NCERT Solutions for Class 6 Maths Chapter 3: Number Play — Complete Guide

Every digit in a number has two values:

Face value: The digit itself, regardless of its position. The face value of $7$ in $3{,}725$ is simply $7$.

Place value: The value the digit represents based on its position. The place value of $7$ in $3{,}725$ is $700$ (because $7$ is in the hundreds place).

The expanded form of a number expresses it as a sum of place values:

Using powers of $10$:

For a two-digit number with tens digit $a$ and units digit $b$:

For a three-digit number $\overline{abc}$:

Understanding this algebraic representation is the key to proving many number tricks.

The digit sum of a number is the sum of all its digits.

Examples:
- Digit sum of $473 = 4 + 7 + 3 = 14$
- Digit sum of $9999 = 9 + 9 + 9 + 9 = 36$

The digit root (or repeated digit sum) is obtained by repeatedly summing the digits until you get a single digit.

Examples:
- Digit root of $473$: $4 + 7 + 3 = 14 \to 1 + 4 = 5$
- Digit root of $9999$: $9 + 9 + 9 + 9 = 36 \to 3 + 6 = 9$

Key property: The digit root of a number equals the remainder when the number is divided by $9$ (with $9$ replacing $0$). This is called casting out nines and has been used for centuries to check arithmetic.

A palindromic number (or palindrome) reads the same forwards and backwards.

Examples: $121$, $1331$, $45654$, $9009$, $7$, $11$, $1001$.

Non-examples: $123$, $100$, $45$.

Key facts about palindromes:
- All single-digit numbers ($1$ to $9$) are palindromes.
- Two-digit palindromes: $11, 22, 33, 44, 55, 66, 77, 88, 99$ (nine in total).
- Three-digit palindromes have the form $\overline{aba}$ where $a \neq 0$. There are $9 \times 10 = 90$ of them.
- The 196 problem asks whether every number eventually becomes a palindrome when you repeatedly reverse and add. This is an unsolved problem in mathematics!

Estimation means finding an approximate answer quickly, without exact calculation.

Rounding is the most common method of estimation:
- To round to the nearest $10$: look at the units digit. If $\geq 5$, round up; if $< 5$, round down.
- To round to the nearest $100$: look at the tens digit.
- To round to the nearest $1000$: look at the hundreds digit.

Examples:
- $347$ rounded to the nearest $10$ is $350$ (because $7 \geq 5$).
- $347$ rounded to the nearest $100$ is $300$ (because $4 < 5$).
- $6{,}782$ rounded to the nearest $1000$ is $7{,}000$ (because $7 \geq 5$).

Estimation in multiplication: Round each number, then multiply.

The exact answer is $3{,}901$, so the estimate is quite close.

Quick tests to check if a number is divisible by small numbers:

• **By $2$:** Last digit is even ($0, 2, 4, 6, 8$).
  - **By $3$:** Sum of digits is divisible by $3$.
  - **By $4$:** Last two digits form a number divisible by $4$.
  - **By $5$:** Last digit is $0$ or $5$.
  - **By $6$:** Divisible by both $2$ and $3$.
  - **By $8$:** Last three digits form a number divisible by $8$.
  - **By $9$:** Sum of digits is divisible by $9$.
  - **By $10$:** Last digit is $0$.
  - **By $11$:** Alternating sum of digits is divisible by $11$.

Example: Is $4{,}356$ divisible by $9$?
Digit sum: $4 + 3 + 5 + 6 = 18$, and $18 \div 9 = 2$. Yes! $\checkmark$

Problem: Write $4{,}738$ in expanded form using place values.

Solution:

Using powers of $10$:

Simplified: $4{,}738 = 4000 + 700 + 30 + 8$.

