MathsICSE Class 8 · Chapter 5

Playing with Numbers ICSE Class 8: General Form, Puzzles and Divisibility Rules

Updated 15 October 2026 · Fact-checked

Playing with Numbers writes a two-digit number as 10a + b and a three-digit number as 100a + 10b + c, where a, b, c are digits. ICSE Class 8 uses these general forms to explain digit-reversal tricks, solve letter puzzles and prove the divisibility tests for 2, 3, 5, 9 and 10.

What Playing with Numbers is about

Every number we write is built from its digits and place values. The number 47 really means $10 \times 4 + 7$ , and 305 means $100 \times 3 + 10 \times 0 + 5$ . Writing a number this way with letters for the digits is called its generalised form:

a two-digit number $\overline{ab} = 10 a + b$ ,
a three-digit number $\overline{ab c} = 100 a + 10 b + c$ ,

where $a, b, c$ are digits from 0 to 9 (and the first digit $a$ is not 0).

The CISCE Class VIII curriculum uses the generalised form to:

explain number puzzles and tricks, such as reversing the digits and adding or subtracting,
solve and create letter puzzles in which each letter stands for a digit,
deduce the divisibility rules for 2, 3, 5, 9 and 10 for two- and three-digit numbers.

The bar in $\overline{ab}$ just reminds you that $ab$ here is a number made of digits, not $a \times b$ . In a school exam you will prove a property in general, apply it to a number, find a missing digit, or solve a letter puzzle with full reasoning.

Weightage: CISCE publishes no question paper, marking scheme or chapter weightage for Class VIII Mathematics: your school sets and marks the exam. Playing with Numbers sits in the Number System theme of the CISCE Curriculum for Upper Primary Classes (VI-VIII).Source: CISCE Curriculum for Upper Primary Classes (VI-VIII), RDCD 2016

Key concepts

Generalised form of numbers

Two-digit: $\overline{ab} = 10 a + b$ , with $a$ from 1 to 9 and $b$ from 0 to 9.
Three-digit: $\overline{ab c} = 100 a + 10 b + c$ , with $a$ from 1 to 9 and $b, c$ from 0 to 9.

The reversed number swaps the order of the digits: the reverse of $\overline{ab}$ is $\overline{ba} = 10 b + a$ , and the reverse of $\overline{ab c}$ is $\overline{c ba} = 100 c + 10 b + a$ .

The point of the general form is that one calculation with letters proves something for every number of that kind at once.

Worked example

Write 47 and 305 in generalised form, and write the reverse of 305 in generalised form.

Try it first, then show the solutionSolution

$47 = 10 \times 4 + 7$ .
$305 = 100 \times 3 + 10 \times 0 + 5$ .
Reverse of 305 is $503 = 100 \times 5 + 10 \times 0 + 3$ .

Answer: $10 \times 4 + 7$ ; $100 \times 3 + 10 \times 0 + 5$ ; $100 \times 5 + 10 \times 0 + 3$

Reversing a two-digit number

Add or subtract a two-digit number and its reverse:

Sum: $(10 a + b) + (10 b + a) = 11 a + 11 b = 11 (a + b)$ . Always a multiple of 11; dividing by 11 gives the sum of the digits.
Difference (when $a > b$ ): $(10 a + b) - (10 b + a) = 9 a - 9 b = 9 (a - b)$ . Always a multiple of 9; dividing by 9 gives the difference of the digits.

These two facts solve most "the digits are reversed" word problems.

Worked example

Check both facts for the number 74.

Try it first, then show the solutionSolution

Sum: $74 + 47 = 121 = 11 \times 11$ , and $7 + 4 = 11$ ✓.
Difference: $74 - 47 = 27 = 9 \times 3$ , and $7 - 4 = 3$ ✓.

Answer: Sum $= 11 \times 11$ ; difference $= 9 \times 3$ .

Three-digit tricks

Reverse and subtract (when $a > c$ ): $\overline{ab c} - \overline{c ba} = (100 a + 10 b + c) - (100 c + 10 b + a) = 99 a - 99 c = 99 (a - c)$ . The middle digit cancels, and the answer is always a multiple of 99 (so also of 9 and 11).
Rotate and add: $\overline{ab c} + \overline{b c a} + \overline{c ab}$ . Each digit appears once in the hundreds, tens and units places, so the sum is $111 (a + b + c)$ . Since $111 = 3 \times 37$ , the sum is always divisible by 37.

