AB * C = CB

Logic Level 1

$\begin{array} { l l l l } & & A & B \\ \times & & & C \\ \hline & & C & B \\ \end{array}$

If $A, B, C$ are distinct digits that satisfy the above cryptogram, what is the value of $B$ ?

The answer is 0.

3 solutions

Nicholas James
Mar 2, 2017

As $AB\times C = CB$ we know that AB must be greater than $10$ (because it has 2 digits), but less than $20$ (or the tens digit of the answer could not be the same as $C$ ). This means that $A=1$ .

We therefore have:

$(10+B)C=10C+B$

$10C+CB=10C+B$

$(C-1)B=0$

So, $C=1$ or $B=0$ . As $A=1$ , $C\neq 1$ .

So $\boxed{B=0}$

That's a great argument that $A = 1$ . I felt that it was obviously true, but didn't really know why / how. Thanks for sharing this approach :)

Chung Kevin - 4 years, 3 months ago

I am sure there is a formal proof, not just the hand-waving appeal to intuition I gave!

Nicholas James - 4 years, 3 months ago

What you had works. The "formal" (and painful) way to express it would be if $A \geq 2$ , then

$\overline{AB} \times C \geq \overline{A0} \times C \geq 20 \times C > 10 \times C + 9 \geq \overline{CB} .$

Note: I cleaned up the presentation of your solution at the end.

Chung Kevin - 4 years, 3 months ago

Isaac Lu
Feb 28, 2017

A, B, and C are distinct digits and AB is a 2-digit number, as well as CB.

Therefore, AB and CB can be rewritten as: $\displaystyle AB = 10A + B$ and $\displaystyle CB = 10C + B$

So, $\displaystyle AB \times C = CB$ is the same as writing $\displaystyle (10A + B)\times C = 10C + B$

$\displaystyle 10AC +BC = 10C + B$

Solving for $\displaystyle B$ , we get this: $\displaystyle B = \frac{10C(1-A)}{C-1}$

Analyzing this equation for $\displaystyle B$ , we can say that the denominator $\displaystyle C-1 > 0$ and in the numerator: $\displaystyle 1-A \ge 0$

We can solve A from the analysis above that $\boxed{\displaystyle 0 \le A \le 1}$

Since the digit is an integer, the possible values of A are 0 and 1. Then we have two cases.

$\displaystyle Case I$ . $\displaystyle A = 0$

$\displaystyle B = \frac{10C(1-(0))}{C-1}$

$\displaystyle B = \frac{10C(1)}{C-1}$

$\displaystyle 10C \div (C-1) = 10 + \frac{10}{C-1}$

$\displaystyle B = \frac{10C(1)}{C-1} = 10 + \boxed{\displaystyle\frac{10}{C-1}}$

Range of values of $\displaystyle \frac{10}{C-1}$ for all $\displaystyle C > 1$ is $\displaystyle (0, 10\big]$

Therefore, the range of values of $\displaystyle B$ is from $10+0$ to $10+10$ or simply 10 to 20.

These range of integral values cannot be the value of B since all of them are 2-digits.

We conclude that case 1 is false.

$\displaystyle Case II.$

$\displaystyle A = 1$

$\displaystyle B = \frac{10C(1-A)}{C-1}$

$\displaystyle B = \frac{10C(0)}{C-1}$

$\boxed{\displaystyle\boxed{\displaystyle B = 0}}$

It "might" have been possible for $C - 1 = 0$ , in which case you made an error of dividing by 0. (The original equation could still make sense, but your resultant equation would not).

It would be better to first show that $C \neq 0, 1$ to conclude that $C - 1 > 0$ and thus $1 - A \geq 0$ . This will allow us to work with the cases of $A = 0$ (which we should reject since this is a leading digit), and $A = 1$ which gives us $B = 0$ (regardless of C).

Chung Kevin - 4 years, 3 months ago

Azadali Jivani
Mar 2, 2017

C A = C ......this shows that A should be equal to 1
C
B = B.......this shows that B should be 0, C may be any digit

Oh, that's nice, looking at the calculations solely in the column.

However, because we might have some carry over, the statements need to be explained further.

Chung Kevin - 4 years, 3 months ago

AB * C = CB

The answer is 0.

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