MathsICSE Class 8 · Playing with Numbers

Generalised Form of Numbers and Digit Puzzles (ICSE Class 8 Maths)

Updated 17 October 2026 · Fact-checked

A two-digit number with tens digit $a$ and ones digit $b$ is $10a + b$; a three-digit number $abc$ is $100a + 10b + c$. Reversing digits turns this into $10b + a$ (or $100c + 10b + a$). Writing numbers this way proves puzzles such as "number + reverse is a multiple of 11".

Write the number as 10a + b and its reverse as 10b + a; adding them always gives 11(a + b), a multiple of 11.

Illustration drawn for QuestionBuddy.

Understand Generalised Form and Number Puzzles

The CISCE curriculum asks you to write two- and three-digit numbers in generalised form and to solve and create puzzles with them. A digit has a place value, so the number $\overline{ab}$ (digits $a$ then $b$ , not the product) equals $10 a + b$ , with $a$ from $1$ to $9$ and $b$ from $0$ to $9$ . Similarly $\overline{ab c} = 100 a + 10 b + c$ .

Reversal facts.

$\overline{ab} + \overline{ba} = (10 a + b) + (10 b + a) = 11 (a + b)$ . The sum is always a multiple of $11$ .
$\overline{ab} - \overline{ba} = 9 (a - b)$ . The difference is a multiple of $9$ .
$\overline{ab c} - \overline{c ba} = 99 (a - c)$ . The middle digit cancels.

Solving digit puzzles. Turn the words into an equation in the digits, using $10 a + b$ . Remember the digit limits ( $0$ to $9$ ; the leading digit is not $0$ ) and use them to pick a unique answer.

Letter-digit (cryptarithm) puzzles. Each letter stands for one digit. Work column by column from the ones place, track the carry, and test each possibility. Checking the final sum by actual addition is essential.

Creating puzzles. Choose a pattern (e.g. reverse and add), prove it with the general form, then give a numerical question. The general form is also the bridge to the divisibility rules of $2, 3, 5, 9, 10$ covered on the next topic page.

Generalised Form and Number Puzzles: formulas to remember

Two-digit number: $\overline{ab} = 10 a + b$; Reverse is $10 b + a$ .
Three-digit number: $\overline{ab c} = 100 a + 10 b + c$; Reverse is $100 c + 10 b + a$ .
Number plus reverse: $\overline{ab} + \overline{ba} = 11 (a + b)$; Always a multiple of 11.
Number minus reverse: $\overline{ab} - \overline{ba} = 9 (a - b), \overline{ab c} - \overline{c ba} = 99 (a - c)$; Multiples of 9 and 99.

How to solve Generalised Form and Number Puzzles questions

Translate each digit statement into an equation using place values.

1Name the digits with letters and write the number in generalised form.
2Translate each condition into an equation (sum of digits, difference, reverse).
3Solve; use the limits $0 \leq digit \leq 9$ and leading digit $\neq = 0$ to pick the valid answer.
4Verify by writing the number and testing every condition.
5For letter puzzles, solve column by column from the ones place with carries.

Generalised Form and Number Puzzles in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Swapping digits on a ticket

Suppose a ticket number has two digits, and a clerk accidentally writes it reversed. The correct number is 27 more than the wrong one, and the digits add to 11.

Find the correct ticket number.

Try it first, then show the solutionSolution

Let digits be $a$ (tens) and $b$ (ones): $9 (a - b) = 27$ , so $a - b = 3$ .
With $a + b = 11$ : $a = 7, b = 4$ .

Answer: 74 (reversed 47; check 74 - 47 = 27).

Memory hook: Reverse-and-subtract always gives a multiple of 9.

Tricks to remember Generalised Form and Number Puzzles

Short memory aids. Say each one aloud once, then use it on a question the same day.

Place value first
Say "tens times ten plus ones" before writing any equation.
Use it when: Every digit problem.
Reverse-add gives 11
Number plus its reverse = 11 times the digit sum.
Use it when: Quick checks and proofs.

Generalised Form and Number Puzzles: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Generalised form · 1 mark · MCQBoard level

A three-digit number has digits $p$ (hundreds), $q$ (tens) and $r$ (ones). Its generalised form is:
(a) $pq r$ (b) $100 p + 10 q + r$ (c) $p + q + r$ (d) $p \times q \times r$

Try it first, then show the solutionSolution

Each digit is multiplied by its place value.

Answer: (b)

Example 2: Digit sum and difference with reverse · 2 marks · Short answerBoard level

A two-digit number has digit sum 11, and it exceeds the number formed by reversing its digits by 27. Find it.

Try it first, then show the solutionSolution

Let the number be $10 a + b$ . Then $(10 a + b) - (10 b + a) = 9 (a - b) = 27$ , so $a - b = 3$ .
With $a + b = 11$ : $2 a = 14$ , $a = 7$ , $b = 4$ .
Check: $74 - 47 = 27$ and $7 + 4 = 11$ .

Answer: 74

Example 3: Proof by example · 3 marks · Short answerHarder board level

Show that the sum of 58 and its reversal is a multiple of 11, and state the quotient in terms of the digits.