Place value of each digit:
- $4$ is in the thousands place $\Rightarrow$ place value $= 4000$
- $7$ is in the hundreds place $\Rightarrow$ place value $= 700$
- $3$ is in the tens place $\Rightarrow$ place value $= 30$
- $8$ is in the ones place $\Rightarrow$ place value $= 8$

Answer: $4{,}738 = 4000 + 700 + 30 + 8$.

Problem: Find the sum of digits of $9 \times 1, 9 \times 2, 9 \times 3, \ldots, 9 \times 10$. What pattern do you notice?

Solution:

Pattern: The digit sum of every multiple of $9$ (from $9$ to $90$) is always $9$.

Why? For $9 \times n$ where $1 \leq n \leq 10$, the two-digit result has tens digit $(n-1)$ and units digit $(10-n)$. Sum $= (n-1) + (10-n) = 9$.

Answer: The digit sum is always $9$.

Problem: Take any two-digit number where the digits are different. Reverse its digits and subtract the smaller from the larger. What do you notice?

Solution:

Let the two-digit number be $\overline{ab}$ where $a > b$.

The number $= 10a + b$.
The reversed number $= 10b + a$.

Difference:

The result is always a **multiple of $9$**.

Examples:
- $72 - 27 = 45 = 9 \times 5$ (and $7 - 2 = 5$)
- $83 - 38 = 45 = 9 \times 5$ (and $8 - 3 = 5$)
- $91 - 19 = 72 = 9 \times 8$ (and $9 - 1 = 8$)
- $64 - 46 = 18 = 9 \times 2$ (and $6 - 4 = 2$)

Key insight: The difference is exactly $9$ times the difference between the two digits.

Answer: The difference is always $9 \times |a - b|$, a multiple of $9$.

Problem: Start with $56$. Reverse its digits and add. Repeat until you get a palindrome.

Solution:

Step 1: $56 + 65 = 121$.

$121$ is a palindrome! (reads the same forwards and backwards)

Let us try another example: Start with $87$.

Step 1: $87 + 78 = 165$.
$165$ is not a palindrome.

Step 2: $165 + 561 = 726$.
$726$ is not a palindrome.

Step 3: $726 + 627 = 1353$.
Not a palindrome.

Step 4: $1353 + 3531 = 4884$.
$4884$ is a palindrome! $\checkmark$

It took $4$ steps starting from $87$.

Answer: $56 \to 121$ (palindrome in $1$ step). $87 \to 4884$ (palindrome in $4$ steps).

Problem: Take a three-digit number $\overline{abc}$ where $a > c$. Reverse it to get $\overline{cba}$ and subtract. What do you notice?

Solution:

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

Difference:

The result is always a **multiple of $99$**!

Examples:
- $752 - 257 = 495 = 99 \times 5$ (and $7 - 2 = 5$)
- $843 - 348 = 495 = 99 \times 5$ (and $8 - 3 = 5$)
- $961 - 169 = 792 = 99 \times 8$ (and $9 - 1 = 8$)

Notice that the middle digit $b$ cancels out completely — the result depends only on the first and last digits!

Answer: The difference is always $99 \times (a - c)$.

Problem: Take a two-digit number, reverse it, and add. What do you notice?

Solution:

Let the number be $\overline{ab} = 10a + b$.
Reversed: $\overline{ba} = 10b + a$.

Sum:

The sum is always a **multiple of $11$**!

Examples:
- $34 + 43 = 77 = 11 \times 7$ (and $3 + 4 = 7$)
- $58 + 85 = 143 = 11 \times 13$ (and $5 + 8 = 13$)
- $29 + 92 = 121 = 11 \times 11$ (and $2 + 9 = 11$)

Answer: The sum is always $11$ times the sum of the digits.

Problem: Take any three-digit number where the first digit is at least $2$ more than the last digit. Reverse it and subtract the smaller from the larger. Now reverse the result and add it to itself. What do you get?

Solution:

Let us try $532$:

Step 1: Reverse $\to 235$. Subtract: $532 - 235 = 297$.
Step 2: Reverse $297 \to 792$. Add: $297 + 792 = 1089$.

