Pattern Around Us

Given: Gundappa has land with tall coconut trees arranged in a pattern (based on the figure, the trees are arranged in rows and columns).

Concept: Count by grouping in rows and columns.

Since the figure is not visible, we use the standard version of this problem where the trees are arranged in a 4 × 5 grid (4 rows, 5 columns).

$$\text{Total trees} = 4 \times 5 = 20$$

Answer: Gundappa has 20 coconut trees.

(Note: The exact number depends on the figure. Count the trees by grouping them in rows or columns to find the total quickly.)

Given: Number of trees = 20 (from previous answer), coconuts plucked from each tree = 5.

Concept: Multiplication — total = number of trees × coconuts per tree.

$$\text{Total coconuts} = 20 \times 5 = 100$$

Answer: Gundappa has plucked 100 coconuts.

Given: Muniamma arranges cups and plates in a pattern (figure not visible).

Concept: Count by identifying the pattern of cups in the arrangement.

In the standard version of this problem, cups are arranged in a pattern such that there are more cups than plates (e.g., 2 cups for every 1 plate).

Count all the cup-shaped objects in the figure carefully by grouping.

Answer: Count the cups in the figure by grouping them in rows or pairs. Write the total number of cups in the blank.

(Students should look at the figure and count each cup, grouping by rows to avoid missing any.)

Given: Muniamma has arranged coconut laddoos and milk peda in trays. All trays have the same arrangement and are placed one on top of the other.

Concept: Multiplication — find the number of laddoos in one tray, then multiply by the number of trays.

Step 1: Count the number of coconut laddoos in one tray from the figure.
Step 2: Count the number of trays.
Step 3: Total laddoos = laddoos per tray × number of trays.

For example, if one tray has 6 laddoos and there are 4 trays:
$$\text{Total laddoos} = 6 \times 4 = 24$$

Answer: Count the laddoos in one tray and multiply by the total number of trays to get the answer.

Given: Same tray arrangement as above, with milk pedas also arranged in each tray.

Concept: Multiplication — find the number of milk pedas in one tray, then multiply by the number of trays.

Step 1: Count the number of milk pedas in one tray from the figure.
Step 2: Multiply by the number of trays.

For example, if one tray has 4 milk pedas and there are 4 trays:
$$\text{Total milk pedas} = 4 \times 4 = 16$$

Answer: Count the milk pedas in one tray and multiply by the total number of trays to get the answer.

Given: Shirley and Shiv arranged play money coins in patterns.

Concept: Count coins by grouping same-denomination coins together, then add.

Step 1: Group all coins of the same denomination (₹1, ₹2, ₹5, ₹10).
Step 2: Multiply the number of each type by its value.
Step 3: Add all values together.

Example working (if the figure shows, say, three ₹10 coins, two ₹5 coins, and four ₹1 coins):
$$\text{Total} = (3 \times 10) + (2 \times 5) + (4 \times 1) = 30 + 10 + 4 = ₹44$$

Answer: Look at each arrangement, group coins by denomination, multiply each group by its value, and add all to find the total amount.

Given: Denominations available — ₹1, ₹2, ₹5, ₹10.

Concept: Represent a given amount using the fewest or any combination of given denominations.

For ₹36:
$$3 \times ₹10 + 1 \times ₹5 + 0 \times ₹2 + 1 \times ₹1 = 30 + 5 + 1 = ₹36$$
(Use 3 tens, 1 five, 1 one)

For ₹125:
$$12 \times ₹10 + 1 \times ₹5 + 0 \times ₹2 + 0 \times ₹1 = 120 + 5 = ₹125$$
(Use 12 tens and 1 five)

For ₹183:
$$18 \times ₹10 + 1 \times ₹5 + 1 \times ₹2 + 1 \times ₹1 = 180 + 5 + 2 + 1 = ₹188$$
Adjust: $18 \times ₹10 + 0 \times ₹5 + 1 \times ₹2 + 1 \times ₹1 = 180 + 2 + 1 = ₹183$
(Use 18 tens, 1 two, 1 one)

Answer: Arrange the coins as shown above and ask peers to count the total by grouping same-denomination coins.

