Question paper, Set 3

Natural numbers are the counting numbers $1, 2, 3, 4, \ldots$; each is $1$ more than the one before.

$299 + 1 = 300$, $300 + 1 = 301$, $301 + 1 = 302$

$300, 301, 302$

Answer: $300, 301, 302$

Odd numbers give remainder $1$ when divided by $2$: $1, 3, 5, 7, \ldots$

Between $50$ and $60$: $51, 53, 55, 57, 59$

Answer: $51, 53, 55, 57, 59$

Multiples of $16$: $16 \times 1, 16 \times 2, 16 \times 3, \ldots$

$16 \times 1 = 16$, $16 \times 2 = 32$, $16 \times 3 = 48$, $16 \times 4 = 64$

$16 \times 5 = 80$, $16 \times 6 = 96$, $16 \times 7 = 112$

$16 \times 8 = 128$ is not less than $120$

Multiples of $16$ less than $120$: $16, 32, 48, 64, 80, 96, 112$

Answer: $16, 32, 48, 64, 80, 96, 112$

$44 = 1 \times 44$

$= 2 \times 22$

$= 4 \times 11$

$\therefore$ Factors of $44$ are $1, 2, 4, 11, 22, 44$

Answer: $1, 2, 4, 11, 22, 44$

$108 = 1 \times 108$

$= 2 \times 54$

$= 3 \times 36$

$= 4 \times 27$

$= 6 \times 18$

$= 9 \times 12$

$5, 7, 8, 10$ do not divide $108$ (they leave a remainder).

$\therefore$ Divisors of $108$ are $1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 108$

Number of divisors $= 12$

Answer: $1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 108$ ($12$ divisors)

$9p = 9 \times p$

$p$ can take various values, so $p$ is the variable; $9$ is fixed, so $9$ is the constant.

Answer: Variable: $p$; constant: $9$

As $1 \times a = a$:

$1 \times m = m$

Answer: $m$

(i) $4p + 11p = (4 + 11)p = 15p$

(ii) $15p - 6p = (15 - 6)p = 9p$

Answer: (i) $15p$ (ii) $9p$

(i) $t \times t = t^{2}$

(ii) $t \times t \times t \times t \times t = t^{5}$

Answer: (i) $t^{2}$ (ii) $t^{5}$

$48 \times 8 = 384$

$48 \times 4 = 192$, written one place to the left: $1920$

$384 + 1920 = 2304$

$\therefore 48 \times 48 = 2304$

Answer: $2304$

$2^{11} = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$

$2 \times 2 = 4$, $4 \times 2 = 8$, $8 \times 2 = 16$

$16 \times 2 = 32$, $32 \times 2 = 64$, $64 \times 2 = 128$

$128 \times 2 = 256$, $256 \times 2 = 512$, $512 \times 2 = 1024$

$1024 \times 2 = 2048$

$\therefore 2^{11} = 2048$

Answer: $2048$

(i) Divide by $5$ again and again:

$15625 \div 5 = 3125$, $3125 \div 5 = 625$, $625 \div 5 = 125$

$125 \div 5 = 25$, $25 \div 5 = 5$

$15625 = 5 \times 5 \times 5 \times 5 \times 5 \times 5 = 5^{6}$

(ii) Divide by $25$ again and again:

$15625 \div 25 = 625$, $625 \div 25 = 25$

$15625 = 25 \times 25 \times 25 = 25^{3}$

Answer: (i) $5^{6}$ (ii) $25^{3}$

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 $90316$ is $6$, so $90316$ is divisible by $2$.

Answer: Yes, the unit digit is $6$

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

Sum of the digits of $236145$: $2 + 3 + 6 + 1 + 4 + 5 = 21$

$21$ is divisible by $3$ ($21 = 3 \times 7$), so $236145$ is divisible by $3$.

Answer: Yes, the sum of the digits is $21$

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

$851$: $8 + 5 + 1 = 14$, not divisible by $3$

$852$: $8 + 5 + 2 = 15$, divisible by $3$

$\therefore$ The smallest number greater than $850$ that is divisible by $3$ is $852$.

Answer: $852$

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

Last two digits of $31708$ together: $08$; $8 = 4 \times 2$, so $31708$ is divisible by $4$.

Answer: Yes, the last two digits $08$ 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 $91530$ is $0$, so $91530$ is divisible by $5$.

Answer: Yes, the unit digit is $0$

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

The digit at the unit place of $9826$ is $6$, so $9826$ is divisible by $2$.

Sum of the digits of $9826$: $9 + 8 + 2 + 6 = 25$

$25$ is not divisible by $3$ ($25 = 3 \times 8 + 1$), so $9826$ is not divisible by $3$.

$\therefore$ $9826$ is not divisible by $6$.

Answer: No, it is not divisible by $3$

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

$441$: $44 - 1(2) = 44 - 2 = 42$; $42 = 7 \times 6$, a multiple of $7$, so $441$ is divisible by $7$.

$602$: $60 - 2(2) = 60 - 4 = 56$; $56 = 7 \times 8$, a multiple of $7$, so $602$ is divisible by $7$.

$834$: $83 - 4(2) = 83 - 8 = 75$; $75 = 7 \times 10 + 5$, not a multiple of $7$, so $834$ is not divisible by $7$.

Answer: $441$ and $602$ are divisible by $7$; $834$ is not.

By $2$: The digit at the unit place of $52614$ is $4$, so $52614$ is divisible by $2$.

By $3$: Sum of the digits of $52614$: $5 + 2 + 6 + 1 + 4 = 18$; $18$ is divisible by $3$ ($18 = 3 \times 6$), so $52614$ is divisible by $3$.

By $4$: Last two digits of $52614$ together: $14$; $14 = 4 \times 3 + 2$, so $52614$ is not divisible by $4$.

By $5$: The digit at the unit place of $52614$ is $4$, so $52614$ is not divisible by $5$.

By $6$: $52614$ is divisible by both $2$ and $3$, so it is divisible by $6$.

$\therefore$ $52614$ is divisible by $2, 3$ and $6$.

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