Sec 1 Maths: Primes, Factors & Multiples (HCF / LCM) — practice questions & worked solutions

For a perfect square and a perfect cube, every index must be a multiple of $6$.

For $2$: index is $2$, need $2^{6}$, so multiply by $2^{4}$.

For $7$: index is $5$, need $7^{6}$, so multiply by $7^{1}$.

$$p=2^{4}\times 7$$

(a)(i) Prime factorisation

$$\begin{aligned}63 &= 3^{2}\times 7\end{aligned}$$
$$\begin{aligned}567 &= 3^{4}\times 7\end{aligned}$$

(ii) Two smallest values of $m$

$\mathrm{HCF}=21=3\times 7$. Since $63=3^{2}\times 7$ and $567=3^{4}\times 7$ already share $3^{2}\times 7=63$, $m$ must contribute exactly $3^{1}\times 7^{1}$ as the common factor. The smallest such values are

$$\begin{aligned}m &= 21\end{aligned}$$
$$\begin{aligned}m &= 42\end{aligned}$$

(iii) Smallest $n$ for $567n$ a perfect cube

$567=3^{4}\times 7$. For $567n$ to be a perfect cube each prime power must be a multiple of $3$:

$$\begin{aligned}3^{4}\times 7 \times n &= 3^{6}\times 7^{3}\end{aligned}$$
$$\begin{aligned}n &= 3^{2}\times 7^{2}\end{aligned}$$

(b) LCM of $P$ and $R$

Since $R$ is a multiple of $P$, the LCM of $P$ and $R$ is

$$\begin{aligned}\mathrm{LCM}(P,R) &= R\end{aligned}$$

(i) Prime factorisation

$$\begin{aligned}216 &= 2^3 \times 3^3 \\ 540 &= 2^2 \times 3^3 \times 5\end{aligned}$$

(ii)(a) HCF — lowest power of each common prime

$$\begin{aligned}\text{HCF} = 2^2 \times 3^3\end{aligned}$$

(ii)(b) LCM — highest power of each prime

$$\begin{aligned}\text{LCM} = 2^3 \times 3^3 \times 5\end{aligned}$$

(iii) Smallest $q$ for $\sqrt[3]{540q}$ a whole number

For $\sqrt[3]{540q}$ to be a whole number, every prime power in $540q$ must be a multiple of $3$.

$$\begin{aligned}540 &= 2^2 \times 3^3 \times 5 \\ q &= 2^1 \times 5^2 \\ q &= 50\end{aligned}$$

Check: $540 \times 50 = 27000 = 2^3 \times 3^3 \times 5^3$, and $\sqrt[3]{27000} = 30$.

(a)(i) Greatest whole number that divides all three (HCF)

HCF takes the lowest power of each common prime.

$$\begin{aligned}\text{HCF} &= 2^{7} \times 3^{3} \times 5^{4}\end{aligned}$$

(a)(ii) LCM

LCM takes the highest power of each prime.

$$\begin{aligned}\text{LCM} &= 2^{13} \times 3^{6} \times 5^{7}\end{aligned}$$

(a)(iii) Smallest $n$ so $n \times \text{LCM}$ is a perfect cube

For a perfect cube each index must be a multiple of $3$.

$$\begin{aligned}\text{LCM} &= 2^{13} \times 3^{6} \times 5^{7}\end{aligned}$$

Need $2$: $13 \to 15$ (add $2^{2}$); $3$: $6$ already a multiple of $3$; $5$: $7 \to 9$ (add $5^{2}$).

$$\begin{aligned}n &= 2^{2} \times 5^{2} = 4 \times 25 = 100\end{aligned}$$

(b) Sweets left in groups of $17$

Leaving $10$ from groups of $11$ and $12$ from groups of $13$ means the number is $1$ short of a multiple of both $11$ and $13$, so $N + 1$ is a multiple of $\text{LCM}(11,13) = 143$.

$$\begin{aligned}N + 1 &= 143k\end{aligned}$$

For $200 \le N \le 400$: $k = 2$ gives $N + 1 = 286$, so $N = 285$.

$$\begin{aligned}285 &= 17 \times 16 + 13\end{aligned}$$

Remainder $= 13$.

