Number Play — NCERT Solutions, AI Tutor & Practice

To check an arrangement, take each child one at a time. For a chosen child, look only at the children standing in front of them and count how many of those are taller.

That count must equal the number the child announced.

For example, if heights (front to back) are $150, 140, 160, 155$ cm and the numbers spoken (front to back) are $0, 1, 0, 2$:

• Child 1: no one in front, taller-in-front $=0$ ✓
• Child 2: front has $150$, taller than $140$ → count $=1$ ✓
• Child 3: front has $150,140$, none taller than $160$ → count $=0$ ✓
• Child 4: front has $150,140,160$; taller than $155$ are $160$ only... $150$ is not taller. So count $=1$, but child said $2$ ✗

Since one child's statement fails, this arrangement is not consistent. A valid arrangement is one where every spoken number matches its actual count.

A palindrome reads the same left-to-right and right-to-left.

2-digit palindromes: both digits must be equal: $11, 22, 33, 44, 55, 66, 77, 88, 99$ There are $9$ of them.

Checking 121: reverse of $121$ is $121$. Same → 121 is a palindrome.

Checking 132: reverse of $132$ is $231$. Since $132 \neq 231$ → 132 is NOT a palindrome.

We repeatedly reverse the number and add it to itself.

Step 1: $39 + 93 = 132$ (reverse of $132$ is $231$, not a palindrome)

Step 2: $132 + 231 = 363$ (reverse of $363$ is $363$) ✓

$363$ is a palindrome.

So starting from $39$ it takes 2 steps to reach the palindrome $363$.

Each digit is used once, so we form all arrangements (permutations) of $1,2,3$: $123,\ 132,\ 213,\ 231,\ 312,\ 321$ There are $3 \times 2 \times 1 = 6$ numbers.

Greatest: put the largest digit in the highest place → $321$.

Smallest: put the smallest digit in the highest place → $123$.

Rule: if even → divide by $2$; if odd → compute $3n+1$.

Starting at $6$:

• $6$ is even: $6 \div 2 = 3$
• $3$ is odd: $3 \times 3 + 1 = 10$
• $10$ is even: $10 \div 2 = 5$
• $5$ is odd: $3 \times 5 + 1 = 16$
• $16$ even: $16 \div 2 = 8$
• $8$ even: $8 \div 2 = 4$
• $4$ even: $4 \div 2 = 2$
• $2$ even: $2 \div 2 = 1$

Sequence: $6 \to 3 \to 10 \to 5 \to 16 \to 8 \to 4 \to 2 \to 1$

For $495$, digits are $4,9,5$.

• Largest arrangement: $954$
• Smallest arrangement: $459$
• Subtract: $954 - 459 = 495$

We get 495 again! This is the Kaprekar constant for 3-digit numbers.

Starting from most 3-digit numbers (with at least two different digits) and repeating this routine, you always end up at $495$, and once there it stays fixed.

We must use each of $0,4,7,8,9$ once.

Largest: arrange digits from biggest to smallest: $98740$

Smallest: we want small digits in front, but $0$ cannot be the leading digit (it would make a 4-digit number). So place the next smallest digit, $4$, first, then $0$: $40789$

So smallest $= 40789$ and largest $= 98740$.

Nearest hundred: look at the tens digit of $6{,}847$, which is $4$. Since $4 < 5$, round down → keep the hundreds digit as is. $6{,}847 \approx 6{,}800$

Nearest thousand: look at the hundreds digit, which is $8$. Since $8 \ge 5$, round up → increase the thousands digit. $6{,}847 \approx 7{,}000$

The number must start near $5000$, so put $5$ or $4$ in the thousands place.

Starting with 5: smallest such number uses smallest remaining digits after $5$: $5149$. Distance from $5000$: $5149 - 5000 = 149$.

Starting with 4: largest such number: $4951$. Distance from $5000$: $5000 - 4951 = 49$.

Since $49 < 149$, the number 4951 is closer to $5000$.

Answer: $4951$ (it is $49$ away from $5000$).

A supercell is greater than each of its immediate neighbours (left and right).

Row: $8,\ 3,\ 12,\ 5,\ 10$

• $8$: neighbour on right is $3$. $8 > 3$ ✓ (edge cell, only one neighbour) → supercell
• $3$: neighbours $8$ and $12$. $3$ is not greater than both ✗
• $12$: neighbours $3$ and $5$. $12 > 3$ and $12 > 5$ ✓ → supercell
• $5$: neighbours $12$ and $10$. Not greater ✗
• $10$: neighbour on left $5$. $10 > 5$ ✓ (edge cell) → supercell

Supercells: $8,\ 12,\ 10$.

Smallest 4-digit number $= 1000$. Largest 4-digit number $= 9999$.

Count of whole numbers from $1000$ to $9999$ including both endpoints: $9999 - 1000 + 1 = 9000$

So there are 9000 four-digit whole numbers in total.

Add the digits of $4{,}829$: $4 + 8 + 2 + 9 = 23$

$23$ is not a single digit, so add again: $2 + 3 = 5$

$5$ is a single digit, so we stop.

Digit sum of 4,829 is $5$.