Let us try $841$:

Step 1: Reverse $\to 148$. Subtract: $841 - 148 = 693$.
Step 2: Reverse $693 \to 396$. Add: $693 + 396 = 1089$.

Let us try $720$:

Step 1: Reverse $\to 027 = 27$. Subtract: $720 - 27 = 693$.
Step 2: Reverse $693 \to 396$. Add: $693 + 396 = 1089$.

The answer is **always $1089$**!

Why? From the earlier result, the subtraction gives $99(a - c)$. When $a - c$ ranges from $2$ to $9$, the possible results are $198, 297, 396, 495, 594, 693, 792, 891$. Each of these, when added to its reverse, gives $1089$.

Answer: The result is always $1089$.

Problem: In a four-digit number, the thousands digit is $3$ more than the units digit, the hundreds digit is twice the units digit, and the tens digit is $1$ less than the hundreds digit. If the units digit is $2$, find the number.

Solution:

Let the units digit $= 2$.

Thousands digit $= 2 + 3 = 5$
Hundreds digit $= 2 \times 2 = 4$
Tens digit $= 4 - 1 = 3$

The number is $5{,}432$.

Verification:
- Thousands ($5$) is $3$ more than units ($2$): $5 - 2 = 3$ $\checkmark$
- Hundreds ($4$) is twice units ($2$): $4 = 2 \times 2$ $\checkmark$
- Tens ($3$) is $1$ less than hundreds ($4$): $4 - 3 = 1$ $\checkmark$

Answer: The number is $5{,}432$.

Problem: A two-digit number has a digit sum of $11$. If the digits are reversed, the new number is $27$ more than the original. Find the number.

Solution:

Let the tens digit be $a$ and units digit be $b$.

Condition 1: $a + b = 11$

Condition 2: The reversed number minus the original number $= 27$:

Solve the system:

Adding (1) and (2): $2b = 14 \Rightarrow b = 7$.
From (1): $a = 11 - 7 = 4$.

The number is $47$.

Verification: Digit sum $= 4 + 7 = 11$ $\checkmark$. Reversed $= 74$. Difference $= 74 - 47 = 27$ $\checkmark$.

Answer: The number is $47$.

Problem: Using the digits $3$, $0$, $7$, and $5$ (each used exactly once), find (a) the largest four-digit number, (b) the smallest four-digit number, (c) their difference.

Solution:

(a) For the largest number, arrange digits in decreasing order:

(b) For the smallest number, arrange digits in increasing order. But the leading digit cannot be $0$:

(c) Difference:

Answer: (a) $7{,}530$ (b) $3{,}057$ (c) $4{,}473$.

Problem: Which of the following are palindromic numbers? $121$, $123$, $1001$, $345$, $4554$, $100$, $7$, $909$.

Solution:

Reverse each number and check if it equals the original:
- $121 \to 121$ $\checkmark$ Palindrome
- $123 \to 321$ $\neq 123$ Not a palindrome
- $1001 \to 1001$ $\checkmark$ Palindrome
- $345 \to 543$ $\neq 345$ Not a palindrome
- $4554 \to 4554$ $\checkmark$ Palindrome
- $100 \to 001 = 1$ $\neq 100$ Not a palindrome
- $7 \to 7$ $\checkmark$ Palindrome (single digit)
- $909 \to 909$ $\checkmark$ Palindrome

Answer: $121$, $1001$, $4554$, $7$, $909$ are palindromes.

Problem: How many (a) two-digit palindromes and (b) three-digit palindromes are there?

Solution:

(a) Two-digit palindromes: These have the form $\overline{aa}$ where $a \neq 0$.
Possible values of $a$: $1, 2, 3, 4, 5, 6, 7, 8, 9$.
There are $9$ two-digit palindromes: $11, 22, 33, 44, 55, 66, 77, 88, 99$.