Taking the lowest power of each common prime factor:

$$\begin{aligned}\text{HCF}&=2\times3\times5\end{aligned}$$
$$\begin{aligned}&=30\end{aligned}$$

The least equal number is the LCM of $1470$ and $300$:

$$\begin{aligned}\text{LCM}&=2^2\times3\times5^2\times7^2\end{aligned}$$
$$\begin{aligned}&=14700\end{aligned}$$

Least number of packs of paperclips:

$$\begin{aligned}\frac{14700}{1470}&=10\end{aligned}$$

(i) Cube root of $A \times B$

$$\begin{aligned}A\times B&=p^6\times q^9\times r^{12}\end{aligned}$$
$$\begin{aligned}\sqrt[3]{A\times B}&=p^2 q^3 r^4\end{aligned}$$

(ii) Smallest $k$ so $C/k$ is a perfect square

For $\dfrac{C}{k}$ to be a perfect square, every prime power must be even. $C=2^1\times7^3\times13^2$, so remove $2^1$ and $7^1$:

$$\begin{aligned}k&=2\times7\end{aligned}$$
$$\begin{aligned}&=14\end{aligned}$$

Length of side:

$$\begin{aligned}\sqrt{2601}&=3\times17\end{aligned}$$
$$\begin{aligned}&=51\text{ cm}\end{aligned}$$

Perimeter:

$$\begin{aligned}4\times51&=204\text{ cm}\end{aligned}$$

(a) Edge of the cube

$$\begin{aligned}\text{edge} &= \sqrt[3]{5000} \\ &= \sqrt[3]{1000 \times 5} \\ &= 10 \sqrt[3]{5} \\ &= 10 \times 1.71 \\ &= 17.1 \text{ cm}\end{aligned}$$

(b)(i) HCF of 36 and 56

Common prime factors: $2^2$.

$$\begin{aligned}\text{HCF} &= 2^2 = 4\end{aligned}$$

(b)(ii) Smallest $k$ so $56/k$ is a perfect square

$56 = 2^3 \times 7 = 2^2 \times (2 \times 7)$. To leave a perfect square, divide out $2 \times 7 = 14$.

$$\begin{aligned}k &= 14\end{aligned}$$

(b)(iii) Smallest $n$ (LCM of 36, 56 and 54)

$$\begin{aligned}36 &= 2^2 \times 3^2 \\ 56 &= 2^3 \times 7 \\ 54 &= 2 \times 3^3 \\ \text{LCM} &= 2^3 \times 3^3 \times 7 \\ &= 1512\end{aligned}$$

$n = 1512$

(a)(i) Value of $x$

The only prime between 75 and 81:

$$x = 79$$

(a)(ii) Value of $h$

Since $79$ is prime, the factors $h-22$ and $h+56$ must be $1$ and $79$. As $h+56 > h-22$:

$$\begin{aligned}h-22 &= 1\\ h &= 23\end{aligned}$$

Check: $h+56 = 79$ and $1 \times 79 = 79$.

(b)(i) Prime factorisation of $127\,008$

$$127\,008 = 2^5 \times 3^4 \times 7^2$$

(b)(ii) Largest cube root and $k$

For the cube root to be largest, $\dfrac{127\,008}{2k}$ must be the largest perfect cube that divides it.

$$\begin{aligned}\frac{127\,008}{2k} &= \frac{2^5 \times 3^4 \times 7^2}{2k}\end{aligned}$$

The largest perfect cube of the form $2^a 3^b 7^c$ dividing $2^4 3^4 7^2 / k$ is $2^3 \times 3^3 = 216$, giving cube root $6$. Then:

$$\begin{aligned}2k &= \frac{127\,008}{216}\\ 2k &= 588\\ k &= 294\end{aligned}$$

Largest value $= 6$, with $k = 294$.

(i) Greatest possible volume (cube edge = HCF)

$$\begin{aligned}48 &= 2^4 \times 3\\ 72 &= 2^3 \times 3^2\\ 252 &= 2^2 \times 3^2 \times 7\end{aligned}$$
$$\begin{aligned}\text{HCF} &= 2^2 \times 3\\ &= 12\end{aligned}$$

Greatest possible volume:

$$\begin{aligned}V &= 12^3\\ &= 1728 \text{ cm}^3\end{aligned}$$

(ii) Least possible number of cubes

$$\begin{aligned}&= \frac{48}{12} \times \frac{72}{12} \times \frac{252}{12}\\ &= 4 \times 6 \times 21\\ &= 504\end{aligned}$$

(a) Smallest whole number

The number is $3$ more than a common multiple of $5$, $6$ and $9$.

$$\begin{aligned}\text{LCM}(5,6,9) &= 90 \\ \text{number} &= 90 + 3 \\ \text{number} &= 93\end{aligned}$$

(b) Largest number of goodie bags (HCF)

$$\begin{aligned}252 &= 2^2 \times 3^2 \times 7 \\ 210 &= 2 \times 3 \times 5 \times 7 \\ \text{HCF} &= 2 \times 3 \times 7 \\ \text{HCF} &= 42\end{aligned}$$

Largest number of goodie bags $= 42$.