Worked example

Find $237 + 372 + 723$ and show it is $111$ times the sum of the digits.

Try it first, then show the solutionSolution

$237 + 372 + 723 = 1332$ .
Digit sum $= 2 + 3 + 7 = 12$ , and $111 \times 12 = 1332$ ✓.

Answer: $1332 = 111 \times 12$

Letter puzzles

In a letter puzzle each letter stands for one digit, different letters stand for different digits, and the first digit of a number is not 0.

Strategy:
1. Start from the units column, where there is no carry coming in.
2. Decide what digit makes the units work and note any carry.
3. Move left, using the carry.
4. Check the finished sum.

If more than one value seems possible, test each one; usually only one fits every column.

Worked example

Solve $\overline{3 A} + 25 = \overline{B 2}$ .

Try it first, then show the solutionSolution

Units: $A + 5$ ends in 2, so $A + 5 = 12$ and $A = 7$ , with carry 1.
Tens: $3 + 2 + 1 = 6$ , so $B = 6$ .
Check: $37 + 25 = 62$ ✓.

Answer: $A = 7$ , $B = 6$

Divisibility by 10, 5 and 2

Write $\overline{ab c} = 100 a + 10 b + c = 10 (10 a + b) + c$ .

The part $10 (10 a + b)$ is always divisible by 10, and therefore by 5 and by 2. So divisibility depends only on the units digit $c$ :

by 10 if $c = 0$ ;
by 5 if $c = 0$ or $5$ ;
by 2 if $c$ is even (0, 2, 4, 6, 8).

The same argument works for a two-digit number $10 a + b$ .

Worked example

Without dividing, decide whether 425 is divisible by 2, 5 and 10.

Try it first, then show the solutionSolution

$425 = 10 \times 42 + 5$ , and $10 \times 42$ is divisible by 2, 5 and 10.
So it depends on the units digit 5: it is odd, so not divisible by 2; it is 5, so divisible by 5 but not by 10.

Answer: Divisible by 5 only.

Divisibility by 9 and 3

Split the general form: $100 a + 10 b + c = 99 a + 9 b + (a + b + c) = 9 (11 a + b) + (a + b + c)$ .

The part $9 (11 a + b)$ is always divisible by 9 (and so by 3). So:

$\overline{ab c}$ is divisible by 9 exactly when the digit sum $a + b + c$ is divisible by 9;
it is divisible by 3 exactly when the digit sum is divisible by 3.

The remainder on dividing by 9 is the same as the remainder of the digit sum. For two-digit numbers, $10 a + b = 9 a + (a + b)$ gives the same rule.

Worked example

Find all possible digits $x$ for which the three-digit number $\overline{5 x 7}$ is divisible by 3.

Try it first, then show the solutionSolution

Digit sum $= 5 + x + 7 = 12 + x$ .
$12 + x$ must be a multiple of 3, with $x$ from 0 to 9: $12, 15, 18, 21$ .
So $x = 0, 3, 6$ or $9$ .

Answer: $x = 0, 3, 6, 9$

Maths in real life: Playing with Numbers

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.

Test for divisibility by 9

Suppose 738 students have to be seated for an assembly in rows of exactly 9, with no gaps.

Is it possible, and if so how many rows are needed?

Try it first, then show the solutionSolution

Digit sum $= 7 + 3 + 8 = 18$ , a multiple of 9, so 738 is divisible by 9.
Rows $= 738 \div 9 = 82$ .

Answer: Yes: 82 rows of 9.

Memory hook: Digit sum a multiple of 9 means the number is too.

Tests for 5 and 10

Imagine you have 425 sweets to pack into bags of 5, or into bags of 10, with none left over.

Which bag size works?

Try it first, then show the solutionSolution

425 ends in 5, so it is divisible by 5: $425 \div 5 = 85$ bags.
It does not end in 0, so bags of 10 would leave some sweets over.

Answer: Bags of 5 work (85 bags); bags of 10 do not.

Memory hook: Ends in 0 or 5 for 5; only 0 for 10.

Digits reversed

Suppose a two-digit locker code has digits adding up to 8, and swapping the digits makes the code 54 more.

What is the code?