Try it first, then show the solutionSolution

$58 + 85 = 143$ .
$143 = 11 \times 13$ and $13 = 5 + 8$ .
In general $(10 a + b) + (10 b + a) = 11 (a + b)$ , so the quotient is the digit sum.

Answer: $143 = 11 \times 13$ ; quotient $= a + b$ .

Example 4: Three-digit reversal · 4 marks · Short answerHarder board level

Show that $\overline{ab c} - \overline{c ba}$ is a multiple of 99. Verify for $832 - 238$ .

Try it first, then show the solutionSolution

$\overline{ab c} - \overline{c ba} = (100 a + 10 b + c) - (100 c + 10 b + a) = 99 a - 99 c = 99 (a - c)$ .
For $832$ : $a = 8$ , $c = 2$ , so $99 \times 6 = 594$ .
Direct: $832 - 238 = 594$ . Verified.

Answer: $99 (a - c)$ ; $832 - 238 = 594 = 99 \times 6$ .

Example 5: Letter addition puzzle · 4 marks · Long answerHOTS / competency

In the addition $3 A + 25 = B 2$ (a two-digit number $3 A$ plus $25$ gives the two-digit number $B 2$ ), find the digits $A$ and $B$ and write the sum.

Try it first, then show the solutionSolution

Ones column: $A + 5$ must end in $2$ , so $A + 5 = 12$ and $A = 7$ (carry $1$ ).
Tens column: $3 + 2 + 1 = 6$ , so $B = 6$ .
Check by place values: $37 + 25 = 62$ . In general $30 + A + 25 = 10 B + 2$ gives $A + 53 = 10 B$ , forcing $A = 7$ , $B = 6$ .

Answer: $A = 7$ , $B = 6$ ; $37 + 25 = 62$ .

Where students lose marks in Generalised Form and Number Puzzles

The slips examiners see most often, and what to write instead.

Mistake: Writing the two-digit number as ab = a x b
The letters are digits, not factors; place values are needed.
Wrong working
The number with digits $a, b$ is $ab$ so reversed minus original $= ba - ab = 0$ .
Right working
$(10 b + a) - (10 a + b) = 9 (b - a)$ .
Do this: Always write 10a + b.
Usually costs: All marks in the puzzle.
Mistake: Ignoring the digit limits
A digit is at most 9, and the leading digit cannot be 0.
Wrong working
$a - b = 3$ , $a + b = 11$ gives $a = 7$ ; also tries $a = 14$ in another case.
Right working
Check every candidate lies in ${0, \dots, 9}$ .
Do this: List allowed digits before accepting an answer.
Usually costs: The final answer mark.

Exam tips

State "let the digits be a and b" at the start; examiners expect it.
Verify by writing the actual number and checking each condition.
In letter puzzles, give the carry for each column.

Practice questions on Generalised Form and Number Puzzles

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The reverse of the two-digit number $10 x + y$ is:
(a) $10 y + x$ (b) $x + y$ (c) $10 x - y$ (d) $x y$
Hint
Swap the place values.
Answer
(a)
Q2 · 2 marks · Short answerBoard level
Write the digits of 63 reversed, and find the sum of 63 and its reverse. Show it equals $11 \times$ (digit sum).
Hint
$63 + 36$ .
Answer
Reverse $= 36$ ; sum $= 99$ . Digit sum $= 9$ and $11 \times 9 = 99$ .
Q3 · 3 marks · Short answerHarder board level
A two-digit number is 4 times the sum of its digits, and its ones digit is 6. Find the number.
Hint
$10 a + 6 = 4 (a + 6)$ .
Answer
$10 a + 6 = 4 a + 24 \Rightarrow 6 a = 18$ , $a = 3$ . The number is $36$ (check: $4 \times 9 = 36$ ).
Q4 · 4 marks · Short answerHarder board level
The sum of a two-digit number and its reverse is 110, and its digits differ by 4. Find the possible numbers.
Hint
Use $11 (a + b) = 110$ .
Answer
$a + b = 10$ and $∣ a - b ∣ = 4$ , so digits are $7$ and $3$ . Numbers: $73$ and $37$ (check: $73 + 37 = 110$ ).
Q5 · 4 marks · Long answerHOTS / competency
For the three-digit number $\overline{pq r}$ with $p > r$ , show that the difference with its reversal does not depend on $q$ . If the difference is 495, find $p - r$ and give an example.
Hint
Expand both numbers; q cancels.
Answer
$\overline{pq r} - \overline{r q p} = 99 (p - r)$ , with no $q$ . $99 (p - r) = 495 \Rightarrow p - r = 5$ . Example: $601 - 106 = 495$ (also $641 - 146 = 495$ ).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Generalised form of the two-digit number with tens digit a and ones digit b?Show answer
10a + b.
2.What is 100a + 10b + c reversed?Show answer
100c + 10b + a.
3.Number + reverse (two digit) equals?Show answer
11(a + b).
4.Number - reverse (two digit) equals?Show answer
9(a - b).

Generalised Form and Number Puzzles: frequently asked questions

Is ab the same as a x b here?

No. In these puzzles ab, written with a bar, means the digits side by side, i.e. 10a + b.

Why can the leading digit not be 0?

Then the number would have fewer digits.