(a)(i) Prime factorisation of $196$

$$196=2^{2}\times7^{2}$$

(a)(ii) Smallest $k$ for $196k$ a perfect cube

For a perfect cube each prime index must be a multiple of $3$:

$$\begin{aligned}196k &= 2^{2}\times7^{2}\times k\\ k &= 2\times7\\ &= 14\end{aligned}$$

(b) Smallest number of identical cubes

For the largest identical cube use the HCF of the dimensions:

$$\begin{aligned}196 &= 2^{2}\times7^{2}\\ 116 &= 2^{2}\times29\\ 80 &= 2^{4}\times5\\ \text{HCF} &= 2^{2}=4\end{aligned}$$

Number of $4$ cm cubes:

$$\begin{aligned}\frac{196}{4}\times\frac{116}{4}\times\frac{80}{4} &= 49\times29\times20\\ &= 28420\end{aligned}$$

(a) Values of $x$ and $y$

$1200 = 2^4 \times 3 \times 5^2$.

$$\begin{aligned}x &= 4 \\ y &= 1\end{aligned}$$

(b) Cubes left over

Largest cube number not exceeding 785 is $9^3 = 729$.

$$\begin{aligned}785 – 729\end{aligned}$$
$$= 56$$

(c) Value of $k$

$$\begin{aligned}7k – 5k &= 2022 \\ 2k &= 2022 \\ k &= 1011\end{aligned}$$

(a) HCF

$$\begin{aligned}\text{HCF} = 2^2 \times 7\end{aligned}$$

(b) Smallest $k$ for $2156k$ a perfect square

$$\begin{aligned}k = 11\end{aligned}$$

(c) Smallest $q$ so $2156q$ is a multiple of $2800$

For $2156q$ to be a multiple of $2800$:

$$\begin{aligned}\text{LCM} = 2^4 \times 5^2 \times 7^2 \times 11\end{aligned}$$
$$\begin{aligned}q &= 2^2 \times 5^2 \\ &= 100\end{aligned}$$

(a) Smallest $m$ so $pm$ is a multiple of $q$

$pm$ must contain $2\times 5^3\times 11$. From $p=2^4\times 3\times 5$, it already has $2$ and one $5$; it lacks $5^2$ and $11$.

$$\begin{aligned}m &= 5^2\times 11 \\ &= 275\end{aligned}$$

(b) Smallest $n$ so $\sqrt{pn}$ is a whole number

$p=2^4\times 3\times 5$. For a perfect square, each prime power must be even. Need an extra $3$ and an extra $5$.

$$\begin{aligned}n &= 3\times 5 \\ &= 15\end{aligned}$$

(a) Largest cube length = HCF(48, 20, 8)

$$\begin{aligned}48 &= 2^4\times 3 \\ 20 &= 2^2\times 5 \\ 8 &= 2^3\end{aligned}$$
$$\begin{aligned}\text{HCF} &= 2^2 \\ &= 4 \text{ cm}\end{aligned}$$

(b) Total number of cubes

$$\begin{aligned}\text{Number of cubes} &= \frac{48}{4}\times\frac{20}{4}\times\frac{8}{4} \\ &= 12\times 5\times 2 \\ &= 120\end{aligned}$$

(a) HCF of $a$ and $b$

Take the lowest power of each common prime:

$$\text{HCF} = 2 \times 5^2 \times 7$$
$$= 350$$

(b) Smallest $m$ for $am$ a perfect cube

$a = 2^5 \times 5^2 \times 7^1$. For a perfect cube each index must be a multiple of $3$:

$$m = 2^1 \times 5^1 \times 7^2$$
$$m = 490$$

$12 = 2^2 \times 3$, $45 = 3^2 \times 5$, $60 = 2^2 \times 3 \times 5$:

$$\text{LCM} = 2^2 \times 3^2 \times 5 = 180 \text{ s}$$
$$= 3 \text{ min}$$

From $8.30$ am to $9.45$ am is $75$ min:

$$\frac{75}{3} = 25$$

Including the first time at $8.30$ am:

$$25 + 1 = 26$$