(b) Three-digit palindromes: These have the form $\overline{aba}$.
- $a$ can be $1$ to $9$ (first digit cannot be $0$): $9$ choices.
- $b$ can be $0$ to $9$: $10$ choices.
- The last digit must equal $a$: no additional choice.

Total: $9 \times 10 = 90$ three-digit palindromes.

Examples: $101, 111, 121, 131, \ldots, 191, 202, 212, \ldots, 999$.

Answer: (a) $9$ (b) $90$.

Problem: Starting from $68$, use the reverse-and-add method to reach a palindrome.

Solution:

Step 1: $68 + 86 = 154$. Not a palindrome.
Step 2: $154 + 451 = 605$. Not a palindrome.
Step 3: $605 + 506 = 1{,}111$. Palindrome! $\checkmark$

It took $3$ steps.

Answer: $68 \to 154 \to 605 \to 1{,}111$.

Problem: A date written as DD/MM/YYYY is palindromic if the $8$-digit number DDMMYYYY is a palindrome. Is $02/02/2020$ (i.e., $02022020$) a palindrome?

Solution:

$02022020$ reversed is $02022020$.

Comparing: $0-2-0-2-2-0-2-0$ forwards equals $0-2-0-2-2-0-2-0$ backwards.

Yes, $02022020$ is a palindrome! $\checkmark$

This date, February 2, 2020, was indeed a palindrome date.

Follow-up: The next palindrome date in this format is $12/02/2021$ ($12022021$). Check: reversed is $12022021$. $\checkmark$

Answer: Yes, $02/02/2020$ is a palindrome date.

Problem: Find (a) the smallest $5$-digit palindrome, (b) the largest $5$-digit palindrome, and (c) the smallest $5$-digit palindrome with all distinct digits.

Solution:

A $5$-digit palindrome has the form $\overline{abcba}$.

(a) Smallest: Make $a$ as small as possible ($a = 1$, since $a \neq 0$), $b = 0$, $c = 0$.

(b) Largest: Make all digits as large as possible: $a = b = c = 9$.

(c) Smallest with all distinct digits: $a = 1$, $b = 0$, $c = 2$ (need $a, b, c$ all different, and digits $a, b, c, b, a$ gives digits $1, 0, 2, 0, 1$ — but $b$ repeats as $0$, and $a$ repeats as $1$). For ALL five digits distinct, we need $a \neq b \neq c$ and the five digits $a, b, c, b, a$ must all differ. But $a$ appears twice and $b$ appears twice, so we can have at most $3$ distinct digits. Five distinct digits is impossible in a $5$-digit palindrome of the form $\overline{abcba}$.

So the question means three distinct values for the positions. Smallest: $10201$.

Answer: (a) $10{,}001$ (b) $99{,}999$ (c) $10{,}201$.

Problem: Find all palindromic numbers between $100$ and $200$ that are divisible by $3$.

Solution:

Three-digit palindromes between $100$ and $200$ have the form $\overline{1b1}$ where $b$ is $0$ to $9$.

These are: $101, 111, 121, 131, 141, 151, 161, 171, 181, 191$.

Check divisibility by $3$ using digit sums:
- $101$: $1+0+1=2$, not divisible by $3$.
- $111$: $1+1+1=3$, divisible by $3$. $\checkmark$
- $121$: $1+2+1=4$, not divisible by $3$.
- $131$: $1+3+1=5$, not divisible by $3$.
- $141$: $1+4+1=6$, divisible by $3$. $\checkmark$
- $151$: $1+5+1=7$, not divisible by $3$.
- $161$: $1+6+1=8$, not divisible by $3$.
- $171$: $1+7+1=9$, divisible by $3$. $\checkmark$
- $181$: $1+8+1=10$, not divisible by $3$.
- $191$: $1+9+1=11$, not divisible by $3$.

Answer: $111$, $141$, $171$.