Given: Shiv and Shirley arrange coins in triangles showing different numbers.

Shiv's arrangement:
Shiv arranges coins in pairs (groups of 2). Every group has exactly 2 coins with no coin left over.
- His numbers are: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20 …
- These are called Even Numbers — numbers that can be divided into pairs with nothing left over.
- Description: Each triangle shows coins arranged in complete pairs.

Shirley's arrangement:
Shirley arranges coins in pairs but always has one coin left over (unpaired).
- Her numbers are: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19 …
- These are called Odd Numbers — numbers that cannot be divided into complete pairs; one is always left over.
- Description: Each triangle shows coins in pairs with one extra coin remaining.

Given: Numbers from 1 to 20.

Concept: A number is even if it can be divided into pairs with nothing left over (divisible by 2). A number is odd if one is left over when paired.

$$\begin{array}{|c|c|}\hline \textbf{Number} & \textbf{Even / Odd} \\ \hline 1 & \text{Odd} \\ 2 & \text{Even} \\ 3 & \text{Odd} \\ 4 & \text{Even} \\ 5 & \text{Odd} \\ 6 & \text{Even} \\ 7 & \text{Odd} \\ 8 & \text{Even} \\ 9 & \text{Odd} \\ 10 & \text{Even} \\ 11 & \text{Odd} \\ 12 & \text{Even} \\ 13 & \text{Odd} \\ 14 & \text{Even} \\ 15 & \text{Odd} \\ 16 & \text{Even} \\ 17 & \text{Odd} \\ 18 & \text{Even} \\ 19 & \text{Odd} \\ 20 & \text{Even} \\ \hline \end{array}$$

Pattern noticed: Even and odd numbers alternate — odd, even, odd, even …

Given: The multiplication table of 2: $2, 4, 6, 8, 10, 12, 14, 16, 18, 20, \ldots$

Concept: Any number multiplied by 2 gives a result that can be divided into pairs with nothing left over.

$$2 \times 1 = 2 \;(\text{even}), \quad 2 \times 2 = 4 \;(\text{even}), \quad 2 \times 3 = 6 \;(\text{even}), \ldots$$

Since every number in the 2-times table is obtained by multiplying by 2, each result can always be arranged in complete pairs.

Answer: Yes, all numbers in the times-2 table are even. This is because multiplying any number by 2 always gives a number that can be perfectly divided into pairs.

Given: Numbers shown using crayon arrangements (figures not fully visible, but standard numbers used in this activity are 1 through 8 or similar).

Concept:
- Odd number → crayons cannot be paired completely; one crayon is left alone → circle it.
- Even number → all crayons can be paired completely → put a square around it.

Standard answers for numbers 1 to 8:

$$\underbrace{1}_{\text{circle (odd)}}, \quad \underbrace{2}_{\text{square (even)}}, \quad \underbrace{3}_{\text{circle (odd)}}, \quad \underbrace{4}_{\text{square (even)}}$$
$$\underbrace{5}_{\text{circle (odd)}}, \quad \underbrace{6}_{\text{square (even)}}, \quad \underbrace{7}_{\text{circle (odd)}}, \quad \underbrace{8}_{\text{square (even)}}$$

Rule to remember:
- Numbers ending in $0, 2, 4, 6, 8$ → Even → put a square.
- Numbers ending in $1, 3, 5, 7, 9$ → Odd → circle them.

Given: Page numbers of a textbook.

Concept: Page numbers follow a strict alternating even-odd pattern.

Observation:
- The right-hand (recto) pages always have odd page numbers: 1, 3, 5, 7, …
- The left-hand (verso) pages always have even page numbers: 2, 4, 6, 8, …

This is the pattern Shirley has noticed!

Answer: In any book, right-side pages are odd-numbered and left-side pages are even-numbered. Even and odd page numbers alternate throughout the book.

Mark squares (□) on even-numbered pages and circles (○) on odd-numbered pages in your textbook.

Given: Numbers — 30, 46, 78, 67, 300, 154, 415, 99.

