Question paper, Set 10

Natural numbers are the counting numbers $1, 2, 3, 4, \ldots$

$0$ is not a natural number.

Less than $8$: $1, 2, 3, 4, 5, 6, 7$

Answer: $1, 2, 3, 4, 5, 6, 7$

Odd numbers give remainder $1$ when divided by $2$: $1, 3, 5, 7, \ldots$; each is $2$ more than the one before.

$397 + 2 = 399$, $399 + 2 = 401$, $401 + 2 = 403$

$399, 401, 403$

Answer: $399, 401, 403$

Multiples of $15$: $15 \times 1, 15 \times 2, 15 \times 3, \ldots$

$15 \times 26 = 390$ is less than $400$

$15 \times 27 = 405$, $15 \times 28 = 420$, $15 \times 29 = 435$

$15 \times 30 = 450$, $15 \times 31 = 465$, $15 \times 32 = 480$

$15 \times 33 = 495$

$15 \times 34 = 510$ is more than $500$

Multiples of $15$ between $400$ and $500$: $405, 420, 435, 450, 465, 480, 495$

Answer: $405, 420, 435, 450, 465, 480, 495$

$52 = 1 \times 52$

$= 2 \times 26$

$= 4 \times 13$

$\therefore$ Factors of $52$ are $1, 2, 4, 13, 26, 52$

Answer: $1, 2, 4, 13, 26, 52$

$198 = 1 \times 198$

$= 2 \times 99$

$= 3 \times 66$

$= 6 \times 33$

$= 9 \times 22$

$= 11 \times 18$

$4, 5, 7, 8, 10, 12, 13, 14$ do not divide $198$ (they leave a remainder).

$\therefore$ Divisors of $198$ are $1, 2, 3, 6, 9, 11, 18, 22, 33, 66, 99, 198$

Number of divisors $= 12$

Answer: $1, 2, 3, 6, 9, 11, 18, 22, 33, 66, 99, 198$ ($12$ divisors)

$r$ and $s$ take various values: variables.

$2$ and $11$ have fixed values: constants.

Answer: $2$ variables ($r, s$) and $2$ constants ($2, 11$)

As $-2 \times a = -2a$:

$-11 \times r = -11r$

Answer: $-11r$

(i) $8r + 9r = (8 + 9)r = 17r$

(ii) $19r - 12r = (19 - 12)r = 7r$

Answer: (i) $17r$ (ii) $7r$

(i) $p + p + p + p + p + p$ is $p$ added $6$ times: $p + p + p + p + p + p = 6p$

(ii) $p \times p \times p \times p \times p \times p$ is $p$ multiplied $6$ times: $p \times p \times p \times p \times p \times p = p^{6}$

Answer: (i) $6p$ (ii) $p^{6}$

$87^{2} = 87 \times 87$

$87 \times 7 = 609$

$87 \times 8 = 696$, written one place to the left: $6960$

$609 + 6960 = 7569$

$\therefore 87^{2} = 7569$

Answer: $7569$

$4^{7} = 4 \times 4 \times 4 \times 4 \times 4 \times 4 \times 4$

$4 \times 4 = 16$, $16 \times 4 = 64$, $64 \times 4 = 256$

$256 \times 4 = 1024$, $1024 \times 4 = 4096$, $4096 \times 4 = 16384$

$\therefore 4^{7} = 16384$

Answer: $16384$

(i) Divide by $2$ again and again:

$512 \div 2 = 256$, $256 \div 2 = 128$, $128 \div 2 = 64$

$64 \div 2 = 32$, $32 \div 2 = 16$, $16 \div 2 = 8$

$8 \div 2 = 4$, $4 \div 2 = 2$

$512 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^{9}$

(ii) Divide by $8$ again and again:

$512 \div 8 = 64$, $64 \div 8 = 8$

$512 = 8 \times 8 \times 8 = 8^{3}$

Answer: (i) $2^{9}$ (ii) $8^{3}$

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 $61301$ is $1$, so $61301$ is not divisible by $2$.

Answer: No, the unit digit is $1$

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

Sum of the digits of $37909$: $3 + 7 + 9 + 0 + 9 = 28$

$28$ is not divisible by $3$ ($28 = 3 \times 9 + 1$), so $37909$ is not divisible by $3$.

Answer: No, the sum of the digits is $28$

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

$2599$: $2 + 5 + 9 + 9 = 25$, not divisible by $3$

$2598$: $2 + 5 + 9 + 8 = 24$, divisible by $3$

$\therefore$ The largest number less than $2600$ that is divisible by $3$ is $2598$.

Answer: $2598$

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

Last two digits of $39146$ together: $46$; $46 = 4 \times 11 + 2$, so $39146$ is not divisible by $4$.

Answer: No, the last two digits $46$ are not 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 $30503$ is $3$, so $30503$ is not divisible by $5$.

Answer: No, the unit digit is $3$

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

The digit at the unit place of $43928$ is $8$, so $43928$ is divisible by $2$.

Sum of the digits of $43928$: $4 + 3 + 9 + 2 + 8 = 26$

$26$ is not divisible by $3$ ($26 = 3 \times 8 + 2$), so $43928$ is not divisible by $3$.

$\therefore$ $43928$ 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 $0$).

$973$: $97 - 3(2) = 97 - 6 = 91$; $91 = 7 \times 13$, a multiple of $7$, so $973$ is divisible by $7$.

$557$: $55 - 7(2) = 55 - 14 = 41$; $41 = 7 \times 5 + 6$, not a multiple of $7$, so $557$ is not divisible by $7$.

$295$: $29 - 5(2) = 29 - 10 = 19$; $19 = 7 \times 2 + 5$, not a multiple of $7$, so $295$ is not divisible by $7$.

Answer: $973$ is divisible by $7$; $557$ and $295$ are not.

By $2$: The digit at the unit place of $89064$ is $4$, so $89064$ is divisible by $2$.

By $3$: Sum of the digits of $89064$: $8 + 9 + 0 + 6 + 4 = 27$; $27$ is divisible by $3$ ($27 = 3 \times 9$), so $89064$ is divisible by $3$.

By $4$: Last two digits of $89064$ together: $64$; $64 = 4 \times 16$, so $89064$ is divisible by $4$.

By $5$: The digit at the unit place of $89064$ is $4$, so $89064$ is not divisible by $5$.

By $6$: $89064$ is divisible by both $2$ and $3$, so it is divisible by $6$.

$\therefore$ $89064$ is divisible by $2, 3, 4$ and $6$.

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