Question paper, Set 8

Just before: $1000 - 1 = 999$

Just after: $1000 + 1 = 1001$

Answer: $999$ and $1001$

$0 \div 2$ leaves remainder $0$, so $0$ is also even.

Even numbers less than $11$: $0, 2, 4, 6, 8, 10$

Answer: $0, 2, 4, 6, 8, 10$

Multiples of $18$: $18 \times 1, 18 \times 2, 18 \times 3, \ldots$

$18 \times 1 = 18$, $18 \times 2 = 36$, $18 \times 3 = 54$, $18 \times 4 = 72$

$18 \times 5 = 90$, $18 \times 6 = 108$, $18 \times 7 = 126$, $18 \times 8 = 144$

First eight multiples of $18$: $18, 36, 54, 72, 90, 108, 126, 144$

Answer: $18, 36, 54, 72, 90, 108, 126, 144$

$38 = 1 \times 38$

$= 2 \times 19$

$\therefore$ Factors of $38$ are $1, 2, 19, 38$

Answer: $1, 2, 19, 38$

$156 = 1 \times 156$

$= 2 \times 78$

$= 3 \times 52$

$= 4 \times 39$

$= 6 \times 26$

$= 12 \times 13$

$5, 7, 8, 9, 10, 11$ do not divide $156$ (they leave a remainder).

$\therefore$ Divisors of $156$ are $1, 2, 3, 4, 6, 12, 13, 26, 39, 52, 78, 156$

Number of divisors $= 12$

Answer: $1, 2, 3, 4, 6, 12, 13, 26, 39, 52, 78, 156$ ($12$ divisors)

$20$ has a fixed value.

So $20$ is a constant.

Answer: Constant

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

$-8 \times p = -8p$

Answer: $-8p$

(i) $z + z + z + z + z$ is $z$ added $5$ times.

(i) $z + z + z + z + z = 5z$

(ii) $18z - 5z = (18 - 5)z = 13z$

Answer: (i) $5z$ (ii) $13z$

(i) $q \times q \times q = q^{3}$

(ii) $q \times q \times q \times q = q^{4}$

Answer: (i) $q^{3}$ (ii) $q^{4}$

$46^{2} = 46 \times 46$

$46 \times 6 = 276$

$46 \times 4 = 184$, written one place to the left: $1840$

$276 + 1840 = 2116$

$\therefore 46^{2} = 2116$

Answer: $2116$

$7^{5} = 7 \times 7 \times 7 \times 7 \times 7$

$7 \times 7 = 49$, $49 \times 7 = 343$, $343 \times 7 = 2401$

$2401 \times 7 = 16807$

$\therefore 7^{5} = 16807$

Answer: $16807$

(i) Divide by $3$ again and again:

$59049 \div 3 = 19683$, $19683 \div 3 = 6561$, $6561 \div 3 = 2187$

$2187 \div 3 = 729$, $729 \div 3 = 243$, $243 \div 3 = 81$

$81 \div 3 = 27$, $27 \div 3 = 9$, $9 \div 3 = 3$

$59049 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^{10}$

(ii) Divide by $9$ again and again:

$59049 \div 9 = 6561$, $6561 \div 9 = 729$, $729 \div 9 = 81$

$81 \div 9 = 9$

$59049 = 9 \times 9 \times 9 \times 9 \times 9 = 9^{5}$

Answer: (i) $3^{10}$ (ii) $9^{5}$

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 $3571$ is $1$, so $3571$ is not divisible by $2$.

The digit at the unit place of $6402$ is $2$, so $6402$ is divisible by $2$.

The digit at the unit place of $8888$ is $8$, so $8888$ is divisible by $2$.

Answer: $6402, 8888$

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

Sum of the digits of $120012$: $1 + 2 + 0 + 0 + 1 + 2 = 6$

$6$ is divisible by $3$ ($6 = 3 \times 2$), so $120012$ is divisible by $3$.

Answer: Yes, the sum of the digits is $6$

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

$4999$: $4 + 9 + 9 + 9 = 31$, not divisible by $3$

$4998$: $4 + 9 + 9 + 8 = 30$, divisible by $3$

$\therefore$ The largest number less than $5000$ that is divisible by $3$ is $4998$.

Answer: $4998$

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

Last two digits of $24918$ together: $18$; $18 = 4 \times 4 + 2$, so $24918$ is not divisible by $4$.

Answer: No, the last two digits $18$ 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 $55554$ is $4$, so $55554$ is not divisible by $5$.

Answer: No, the unit digit is $4$

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

The digit at the unit place of $26118$ is $8$, so $26118$ is divisible by $2$.

Sum of the digits of $26118$: $2 + 6 + 1 + 1 + 8 = 18$

$18$ is divisible by $3$ ($18 = 3 \times 6$), so $26118$ is divisible by $3$.

$\therefore$ $26118$ is divisible by $6$.

Answer: Yes, it is divisible by both $2$ and $3$

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

$589$: $58 - 9(2) = 58 - 18 = 40$; $40 = 7 \times 5 + 5$, not a multiple of $7$, so $589$ is not divisible by $7$.

$423$: $42 - 3(2) = 42 - 6 = 36$; $36 = 7 \times 5 + 1$, not a multiple of $7$, so $423$ is not divisible by $7$.

$245$: $24 - 5(2) = 24 - 10 = 14$; $14 = 7 \times 2$, a multiple of $7$, so $245$ is divisible by $7$.

Answer: $245$ is divisible by $7$; $589$ and $423$ are not.

By $2$: The digit at the unit place of $43580$ is $0$, so $43580$ is divisible by $2$.

By $3$: Sum of the digits of $43580$: $4 + 3 + 5 + 8 + 0 = 20$; $20$ is not divisible by $3$ ($20 = 3 \times 6 + 2$), so $43580$ is not divisible by $3$.

By $4$: Last two digits of $43580$ together: $80$; $80 = 4 \times 20$, so $43580$ is divisible by $4$.

By $5$: The digit at the unit place of $43580$ is $0$, so $43580$ is divisible by $5$.

By $6$: $43580$ is not divisible by both $2$ and $3$, so it is not divisible by $6$.

$\therefore$ $43580$ is divisible by $2, 4$ and $5$.

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