Question paper, Set 5

Counting starts from $1$, so $0$ is not a natural number.

Whole numbers are the natural numbers along with $0$, so $0$ is a whole number.

Answer: $0$ is not a natural number; $0$ is a whole number.

$79 = 2 \times 39 + 1$

Remainder is $1$, so $79$ is an odd number.

Answer: Remainder $1$; $79$ is odd

Multiples of $23$: $23 \times 1, 23 \times 2, 23 \times 3, \ldots$

$23 \times 1 = 23$, $23 \times 2 = 46$, $23 \times 3 = 69$, $23 \times 4 = 92$

$23 \times 5 = 115$, $23 \times 6 = 138$, $23 \times 7 = 161$, $23 \times 8 = 184$

First eight multiples of $23$: $23, 46, 69, 92, 115, 138, 161, 184$

Answer: $23, 46, 69, 92, 115, 138, 161, 184$

$50 = 1 \times 50$

$= 2 \times 25$

$= 5 \times 10$

$\therefore$ Factors of $50$ are $1, 2, 5, 10, 25, 50$

Answer: $1, 2, 5, 10, 25, 50$

$132 = 1 \times 132$

$= 2 \times 66$

$= 3 \times 44$

$= 4 \times 33$

$= 6 \times 22$

$= 11 \times 12$

$5, 7, 8, 9, 10$ do not divide $132$ (they leave a remainder).

$\therefore$ Divisors of $132$ are $1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66, 132$

Number of divisors $= 12$

Answer: $1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66, 132$ ($12$ divisors)

A symbol which has a fixed value is called a constant.

$15$ and $4$ have fixed values; $b, t, z$ are variables.

Answer: $15, 4$

As $2 \times a = 2a$, a number times a letter is written without the $\times$ sign.

So $12t = 12 \times t$

Answer: $12 \times t$

(i) $y + y + y + y + y + y + y$ is $y$ added $7$ times.

(i) $y + y + y + y + y + y + y = 7y$

(ii) $13y - 8y = (13 - 8)y = 5y$

Answer: (i) $7y$ (ii) $5y$

$n^{4} = n \times n \times n \times n$

Read it as '$n$ power $4$'.

Answer: $n \times n \times n \times n$, read as '$n$ power $4$'

$31^{2} = 31 \times 31$

$31 \times 1 = 31$

$31 \times 3 = 93$, written one place to the left: $930$

$31 + 930 = 961$

$\therefore 31^{2} = 961$

Answer: $961$

$15^{3} = 15 \times 15 \times 15$

$15 \times 15 = 225$, $225 \times 15 = 3375$

$\therefore 15^{3} = 3375$

Answer: $3375$

(i) Divide by $6$ again and again:

$1296 \div 6 = 216$, $216 \div 6 = 36$, $36 \div 6 = 6$

$1296 = 6 \times 6 \times 6 \times 6 = 6^{4}$

(ii) Divide by $36$ again and again:

$1296 \div 36 = 36$

$1296 = 36 \times 36 = 36^{2}$

Answer: (i) $6^{4}$ (ii) $36^{2}$

A number is divisible by $2$ if the digit at its unit place is $0, 2, 4, 6$ or $8$.

The digit at the unit place of $5280$ is $0$, so $5280$ is divisible by $2$.

The digit at the unit place of $637$ is $7$, so $637$ is not divisible by $2$.

The digit at the unit place of $914$ is $4$, so $914$ is divisible by $2$.

Answer: $5280, 914$

A number is divisible by $3$ if the sum of all its digits is divisible by $3$.

Sum of the digits of $40026$: $4 + 0 + 0 + 2 + 6 = 12$

$12$ is divisible by $3$ ($12 = 3 \times 4$), so $40026$ is divisible by $3$.

Answer: Yes, the sum of the digits is $12$

A number is divisible by $3$ if the sum of all its digits is divisible by $3$.

$2041$: $2 + 0 + 4 + 1 = 7$, not divisible by $3$

$2042$: $2 + 0 + 4 + 2 = 8$, not divisible by $3$

$2043$: $2 + 0 + 4 + 3 = 9$, divisible by $3$

$\therefore$ The smallest number greater than $2040$ that is divisible by $3$ is $2043$.

Answer: $2043$

A number is divisible by $4$ if its last two digits together are divisible by $4$.

Last two digits of $92372$ together: $72$; $72 = 4 \times 18$, so $92372$ is divisible by $4$.

Answer: Yes, the last two digits $72$ are divisible by $4$

A number is divisible by $5$ if the digit at its unit place is $0$ or $5$.

The digit at the unit place of $63005$ is $5$, so $63005$ is divisible by $5$.

Answer: Yes, the unit digit is $5$

A number is divisible by $6$ if it is divisible by both $2$ and $3$.

The digit at the unit place of $3573$ is $3$, so $3573$ is not divisible by $2$.

Sum of the digits of $3573$: $3 + 5 + 7 + 3 = 18$

$18$ is divisible by $3$ ($18 = 3 \times 6$), so $3573$ is divisible by $3$.

$\therefore$ $3573$ is not divisible by $6$.

Answer: No, it is not divisible by $2$

A number is divisible by $7$ if the remaining part minus twice the unit digit is a multiple of $7$ (or $0$).

$168$: $16 - 8(2) = 16 - 16 = 0$; it is equal to $0$, so $168$ is divisible by $7$.

$936$: $93 - 6(2) = 93 - 12 = 81$; $81 = 7 \times 11 + 4$, not a multiple of $7$, so $936$ is not divisible by $7$.

$518$: $51 - 8(2) = 51 - 16 = 35$; $35 = 7 \times 5$, a multiple of $7$, so $518$ is divisible by $7$.

Answer: $168$ and $518$ are divisible by $7$; $936$ is not.

By $2$: The digit at the unit place of $81930$ is $0$, so $81930$ is divisible by $2$.

By $3$: Sum of the digits of $81930$: $8 + 1 + 9 + 3 + 0 = 21$; $21$ is divisible by $3$ ($21 = 3 \times 7$), so $81930$ is divisible by $3$.

By $4$: Last two digits of $81930$ together: $30$; $30 = 4 \times 7 + 2$, so $81930$ is not divisible by $4$.

By $5$: The digit at the unit place of $81930$ is $0$, so $81930$ is divisible by $5$.

By $6$: $81930$ is divisible by both $2$ and $3$, so it is divisible by $6$.

$\therefore$ $81930$ is divisible by $2, 3, 5$ and $6$.

Answer: Divisible by $2, 3, 5$ and $6$; not by $4$