Factors, multiples and primes

Multiples of 9: 9, 18, 27, 36, …

The only one between 20 and 30 is $27$.

Multiples of 7: $28, 35, 42, \dots$

The only one between 30 and 40 is $35$.

Factor pairs of 16: $1 \times 16$, $2 \times 8$, $4 \times 4$.

So the factors are 1, 2, 4, 8 and 16. Any two of these, e.g. 2 and 8.

Find the factor pairs of 20: $1 \times 20$, $2 \times 10$, $4 \times 5$.

So the factors are 1, 2, 4, 5, 10, 20. Any three of these will do, e.g. 2, 4, 5.

Factor pairs of 18: $1 \times 18$, $2 \times 9$, $3 \times 6$.

So the factors are 1, 2, 3, 6, 9 and 18. Any two of these, e.g. 2 and 9.

Multiples of 25 are 25, 50, 75, 100, ...

Only 100 is in the list ($25 \times 4 = 100$).

Factors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60.

The ones between 8 and 14 are 10 and 12, so either answer works.

Multiples of 5 always end in 0 or 5.

So any number from 100 to 995 ending in 0 or 5 works, for example $725$.

Check each number: $18 = 3 \times 6$, and none of 5, 11, 22 or 29 divides exactly by 3.

Answer: 18

Factors of 12 are the numbers that divide into it with no remainder: 1, 2, 3, 4, 6 and 12.

Any two of these, for example 3 and 4, are correct.

(a) $5 \times 5 = 25$, so 25 is square.

(b) $8 \times 3 = 24$, so 24 is a multiple of 8.

Find pairs that multiply to 35: $1 \times 35$ and $5 \times 7$.

So the factors are 1, 5, 7 and 35. Any two of these will do, e.g. 5 and 7.

Check each number: $18 = 3 \times 6$. None of 5, 11, 22 or 29 divides exactly by 3.

So the multiple of 3 is 18.

Check each odd number from 21 to 29.

21 = 3 × 7, 25 = 5 × 5, 27 = 3 × 9, so these are not prime.

23 and 29 have no factors other than 1 and themselves, so either is correct.

Factor pairs of 15: $1 \times 15$ and $3 \times 5$.

So the factors are 1, 3, 5 and 15; any two of these will do, e.g. 3 and 5.

The factor pairs of 12 are $1\times12$, $2\times6$ and $3\times4$.

So any two of 1, 2, 3, 4, 6, 12 will do, for example 3 and 4.

The 6 times table goes 6, 12, 18, 24, ...

The only one in the list is 18.

The 8 times table: 40, 48, 56, 64, ...

The ones between 41 and 60 are 48 and 56; either will do.

The factors of 10 are 1, 2, 5 and 10.

The only one in the list is 5.

Multiples of 7: 7, 14, 21, ...

7 is odd, so the first even one is 14.

Check each number: $24 \div 1, 24 \div 3, 24 \div 6, 24 \div 12, 24 \div 24$ all give whole numbers.

$24 \div 9 = 2.66...$, so 9 is not a factor.

(i) $12 = 4 \times 3$, so 12 is a multiple of 4.

(ii) 2 and 5 each have exactly two factors, so either is prime. (1 is not prime.)

(a) No. $10^2 = 10 \times 10 = 100$, not 20. He has worked out $10 \times 2$. The squares near 20 are 16 and 25.

(b) $12 \div 4 = 3$, and 3 is odd, so the claim is false.

Check each odd number from 21 to 39 for factors:

21, 25, 27, 33, 35, 39 all have factors other than 1 and themselves.

The primes are $23, 29, 31, 37$, so any three of these.

Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36

The odd ones are 1, 3 and 9. Of these, 3 and 9 are multiples of 3.

So the numbers are 3 and 9.

(a) Check 21 to 29: $21 = 3 \times 7$, $25 = 5 \times 5$, $27 = 3 \times 9$, and the even ones divide by 2. Only 23 and 29 are prime.

(b) Yes. Any even number bigger than 2 can be divided by 2, so it has at least three factors and cannot be prime.

Find the factor pairs of 30:

$1 \times 30$, $2 \times 15$, $3 \times 10$, $5 \times 6$.

So the factors are $1, 2, 3, 5, 6, 10, 15, 30$.

$63 = 3 \times 3 \times 7$

$105 = 3 \times 5 \times 7$

Common prime factors: $3 \times 7 = 21$

$60 = 2 \times 30 = 2 \times 2 \times 15 = 2 \times 2 \times 3 \times 5$

So $60 = 2^2 \times 3 \times 5$

Try primes $p$ and check whether $32 - p$ is prime:

$32 - 3 = 29$ (prime) ✓

$32 - 13 = 19$ (prime) ✓

So 3 and 29 (or 13 and 19).

