Numbers in generalized form

Place value as algebra. A 2-digit number is built from a tens digit and a units digit. If the tens digit is $a$ and the units digit is $b$, the number is not the two symbols side by side — it is ten times the tens digit plus the units digit:

where: $\overline{ab}$ denotes the 2-digit number with tens digit $a$ and units digit $b$ (the bar is notation for "the number formed by these digits," not multiplication); $a \in \{1,\dots,9\}$ since a leading digit cannot be zero; $b \in \{0,\dots,9\}$.

Extending to 3 digits. The same reasoning stacks one more place value on top. A 3-digit number with hundreds digit $a$, tens digit $b$, and units digit $c$ is

where: $100a$ is the hundreds contribution, $10b$ the tens contribution, $c$ the units contribution. The same pattern extends to 4 digits as $1000a+100b+10c+d$, though the syllabus mainly works with 2- and 3-digit numbers, since those already carry every technique you need.

Reversing digits. Swapping the digits of $\overline{ab}=10a+b$ gives $\overline{ba}=10b+a$. Because both are algebraic expressions in the same two letters, you can add or subtract them like any other algebra:

where: $a$ and $b$ are the original digits, so the sum of a 2-digit number and its digit-reversal is always a multiple of 11, and (for $a>b$) their difference is always a multiple of 9 — regardless of which digits you actually pick.

Why the digit ranges matter. The restriction $a\in\{1,\dots,9\}$ for the leading digit is not a technicality to memorise — it is what keeps $\overline{ab}$ genuinely 2-digit. If you allowed $a=0$, the expression $10a+b$ would collapse to just $b$, a 1-digit number wearing a fake tens place. The same logic fixes $a\in\{1,\dots,9\}$ in the 3-digit form while $b$ and $c$, sitting in the tens and units places, are free to be any digit from 0 to 9, including 0, without breaking the number's digit count. The figure below shows how the same digit pair, reversed, always lands its sum on a multiple of 11 on a number line — a fact you can now prove instead of just notice.

Problem. A 2-digit number is written as $\overline{ab}$. When its digits are reversed, the new number is 27 more than the original number, and the sum of its digits is 9. Find the original number.

Step 1. Set up the generalized-form expressions. Let the tens digit be $a$ and the units digit be $b$, so the number is $10a+b$ and the reversed number is $10b+a$.

Step 2. Translate the digit-sum condition into an equation. The digits add to 9, so $a+b=9$.

Step 3. Translate the second condition into an equation. The reversed number is 27 more than the original: $10b+a = (10a+b)+27$. Rearranging, $10b+a-10a-b=27$, which simplifies to $9b-9a=27$, or $b-a=3$.

Step 4. Solve the pair of equations $a+b=9$ and $b-a=3$ together. Adding them eliminates $a$: $2b=12$, so $b=6$. Substituting back gives $a=9-6=3$.

Step 5. Assemble and check the answer. The original number is $10(3)+6=36$. Reversed, it is 63, and indeed $63=36+27$; the digit sum is $3+6=9$. Both conditions hold, so the original number is 36.

Competitive stretch

The same generalized-form method extends to three digits. For a number $\overline{abc}=100a+10b+c$ with $a=2b$, $c=b+1$, and digit sum $13$, substituting the first two conditions into $a+b+c=13$ gives $2b+b+(b+1)=13$, so $4b=12$ and $b=3$. Then $a=2(3)=6$ and $c=3+1=4$, giving $\overline{abc}=100(6)+10(3)+4=\mathbf{634}$. Checking: $a=6=2\times3$, $c=4=3+1$, digit sum $6+3+4=13$ — all conditions confirmed. Notice the pattern: every extra digit condition just becomes one more linear equation, so a 3-digit puzzle is solved exactly the way a 2-digit one is, only with one extra letter to track.

• Writing $\overline{ab}$ as the product $a\times b$ instead of $10a+b$. The bar notation means "the number with these digits," not multiplication — always expand it by place value first.
• Forgetting the restriction $a\neq0$ for the leading digit; in $10a+b$ or $100a+10b+c$, the first digit cannot be zero, or the number would really have fewer digits.
• Mixing up digit values with the number itself — thinking 47 equals $4+7$ instead of $10(4)+7$.
• Sign errors when subtracting a reversed number: $\overline{ab}-\overline{ba}=9(a-b)$ is only positive when $a>b$; if $b>a$, the difference is negative unless you swap which number you subtract from which.

Ask the tutor: Want to practice turning a tricky "reverse the digits" or "digit sum" word problem into equations step by step? Ask the AI tutor for a few more examples.