Problem: Is the sum of two palindromes always a palindrome? Is the product of two palindromes always a palindrome?

Solution:

Sum: Not always. $121 + 131 = 252$ (palindrome). But $11 + 22 = 33$ (palindrome), while $99 + 11 = 110$ (not a palindrome).

Product: Not always. $11 \times 11 = 121$ (palindrome). But $11 \times 22 = 242$ (palindrome), while $11 \times 33 = 363$ (palindrome), and $11 \times 99 = 1089$ (not a palindrome).

Answer: Neither the sum nor the product of two palindromes is necessarily a palindrome.

Problem: Start with $196$. Apply the reverse-and-add process for $3$ steps. Do you get a palindrome?

Solution:

Step 1: $196 + 691 = 887$. Not a palindrome.
Step 2: $887 + 788 = 1{,}675$. Not a palindrome.
Step 3: $1{,}675 + 5{,}761 = 7{,}436$. Not a palindrome.

$196$ is famous in mathematics! After hundreds of millions of reverse-and-add steps (computed by powerful computers), no palindrome has been found. It is conjectured that $196$ NEVER produces a palindrome, but this has not been proven.

Answer: No palindrome is reached in $3$ steps. $196$ is a famous unsolved case.

Problem: Complete the magic square using numbers $1$ to $9$:

Solution:

Step 1: Find the magic constant. The sum of all numbers $1$ to $9$ is:

Since there are $3$ rows and each has the same sum:

Step 2: The standard $3 \times 3$ magic square with numbers $1$ to $9$ is:

Step 3: Verify all sums equal $15$:
- Rows: $2+7+6=15$, $9+5+1=15$, $4+3+8=15$ $\checkmark$
- Columns: $2+9+4=15$, $7+5+3=15$, $6+1+8=15$ $\checkmark$
- Diagonals: $2+5+8=15$, $6+5+4=15$ $\checkmark$

Answer: Magic square completed with magic constant $15$.

Problem: In a $3 \times 3$ magic square using numbers $1$ to $9$, which number must go in the center?

Solution:

The center number appears in one row, one column, and both diagonals — that is $4$ lines.

The magic constant is $15$. The center number $c$ plus two other numbers must equal $15$ for each of these $4$ lines.

Summing all $4$ lines through the center:

But each pair of opposite numbers is counted once, and the center is counted $4$ times. The sum of all $9$ numbers is $45$, and each of the $8$ non-center numbers appears in exactly $2$ of these $4$ lines.

Alternatively: the three numbers in the middle row, middle column, and two diagonals together use all $9$ numbers plus $3$ extra copies of the center.

Simpler approach: For the standard $3 \times 3$ magic square with $1$-$9$, the center must be $5$ (the middle value). This can be proven by considering the $4$ lines through the center.

Answer: The center must be $5$.

Problem: Arrange the numbers $2, 4, 6, 8, 10, 12, 14, 16, 18$ in a $3 \times 3$ magic square.

Solution:

Step 1: Find the magic constant.
Sum $= 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 = 90$.
Magic constant $= \frac{90}{3} = 30$.

Step 2: Notice these are the numbers $1$ to $9$ each multiplied by $2$. So take the standard magic square and multiply every entry by $2$:

Step 3: Verify:
- Rows: $4+14+12=30$, $18+10+2=30$, $8+6+16=30$ $\checkmark$
- Columns: $4+18+8=30$, $14+10+6=30$, $12+2+16=30$ $\checkmark$
- Diagonals: $4+10+16=30$, $12+10+8=30$ $\checkmark$

Answer: Magic square with magic constant $30$.

Problem: Fill in the blanks in this addition:

Solution:

Let the missing digits be $x$ and $y$:

Ones column: $7 + 2 = 9$ $\checkmark$ (no carry)

Tens column: $x + y = 2$ (if no carry from ones) or $x + y = 12$ (if there is a carry to hundreds).

