# solve for the values of A, B, and C. [closed]

I need help with this problem, please.

If we have a number $ABC$ written in "decimal form" that is, its value is:

$$100A+10B+C.$$

If each letter represents the same digit (and different letters mean different digits), solve for the values of $A$, $B$, and $C$.

$$ABC+AB+A=300.$$

## closed as off-topic by Martin R, user99914, Aqua, Guy Fsone, user223391 Nov 11 '17 at 18:34

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• $271+27+2=300$ is one solution – Dr. Sonnhard Graubner Nov 10 '17 at 21:02
• \begin{align} 100a + 10b + c + 10a + b + a &= 300\\ 111a+11b +c &= 300\\ a+b+c &\equiv 0 \mod 10\\ 2b + c &\equiv 0 \mod 3\\ \end{align} Also, note that $a \lt 3$ – John Lou Nov 10 '17 at 21:31

## 4 Answers

As you've stated the number $ABC$, written in decimal form can be expressed as: $A × 100 + B × 10 + C × 1$

Therefore,

$ABC + AB + A = (A × 100 + B × 10 + C × 1) + (A × 10 + B × 1) + (A × 1)$

$= 111A + 11B + C = 300$

Now, check for yourself that $A$ must be $2$ (What happens if $A > 2$? If $A < 2$?)

So we see that $11B + C = 78$

Similarly to above, we see that $B=7$, finally giving us that $C=1$

Thus, your solution, as Dr. Graubner stated above, is $271$

First we can see that the only choice for A is 2 (if it's 3, the sum is too large, if it's 1 then AB is less than 20 and the sum is too small).

So we have $100A+10B+C+10A+B+A=222+11B+C=300$. After simplifying, we get $11B+C=78$ and it's not difficult to see that the only solutions are $B=7$, $C=1$

Just do it.

$ABC + AB + A = (100A + 10B + C) + (10A + B) + A = 111A + 11B + C = 300$.

$C$ is at most $9$ and $11B$ is at most $99$ so $111A$ is at least $300 - 108=196$ so $196 \le 111A \le 300$ and $\frac {196}{111} \le A \le \frac {300}{111}$.

So $A = 2$ and $11B+ C= 300 -222 = 78$.

$C$ is at most $9$ so $11B$ is at least $69$. So $69\le 11B \le 78$ so $\frac {69}{11} \le B \le \frac {78}{11}$ so $B = 7$.

So $C = 78 - 77= 1$

So $271 + 27 + 2 = 300$.

Obviously $A=1\text{ or } 2$ and since $198+19+1\lt300$ we have $A=2$

\begin{align}\frac{ABC+\\\space\space AB\\\space {} \space \space A}{300}\end{align} It follows $A+B+C\equiv 0\pmod{10}\Rightarrow A+B+C=10 \text{ or }20\Rightarrow B+C=8$. There are only four possibilities $(B,C)=(7,1),(1,7),(5,3),(3,5)$ from which only $(7,1)$ fits. The only solution is $$271$$