Try it first, then show the solutionSolution

Let the code be $10 a + b$ . Then $a + b = 8$ and $9 (b - a) = 54$ , so $b - a = 6$ .
Adding: $2 b = 14$ , $b = 7$ and $a = 1$ . Code $= 17$ (check: $71 - 17 = 54$ ).

Answer: 17

Memory hook: Reversal changes a two-digit number by 9 times the digit difference.

Topics in this chapter

Playing with Numbers formula sheet

Two-digit number: $\overline{ab} = 10 a + b$; a from 1 to 9, b from 0 to 9.
Three-digit number: $\overline{ab c} = 100 a + 10 b + c$; a ≠ 0.
Two-digit sum with reverse: $\overline{ab} + \overline{ba} = 11 (a + b)$; Always a multiple of 11.
Two-digit difference with reverse: $\overline{ab} - \overline{ba} = 9 (a - b)$; a > b; multiple of 9.
Three-digit difference with reverse: $\overline{ab c} - \overline{c ba} = 99 (a - c)$; a > c; multiple of 99.
Test for 9 and 3: $100 a + 10 b + c = 9 (11 a + b) + (a + b + c)$; Look at the digit sum.

How the school exam asks Playing with Numbers

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Generalised form	Write 908 in generalised form; write the reverse of a number in generalised form.	Multiply each digit by its place value.
Prove a reversal property	Show that the sum of a two-digit number and its reverse is divisible by 11.	Use letters, simplify, and factor out the 11 (or 9, or 99).
Find a missing digit	If 7x2 is divisible by 9, find x.	Write the digit sum, list multiples of 9 (or 3) it can equal, keep values 0 to 9.
Letter puzzles	Find the digits A and B in AB × 6 = BBB.	Start from the units column; track carries; test each possibility; check.
Digit word problems	The digits of a two-digit number add to 9; reversing them increases the number by 27. Find it.	Let the number be 10a + b, make two equations and solve.

Tricks to remember Playing with Numbers

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

Add → 11, subtract → 9
Two-digit number plus its reverse is a multiple of 11; minus its reverse is a multiple of 9.
Use it when: Reversal problems.
Middle digit vanishes
In abc − cba the middle digit cancels, leaving 99(a − c).
Use it when: Three-digit reversal.
Last digit for 2, 5, 10; digit sum for 3, 9
The units digit decides 2, 5 and 10. The digit sum decides 3 and 9.
Use it when: Choosing the right test fast.
Units first in letter puzzles
No carry enters the units column, so it is the safest place to start.
Use it when: Cryptarithms.

Playing with Numbers: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: From general form to number · 1 mark · MCQExam level

$100 \times 6 + 10 \times 0 + 9$ is the number:
(a) 69
(b) 609
(c) 690
(d) 6009

Try it first, then show the solutionSolution

Hundreds digit 6, tens digit 0, units digit 9.
The number is 609.

Answer: (b)

Example 2: Sum with the reverse · 3 marks · Short answerExam level

Show that $63 + 36$ is divisible by 11 and that the quotient equals the sum of the digits of 63. Then prove this for every two-digit number.

Try it first, then show the solutionSolution

$63 + 36 = 99 = 11 \times 9$ , and $6 + 3 = 9$ ✓.
In general: $(10 a + b) + (10 b + a) = 11 a + 11 b$ .
$= 11 (a + b)$ , so the sum is divisible by 11 with quotient $a + b$ .

Answer: $99 = 11 \times 9$ ; in general $11 (a + b)$ .

Example 3: Missing digit for 9 · 3 marks · Short answerExam level

If the three-digit number $\overline{7 x 2}$ is divisible by 9, find the possible values of $x$ .

Try it first, then show the solutionSolution

Digit sum $= 7 + x + 2 = 9 + x$ .
With $x$ from 0 to 9, $9 + x$ is between 9 and 18; the multiples of 9 there are 9 and 18.
So $x = 0$ or $x = 9$ (numbers 702 and 792).

Answer: $x = 0$ or $9$

Example 4: A letter sum · 3 marks · Short answerHarder exam level

Find the digits $A$ and $B$ if $\overline{A 1} + \overline{1 B} = \overline{B 0}$ .

Try it first, then show the solutionSolution

Units: $1 + B$ ends in 0, so $1 + B = 10$ , $B = 9$ , carry 1.
Tens: $A + 1 + 1 = B = 9$ , so $A = 7$ .
Check: $71 + 19 = 90$ ✓.