Hundreds column: $3 + 4 = 7$, but we need $8$, so there must be a carry of $1$ from the tens column.

This means $x + y \geq 10$, so $x + y = 12$ (with carry $1$ to hundreds: $3 + 4 + 1 = 8$ $\checkmark$).

The tens digit of the sum: $(x + y) - 10 = 12 - 10 = 2$ $\checkmark$ (matches the $2$ in $829$).

Possible values: $x + y = 12$ with $x, y$ single digits (0-9).
Options: $(3,9), (4,8), (5,7), (6,6), (7,5), (8,4), (9,3)$.

All are valid. One solution is $x = 5, y = 7$: $357 + 472 = 829$. $\checkmark$

Answer: One solution: $357 + 472 = 829$ (with $x = 5, y = 7$).

Problem: A number trick says: Think of a number, double it, add $10$, divide by $2$, subtract the original number. The answer is always $5$. Why?

Solution:

Let the number be $n$.

Step 1: Double it: $2n$.
Step 2: Add $10$: $2n + 10$.
Step 3: Divide by $2$: $\frac{2n + 10}{2} = n + 5$.
Step 4: Subtract the original: $(n + 5) - n = 5$.

The $n$ cancels out, so the result is always $5$, regardless of the starting number.

Answer: The algebraic proof shows the original number cancels out, always leaving $5$.

Problem: Take any four-digit number where not all digits are the same. Arrange its digits in descending order and ascending order, and subtract. Repeat. What happens?

Solution:

Let us start with $3187$:

Step 1: Descending $8731$ $-$ Ascending $1378$ $= 7353$.
Step 2: $7553 - 3557 = 3996$.
Step 3: $9963 - 3699 = 6264$.
Step 4: $6642 - 2466 = 4176$.
Step 5: $7641 - 1467 = 6174$.
Step 6: $7641 - 1467 = 6174$.

The process reaches $6174$ and stays there! This number is called Kaprekar's constant.

Remarkable fact: Every four-digit number (with not all digits equal) reaches $6174$ within $7$ steps.

Answer: The process always converges to $6174$ (Kaprekar's constant).

Problem: Check whether $85{,}294$ is divisible by $11$ using the alternating sum rule.

Solution:

The divisibility rule for $11$: compute the alternating sum of digits (subtract and add alternately from right to left).

Digits of $85{,}294$: $8, 5, 2, 9, 4$.

Alternating sum from right: $4 - 9 + 2 - 5 + 8 = 0$.

Since $0$ is divisible by $11$, $85{,}294$ is divisible by $11$.

Verification: $85{,}294 \div 11 = 7{,}754$. $\checkmark$

Answer: Yes, $85{,}294$ is divisible by $11$.

Problem: In the addition $\text{SEND} + \text{MORE} = \text{MONEY}$, each letter stands for a different digit. Find the value of $M$.

Solution:

Since SEND and MORE are both four-digit numbers and MONEY is a five-digit number, there must be a carry from the thousands column.

This means $S + M + \text{carry} \geq 10$.

Since $M$ is the leading digit of MONEY and it comes from a carry, $M = 1$.

(The full solution is $S=9, E=5, N=6, D=7, M=1, O=0, R=8, Y=2$: $9567 + 1085 = 10652$.)

Answer: $M = 1$.

Problem: Round each number to the nearest $10$: (a) $73$ (b) $145$ (c) $238$ (d) $595$

Solution:

Look at the units digit:

(a) $73$: Units digit $= 3 < 5$, round down $\to 70$.
(b) $145$: Units digit $= 5 \geq 5$, round up $\to 150$.
(c) $238$: Units digit $= 8 \geq 5$, round up $\to 240$.
(d) $595$: Units digit $= 5 \geq 5$, round up $\to 600$.

Answer: (a) $70$ (b) $150$ (c) $240$ (d) $600$.