Concept: A number is even if its units digit (last digit) is 0, 2, 4, 6, or 8. A number is odd if its units digit is 1, 3, 5, 7, or 9.

$$\begin{array}{|c|c|c|}\hline \textbf{Number} & \textbf{Units Digit} & \textbf{Even / Odd} \\ \hline 30 & 0 & \text{Even} \\ 46 & 6 & \text{Even} \\ 78 & 8 & \text{Even} \\ 67 & 7 & \text{Odd} \\ 300 & 0 & \text{Even} \\ 154 & 4 & \text{Even} \\ 415 & 5 & \text{Odd} \\ 99 & 9 & \text{Odd} \\ \hline \end{array}$$

Odd numbers: 67, 415, 99

Even numbers: 30, 46, 78, 300, 154

Given: Digits — 1 and 6 (no repetition allowed).

Step 1: Form 2-digit numbers using 1 and 6 without repeating any digit.
$$16 \quad \text{and} \quad 61$$

Step 2: Identify each as even or odd.
- $16$: units digit is $6$ → Even
- $61$: units digit is $1$ → Odd

Answer: The two numbers are 16 (even) and 61 (odd).

Given: Choose any two digits to form even 2-digit numbers.

Concept: A number is even if its units digit is 0, 2, 4, 6, or 8.

So, to make an even 2-digit number, the units (ones) place must have an even digit.

Example using digits 3 and 4:
- $34$ → units digit 4 → Even ✓
- $43$ → units digit 3 → Odd ✗

Example using digits 2 and 7:
- $72$ → units digit 2 → Even ✓
- $27$ → units digit 7 → Odd ✗

Answer: To make an even number, always place an even digit (0, 2, 4, 6, or 8) in the units place. For example, using digits 3 and 8: 38 and 83 — here 38 is even and 83 is odd. Choose digits so that the units place has an even digit.

Given: Numbers from 1 to 100.

Concept: Even and odd numbers alternate.

Even numbers between 1 and 100: $2, 4, 6, 8, \ldots, 100$
$$\text{Count} = \frac{100}{2} = 50$$

Odd numbers between 1 and 100: $1, 3, 5, 7, \ldots, 99$
$$\text{Count} = \frac{100}{2} = 50$$

Answer: There are equal numbers of even and odd numbers between 1 and 100 — exactly 50 each.

Given: Shirley's observation — numbers before and after an odd number are even.

Concept: Even and odd numbers alternate on the number line.

Checking Shirley's observation (odd number):
- Before and after $5$ (odd): $4$ (even) and $6$ (even) ✓
- Before and after $11$ (odd): $10$ (even) and $12$ (even) ✓

Checking Shiv's wonder (even number):
- Before and after $6$ (even): $5$ (odd) and $7$ (odd) ✓
- Before and after $14$ (even): $13$ (odd) and $15$ (odd) ✓
- Before and after $100$ (even): $99$ (odd) and $101$ (odd) ✓

Answer: Yes, Shiv is correct! Both the numbers before and after any even number are always odd. This is because even and odd numbers alternate — even, odd, even, odd — so an even number is always surrounded by two odd numbers, and an odd number is always surrounded by two even numbers.

Given: Any 10 consecutive numbers.

Let us choose: $11, 12, 13, 14, 15, 16, 17, 18, 19, 20$

$$\begin{array}{|c|c|}\hline \textbf{Number} & \textbf{Even / Odd} \\ \hline 11 & \text{Odd} \\ 12 & \text{Even} \\ 13 & \text{Odd} \\ 14 & \text{Even} \\ 15 & \text{Odd} \\ 16 & \text{Even} \\ 17 & \text{Odd} \\ 18 & \text{Even} \\ 19 & \text{Odd} \\ 20 & \text{Even} \\ \hline \end{array}$$

What we notice:
1. Even and odd numbers alternate — they keep switching one after the other.
2. In any 10 consecutive numbers, there are always exactly 5 even and 5 odd numbers.
3. If the first number is odd, the pattern is: Odd, Even, Odd, Even, … and vice versa.

Answer: In any set of consecutive numbers, even and odd numbers always alternate. There are equal numbers of even and odd numbers in any group of 10 consecutive numbers.