Answer: $A = 7$ , $B = 9$

Example 5: Reverse and subtract · 4 marks · Long answerHarder exam level

Prove that $\overline{ab c} - \overline{c ba}$ (with $a > c$ ) is always divisible by 99. If the difference is 495, what is $a - c$ ?

Try it first, then show the solutionSolution

$\overline{ab c} - \overline{c ba} = (100 a + 10 b + c) - (100 c + 10 b + a)$ .
$= 99 a - 99 c = 99 (a - c)$ , which is divisible by 99.
If the difference is 495, then $99 (a - c) = 495$ .
$a - c = 495 \div 99 = 5$ .

Answer: Divisible by 99; $a - c = 5$ .

Example 6: A product puzzle · 4 marks · Long answerHarder exam level

Find the digits $A$ and $B$ if $\overline{A B} \times 6 = \overline{B B B}$ .

Try it first, then show the solutionSolution

$\overline{B B B} = 111 B$ , so $\overline{A B} = \frac{111 B}{6} = \frac{37 B}{2}$ . So $B$ must be even.
$B = 2$ : $\overline{A B} = 37$ , but its units digit is 7, not 2 ✗. $B = 4$ : $\overline{A B} = 74$ , units digit 4 ✓.
$B = 6$ gives 111 and $B = 8$ gives 148, which are not two-digit ✗. ( $B = 0$ gives 0 ✗.)
So $A = 7$ , $B = 4$ ; check $74 \times 6 = 444$ ✓.

Answer: $A = 7$ , $B = 4$

Example 7: Rotate and add · 5 marks · Long answerHOTS / competency

For any three-digit number $\overline{ab c}$ , prove that $\overline{ab c} + \overline{b c a} + \overline{c ab}$ is divisible by 37. Verify for 248 and find the quotient.

Try it first, then show the solutionSolution

Hundreds: $100 (a + b + c)$ ; tens: $10 (b + c + a)$ ; units: $(c + a + b)$ .
Sum $= 111 (a + b + c) = 3 \times 37 \times (a + b + c)$ , so it is divisible by 37.
For 248: $248 + 482 + 824 = 1554$ .
$1554 \div 37 = 42$ .
Check: $3 \times (2 + 4 + 8) = 3 \times 14 = 42$ ✓.

Answer: Divisible by 37; for 248 the quotient is 42.

Case-study practice (Section B style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 10-mark question split into 3 + 3 + 4 marks.

The magic number trick

Suppose at a school fair a student runs this trick: "Think of a two-digit number. Add it to the number with its digits reversed. Divide the result by 11. Tell me your answer and I will tell you something about your number."

(i) · 3 marks
Do the trick with 58. What answer do you get, and how is it related to 58?
Answer
Answer: 13, which is $5 + 8$
$58 + 85 = 143$ ; $143 \div 11 = 13 = 5 + 8$ .
(ii) · 3 marks
Using the general form, prove that the answer is always the sum of the digits.
Answer
Answer: $a + b$
$(10 a + b) + (10 b + a) = 11 (a + b)$ ; dividing by 11 gives $a + b$ .
(iii) · 4 marks
A player says the answer is 14, and the tens digit of the number was 9. Find the number.
Answer
Answer: 95
$a + b = 14$ with $a = 9$ , so $b = 5$ . The number is 95. Check: $95 + 59 = 154$ , $154 \div 11 = 14$ ✓.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Prove that the difference between a two-digit number $\overline{ab}$ ( $a > b$ ) and its reverse is divisible by 9. (3 marks)

Step	Marks
Number $= 10 a + b$ ; reverse $= 10 b + a$	1
Difference $= 9 a - 9 b$	1
$= 9 (a - b)$ , so divisible by 9	1

Using the general form, prove that a three-digit number is divisible by 9 if the sum of its digits is divisible by 9. (4 marks)

Step	Marks
$\overline{ab c} = 100 a + 10 b + c$	1
$= 99 a + 9 b + (a + b + c)$	1
$= 9 (11 a + b) + (a + b + c)$ , and the first part is divisible by 9	1
So if $a + b + c$ is divisible by 9, the whole number is	1