Problem: Estimate $487 + 312 + 176$ by rounding each number to the nearest hundred.

Solution:

Round each number:

Estimated sum: $500 + 300 + 200 = 1{,}000$.

Exact sum: $487 + 312 + 176 = 975$.

The estimate ($1{,}000$) is close to the exact answer ($975$). The error is only $25$.

Answer: Estimated sum $\approx 1{,}000$ (exact: $975$).

Problem: Estimate $38 \times 72$ by rounding to the nearest ten.

Solution:

Estimated product: $40 \times 70 = 2{,}800$.

Exact product: $38 \times 72 = 2{,}736$.

The estimate is off by $64$, which is about $2.3\%$ — quite good for a quick mental calculation!

Answer: Estimated product $\approx 2{,}800$.

Problem: A student calculates $47 \times 83 = 39{,}01$. Use estimation to check if this answer is reasonable.

Solution:

Estimate: $47 \approx 50$ and $83 \approx 80$.

The student's answer is $39{,}01$ (which seems to be $3{,}901$). Our estimate is $4{,}000$, so $3{,}901$ is reasonable.

But if the student meant $39{,}01$ as $39{,}010$, that would be way too large (nearly $10$ times the estimate). In that case, the student likely misplaced a digit.

Actual answer: $47 \times 83 = 3{,}901$.

Answer: The estimate of $4{,}000$ confirms that $3{,}901$ is reasonable.

Problem: A school has $38$ classrooms, each with approximately $42$ students. Estimate the total number of students.

Solution:

Estimate: $40 \times 40 = 1{,}600$ students.

Alternative (better) estimate: $38 \approx 40$ and keep $42$:

Exact: $38 \times 42 = 1{,}596$.

For planning purposes (ordering supplies, arranging transport), an estimate of $1{,}600$ is perfectly adequate.

Answer: Approximately $1{,}600$ students.

Problem: India's population is approximately $1{,}420{,}000{,}000$. If each person generates about $0.5$ kg of waste per day, estimate the total daily waste.

Solution:

Daily waste $\approx 1.4 \times 10^9 \times 0.5 = 0.7 \times 10^9 = 700{,}000{,}000$ kg $= 700{,}000$ tonnes.

That is about $7{,}00{,}000$ tonnes per day!

Answer: Approximately $700{,}000$ tonnes of waste per day.

Problem: Complete the table showing what happens when you add or multiply even and odd numbers:

Solution:

Addition:
| $+$ | Even | Odd |
|-----|------|-----|
| Even | Even | Odd |
| Odd | Odd | Even |

Multiplication:
| $\times$ | Even | Odd |
|----------|------|-----|
| Even | Even | Even |
| Odd | Even | Odd |

Why?
- Even + Even = Even: $2a + 2b = 2(a+b)$
- Odd + Odd = Even: $(2a+1) + (2b+1) = 2(a+b+1)$
- Even $\times$ Odd = Even: $2a \times (2b+1) = 2 \times a(2b+1)$
- Odd $\times$ Odd = Odd: $(2a+1)(2b+1) = 4ab + 2a + 2b + 1 = 2(2ab+a+b) + 1$

Answer: Tables completed as above.

Problem: Check which of these numbers are divisible by $3$ and by $9$: $234$, $567$, $891$, $1{,}233$, $4{,}518$.

Solution:

Compute digit sums:
- $234$: $2+3+4 = 9$. Divisible by $9$ (hence also by $3$). $\checkmark$
- $567$: $5+6+7 = 18$. Divisible by $9$ and $3$. $\checkmark$
- $891$: $8+9+1 = 18$. Divisible by $9$ and $3$. $\checkmark$
- $1{,}233$: $1+2+3+3 = 9$. Divisible by $9$ and $3$. $\checkmark$
- $4{,}518$: $4+5+1+8 = 18$. Divisible by $9$ and $3$. $\checkmark$