Subtract Any Three-Digit Number From Its Reverse and 99 Divides It

Because the middle digit cancels out and what is left is $99$ times something.

Take $742$ and its reverse $247$:

$$742-247=495=99\times 5$$

The $5$ is no accident — it is the difference between the first and last digits, $7-2$. Try any other three-digit number and the same thing happens, because the tens digit contributes $10b$ to both numbers and vanishes in the subtraction.

One line of algebra proves it for every case at once, which is what the generalised form of a number is for. This page covers the first part of the ICSE Class 8 Mathematics chapter on playing with numbers.

Replace each digit by a letter and multiply it by its place value.

A two-digit number with tens digit $a$ and units digit $b$ is written $\overline{ab}$, and

$$\overline{ab}=10a+b$$

Its reverse is

$$\overline{ba}=10b+a$$

A three-digit number with digits $a$, $b$, $c$ from the left:

$$\overline{abc}=100a+10b+c$$

and its reverse is

$$\overline{cba}=100c+10b+a$$

Checking the notation on a real number. For $742$ we have $a=7$, $b=4$, $c=2$, so

$$100(7)+10(4)+2=700+40+2=742$$

Correct.

Why the bar matters. $\overline{ab}$ means the two-digit number whose digits are $a$ and $b$; $ab$ without the bar means $a$ multiplied by $b$. These are completely different things — for $a=7$ and $b=4$, $\overline{ab}=74$ while $ab=28$ — and mixing them up makes every subsequent line meaningless.

The constraints on the letters. Each letter stands for a single digit, so $0\le a,b,c\le 9$. And the leading digit cannot be $0$, since a three-digit number cannot begin with zero — it would simply be a two-digit number instead. Those two conditions are what make the puzzles in the last section solvable.

What generalised form buys you. Testing $742$ tells you about $742$. Working with $100a+10b+c$ tells you about every three-digit number in one go — so a single line of algebra replaces nine hundred separate checks. That is the whole reason this notation is introduced.

Several, and each is proved by expanding, then factorising until a divisor appears.

Result 1 — two digits, reverse and subtract.

$$(10a+b)-(10b+a)=9a-9b=9(a-b)$$

So the difference is always divisible by **$9$**. Checking with $72$ and $27$: $72-27=45=9\times 5$, and $a-b=5$. Correct.

Result 2 — two digits, reverse and add.

$$(10a+b)+(10b+a)=11a+11b=11(a+b)$$

Always divisible by **$11$**. Checking: $72+27=99=11\times 9$, and $a+b=9$. Correct.

Result 3 — three digits, reverse and subtract.

$$(100a+10b+c)-(100c+10b+a)=99a-99c=99(a-c)$$

Always divisible by **$99$, and therefore by $9$ and by $11$** as well.

Checking with $742$ and $247$: $742-247=495$, and $99(7-2)=99\times 5=495$. Correct.

Notice what the algebra reveals: the **middle digit $b$ plays no part whatever**. Every three-digit number beginning with $7$ and ending with $2$ gives the same difference of $495$, whatever its tens digit — which is a stronger statement than the original claim and could not be guessed from one example.

Result 4 — the three cyclic rearrangements. Take a three-digit number and form the two numbers got by moving each digit one place round:

$$\overline{abc}+\overline{bca}+\overline{cab}$$

Expanding all three:

$$(100a+10b+c)+(100b+10c+a)+(100c+10a+b)$$

Collecting the terms for each letter gives $111a+111b+111c$, so

$$\overline{abc}+\overline{bca}+\overline{cab}=111(a+b+c)$$

And since $111=3\times 37$, the sum is always divisible by **$3$, by $37$ and by $111$**.

Checking with $742$, $427$ and $274$:

$$742+427+274=1443$$

and $111\times (7+4+2)=111\times 13=1443$. Correct. Dividing by $37$ gives $39$, confirming that divisor too.

A boundary case — where there is no neat rule. Adding a three-digit number to its reverse gives

$$(100a+10b+c)+(100c+10b+a)=101(a+c)+20b$$

which does not factorise into a single divisor, because the $b$ term survives. So the three-digit add case has no rule matching the two-digit one — and knowing when a pattern fails is as much a result as knowing when it holds.

A cryptarithm is an arithmetic problem in which letters stand for digits. Solving one means finding which digit each letter is.

The rules of the game:

- Each letter stands for one digit from $0$ to $9$.
- Different letters stand for different digits, and the same letter always means the same digit.
- The leading digit of a number cannot be $0$.
- Work from the units column leftwards, carrying as usual.

Worked example 1 — an addition. Find $A$ and $B$:

3 A

+ 2 5

-------

B 2

- Units column: $A+5$ must end in $2$. Since $A$ is a single digit, $A+5=12$, giving $A=7$ with a **carry of $1$.
- Tens column**: $3+2+1=6$, so $B=6$.

Checking: $37+25=62$. Correct.

Worked example 2 — a multiplication. Find $A$, $B$ and $C$:

A B

× 3

-------

C A B

Use generalised form. The two-digit number is $10A+B$, and the three-digit answer is $100C+10A+B$:

$$3(10A+B)=100C+10A+B$$

$$30A+3B=100C+10A+B$$

$$20A+2B=100C\Rightarrow 10A+B=50C$$

So the two-digit number $\overline{AB}$ equals $50C$. Since it must have two digits, $C$ can only be $1$, giving $\overline{AB}=50$.

So $A=5$, $B=0$, $C=1$. Checking: $50\times 3=150$. Correct.

Worked example 3 — a single letter. Find $A$:

1 A

× A

-------

9 A

In generalised form:

$$A(10+A)=90+A$$

$$10A+A^2=90+A\Rightarrow A^2+9A-90=0$$

$$(A+15)(A-6)=0$$

Since $A$ is a digit it cannot be $-15$, so $A=6$. Checking: $16\times 6=96$. Correct.

Worked example 4 — a puzzle with many answers. Find $A$ and $B$:

A B

+ B A

-------

1 3 2

By Result 2 of the previous section, $\overline{AB}+\overline{BA}=11(A+B)$, so

$$11(A+B)=132\Rightarrow A+B=12$$

That is all the puzzle tells you. Any pair of digits adding to $12$ works: $48+84=132$, and $57+75=132$, and $39+93=132$.

So a cryptarithm need not have a unique answer. When a puzzle reduces to a single equation in two unknowns, the honest answer is to state the condition and give examples — and claiming one particular pair as the answer would be wrong.

The strategy that cracks most of these. Start wherever the information is tightest — usually the units column, or a column with a known digit. Then use the fact that a carry can only be $0$ or $1$ in an addition, which is often enough to pin a letter down completely.