Hence, \(\displaystyle 5\times A\,+\,2\) will end in 2 or 7 ... not 5.

Therefore: \(\displaystyle \:B \,\neq\,5\) and \(\displaystyle B\,=\,6.\)

Then we have:
. . \(\displaystyle \L\begin{array}{cccc} & A & A & 6 \\ & \times & & 6 \\ \hline
C & 6 & 5 & 6 \end{array}\)

The first multiplication is: \(\displaystyle \:6^2\:=\:36\) . . . with 3 to "carry."

The next multiplication is: \(\displaystyle \,6\times A\,+\,3\) which must end in 5.
. . So, \(\displaystyle 6\times A\) must end in 2.
This happens for: \(\displaystyle \,A \,=\,2\text{ or }7\)

If \(\displaystyle A\,=\,2\), the problem becomes: \(\displaystyle \,226\,\times\,6\:=\:1356\)
. . But the second digit of the product must be 6.

Hence: \(\displaystyle \,A\,=\,7\) and the solution is:

. . \(\displaystyle {\color{blue}\L\begin{array}{cccc} & 7 & 7 & 6 \\ & \times & & 6 \\ \hline
4 & 6 & 5 & 6 \end{array}}\)

 

[Mustang1910]

Hi soroban,
Thanks so much for your help! Have a great day.