(a) 12 is even, but $3 \times 4 = 12$, so 3 is an odd factor of 12.

(b) 23 is odd, but its digit 2 is even.

A card with both marks is a multiple of 3 and of 5, so a multiple of 15.

Multiples of 15 up to 100: 15, 30, 45, 60, 75, 90.

That is 6 cards.

(a) Try 18: $18 \div 2 = 9$, which ends in 9, not 4. (38 → 19 works too.)

(b) Alice is wrong. 27 ends in 7 but $27 = 3 \times 9$, so it is not prime. (57 = 3 × 19 also works.)

Square numbers: 1, 4, 9, 16, 25, 36.

$29 = 25 + 4 = 5^2 + 2^2$

$37 = 36 + 1 = 6^2 + 1^2$

So yes, Nadia is correct.

(a) $90 = 2 \times 45 = 2 \times 3 \times 15 = 2 \times 3 \times 3 \times 5 = 2 \times 3^2 \times 5$

(b) $A = 2^2 \times 3 = 12$, $B = 2 \times 3^2 = 18$.

LCM uses the highest power of each prime: $2^2 \times 3^2 = 36$

(a) $126 = 2 \times 63 = 2 \times 7 \times 9 = 2 \times 3^2 \times 7$

(b) $210 = 2 \times 3 \times 5 \times 7$

Common primes: $2 \times 3 \times 7 = 42$

(a) $156 = 2 \times 78 = 2 \times 2 \times 39 = 2 \times 2 \times 3 \times 13$, so $156 = 2^2 \times 3 \times 13$

(b) $130 = 2 \times 5 \times 13$

Shared primes: $2 \times 13 = 26$

Multiples of 30: 30, 60, 90, 120

Multiples of 24: 24, 48, 72, 96, 120

LCM $= 120$, so $120 \div 30 = 4$ packs and $120 \div 24 = 5$ boxes.

$500 = 2 \times 250 = 2 \times 2 \times 125$

$125 = 5 \times 5 \times 5$

So $500 = 2^2 \times 5^3$

$124 = 2 \times 62$

$62 = 2 \times 31$, and 31 is prime.

So $124 = 2 \times 2 \times 31 = 2^2 \times 31$

$24 = 2^3 \times 3$ and $56 = 2^3 \times 7$

LCM $= 2^3 \times 3 \times 7 = 168$

(Check: multiples of 56 are 56, 112, 168, and $168 = 7 \times 24$.)

(a) $60 = 2^2 \times 3 \times 5$ and $84 = 2^2 \times 3 \times 7$.

HCF $= 2^2 \times 3 = 12$.

(b) $24 = 2^3 \times 3$ and $40 = 2^3 \times 5$.

LCM $= 2^3 \times 3 \times 5 = 120$.

(a) $84 = 2 \times 42 = 2 \times 2 \times 21 = 2 \times 2 \times 3 \times 7$

(b) $60 = 2^2 \times 3 \times 5$ and $84 = 2^2 \times 3 \times 7$

LCM $= 2^2 \times 3 \times 5 \times 7 = 420$

$72 = 2^3 \times 3^2$ and $90 = 2 \times 3^2 \times 5$

Shared primes: $2 \times 3^2 = 18$

(a) $40 = 2^3 \times 5$ and $56 = 2^3 \times 7$

LCM $= 2^3 \times 5 \times 7 = 280$

(b) Take the lowest power of each shared prime: $2^2 \times 3 \times 5 = 60$

(a) Take the lowest power of each shared prime: $3^2 \times 7 = 63$

(b) Take the highest power of every prime: $2^2 \times 3^4 \times 7^2 = 4 \times 81 \times 49 = 15876$

m must be a multiple of both 5 and 7, so try $m = 35$.

n must be a multiple of both 2 and 7, so try $n = 14$.

$35 = 5 \times 7$ and $14 = 2 \times 7$, so the HCF is 7.

$108 = 2^2 \times 3^3$

$120 = 2^3 \times 3 \times 5$

Take the highest power of each prime:

LCM $= 2^3 \times 3^3 \times 5 = 8 \times 27 \times 5 = 1080$

$20 = 2^2 \times 5$, $45 = 3^2 \times 5$, $120 = 2^3 \times 3 \times 5$

LCM $= 2^3 \times 3^2 \times 5 = 360$ seconds

One hour $= 3600$ seconds, and $3600 \div 360 = 10$

So they flash together 10 times (11 if you count the start as well).

Every number in the list is a power of 98.

Each one divides the largest, $98^{91}$, so the lowest common multiple is $98^{91}$.