A two-digit number is 4 times the sum of its digits. If 18 is added to it, the digits are reversed. Find the number. (5 marks)

Step	Marks
Let the number be $10 a + b$ ; then $10 a + b = 4 (a + b)$	1
Simplify: $6 a = 3 b$ , so $b = 2 a$	1
$10 a + b + 18 = 10 b + a \Rightarrow 9 (b - a) = 18 \Rightarrow b - a = 2$	1
Then $2 a - a = 2$ , so $a = 2$ and $b = 4$	1
Number $= 24$ (check: $24 = 4 \times 6$ ; $24 + 18 = 42$ )	1

Where students lose marks in Playing with Numbers

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

Mistake: Reading $\overline{ab}$ as $a \times b$
The digits $a$ and $b$ have place values; $\overline{ab}$ means $10 a + b$ .
Wrong working
$\overline{ab} + \overline{ba} = ab + ba = 2 ab$
Right working
$(10 a + b) + (10 b + a) = 11 (a + b)$
Do this: Always expand with place values before simplifying.
Usually costs: The whole proof.
Mistake: Checking only one example as a proof
Showing $74 + 47 = 121$ does not prove the result for every number.
Do this: Use the general form with letters for a proof; examples are only for checking.
Usually costs: Most of the marks for a 'prove' question.
Mistake: Missing a possible digit
A digit sum can equal more than one multiple of 3 or 9.
Wrong working
$\overline{5 x 7}$ divisible by 3, so $x = 3$
Right working
$12 + x \in {12, 15, 18, 21}$ , so $x = 0, 3, 6, 9$
Do this: List every multiple in the possible range of the digit sum.
Usually costs: 1 to 2 marks.
Mistake: Forgetting the carry in letter puzzles
A column that adds to 10 or more passes 1 to the next column.
Do this: Write the carry above the next column every time.
Usually costs: The answer and the check.
Mistake: Letting the first digit be 0
A two-digit number cannot start with 0, so $a \neq = 0$ .
Do this: State the range of each digit before solving.
Usually costs: 1 mark for an invalid solution.

Practice questions on Playing with Numbers

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 · MCQExam level
Which number is divisible by 5 but not by 10?
(a) 340
(b) 342
(c) 345
(d) 350
Hint
Look at the units digit.
Answer
(c)
Q2 · 1 mark · Very short answerExam level
Write 908 in generalised form.
Hint
Place values 100, 10, 1.
Answer
$100 \times 9 + 10 \times 0 + 8$ .
Q3 · 2 marks · Short answerExam level
Without dividing, test 738 for divisibility by 2, 3, 5, 9 and 10.
Hint
Units digit 8; digit sum 18.
Answer
Divisible by 2 (even), 3 and 9 (digit sum 18); not by 5 or 10 (units digit 8).
Q4 · 2 marks · Short answerHarder exam level
If the three-digit number $\overline{5 y 4}$ is divisible by 3, find all possible values of $y$ .
Hint
Digit sum $9 + y$ .
Answer
$9 + y \in {9, 12, 15, 18}$ , so $y = 0, 3, 6, 9$ .
Q5 · 1 mark · Assertion-reasonHarder exam level
Assertion (A): For any two-digit number $\overline{ab}$ with $a > b$ , $\overline{ab} - \overline{ba}$ is divisible by 9.
Reason (R): $(10 a + b) - (10 b + a) = 9 (a - b)$ .
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
Is (R) the proof of (A)?
Answer
(c)
Q6 · 3 marks · Short answerHarder exam level
Find the digit $A$ if $\overline{1 A} \times A = \overline{9 A}$ .
Hint
The units digit of $A \times A$ must be $A$ .
Answer
$A \times A$ ends in $A$ for $A = 0, 1, 5, 6$ . Test: $10 \times 0 = 0$ ✗, $11 \times 1 = 11$ ✗, $15 \times 5 = 75$ ✗, $16 \times 6 = 96$ ✓. $A = 6$ .
Q7 · 3 marks · Short answerHarder exam level
The sum of the digits of a two-digit number is 9. When the digits are reversed, the number increases by 27. Find the number.
Hint
$9 (b - a) = 27$ .
Answer
$a + b = 9$ and $b - a = 3$ , so $b = 6$ , $a = 3$ . Number $= 36$ .
Q8 · 2 marks · Short answerHarder exam level
The three-digit number $\overline{24 x}$ is divisible by 9. Find $x$ and show that the number is also divisible by 3.
Hint
Digit sum $6 + x$ .
Answer
$6 + x = 9$ gives $x = 3$ (18 would need $x = 12$ ). The number is 243; digit sum 9 is divisible by 3, so 243 is divisible by 3.
Q9 · 3 marks · Short answerHarder exam level
Without dividing, find the remainder when 457 is divided by 9. Explain using the general form.
Hint
$457 = 9 (11 \times 4 + 5) + (4 + 5 + 7)$ .
Answer
$457 = 9 \times 49 + 16$ , and $16 = 9 + 7$ . Remainder 7, the same as the remainder of the digit sum 16.
Q10 · 3 marks · Short answerHarder exam level
Solve the letter puzzle $\overline{A B} + 37 = \overline{9 A}$ .
Hint
Try carry 0 and carry 1 from the units column.
Answer
Tens: $A + 3 + carry = 9$ . With no carry $A = 6$ and $B + 7 = 6$ is impossible. With carry 1, $A = 5$ and $B + 7 = 15$ , $B = 8$ . Check $58 + 37 = 95$ ✓.
Q11 · 4 marks · Long answerHOTS / competency
A three-digit number has middle digit 3. When its digits are reversed, the new number is 594 less. The sum of its first and last digits is 10. Find the number.
Hint
$99 (a - c) = 594$ .
Answer
$a - c = 6$ and $a + c = 10$ , so $a = 8$ , $c = 2$ . Number $= 832$ (check $832 - 238 = 594$ ).
Q12 · 3 marks · Short answerHOTS / competency
Prove that a two-digit number $10 a + b$ is divisible by 3 exactly when $a + b$ is divisible by 3.
Hint
Split 10a as 9a + a.
Answer
$10 a + b = 9 a + (a + b)$ . Since $9 a$ is always divisible by 3, $10 a + b$ is divisible by 3 exactly when $a + b$ is.

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.Write the general form of a three-digit number.Show answer
100a + 10b + c.
2.What is the reverse of 10a + b?Show answer
10b + a.
3.A two-digit number plus its reverse is always divisible by which number?Show answer
11.
4.abc − cba (a > c) is always divisible by which number?Show answer
99.
5.When is a number divisible by 5?Show answer
When its units digit is 0 or 5.
6.Is 459 divisible by 9?Show answer
Yes: 4 + 5 + 9 = 18.
7.Remainder when 457 is divided by 9?Show answer
7 (digit sum 16, and 16 leaves 7).
8.abc + bca + cab equals what?Show answer
111(a + b + c).

Best order to study it

Learn the generalised form of two- and three-digit numbers and their reverses.
Prove the two-digit reversal results (11 and 9), then the three-digit ones (99 and 111).
Deduce the tests for 10, 5 and 2, then for 9 and 3, from the general form.
Practise missing-digit questions.
Finish with letter puzzles and digit word problems.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you write any number in generalised form?
Can you prove ab + ba = 11(a + b) and ab − ba = 9(a − b)?
Can you prove the test for 9 from 100a + 10b + c?
Do you list every possible value when finding a missing digit?
Do you start letter puzzles from the units column and track carries?
Do you check that the first digit is not 0?

Playing with Numbers: frequently asked questions

Why do we need the general form if the rules are already known?

The general form shows why a rule works for every number, not just the examples you tried. That is what 'deduce the divisibility rules' in the syllabus means.

Does the test for 9 tell me the remainder too?

Yes. Because 100a + 10b + c = 9(11a + b) + (a + b + c), the remainder on dividing the number by 9 is the same as the remainder on dividing its digit sum by 9.

Can two letters in a puzzle stand for the same digit?

No. In these puzzles different letters stand for different digits, and the same letter always stands for the same digit.

Why is the test for 3 the same as the test for 9?

Because 9 is a multiple of 3: the part 9(11a + b) is divisible by 3 as well, so again only the digit sum matters.

Which divisibility rules are in the Class VIII syllabus?

The rules for 2, 3, 5, 9 and 10, deduced for two- and three-digit numbers written in general form.

Based on: CISCE Curriculum for Upper Primary Classes (VI-VIII), Mathematics (Class VIII); CISCE RDCD: Curriculum Pre-School to Class VIII