HCF and LCM

Factors, Multiples, HCF and LCM

A factor of a number divides it exactly. The factors of 18 are 1, 2, 3, 6, 9 and 18. A multiple is what you get by multiplying the number by 1, 2, 3, and so on. The multiples of 18 are 18, 36, 54, 72, …

The HCF (highest common factor, also called the GCD or greatest common divisor) of some numbers is the biggest number that divides all of them. The LCM (lowest common multiple) is the smallest number that all of them divide into.

For 12 and 18, the common factors are 1, 2, 3 and 6, so the HCF is 6. The common multiples are 36, 72, 108, …, so the LCM is 36.

By prime factorisation

Break each number into primes. Then:

• HCF: take only the primes common to all the numbers, each with its smallest power.
• LCM: take every prime that appears anywhere, each with its biggest power.

Take 72, 120 and 180:

$$72 = 2^3 \times 3^2 \qquad 120 = 2^3 \times 3 \times 5 \qquad 180 = 2^2 \times 3^2 \times 5$$

For the HCF, 2 and 3 appear in all three. The smallest power of 2 is $2^2$ and the smallest power of 3 is $3^1$, so the HCF is $2^2 \times 3 = 12$. The 5 is missing from 72, so it stays out.

For the LCM, the primes are 2, 3 and 5 with biggest powers $2^3$, $3^2$ and $5^1$, so the LCM is 8 × 9 × 5 = 360.

When the options are written in prime-power form, you don't even need to multiply. Just compare the powers.

Facts that kill wrong options

• The HCF divides every one of the numbers, and every number divides the LCM. So the HCF always divides the LCM.
• The HCF can never be bigger than the smallest number, and the LCM can never be smaller than the largest.
• If one number divides the other (like 15 and 45), the HCF is the smaller number and the LCM is the bigger one.
• Co-prime numbers have HCF 1, so their LCM is simply their product. The LCM of 8 and 15 is 120.
• Two consecutive numbers always have HCF 1. Two consecutive even numbers always have HCF 2.

The Division Method and HCF of Big Numbers

Prime factors get slow when the numbers are big or hide a large prime. The division method (Euclid's method) is quicker:

1. Divide the bigger number by the smaller one.
2. Now divide the previous divisor by the remainder.
3. Keep going until the remainder is 0. The last divisor is the HCF.

HCF of 1517 and 1961:

The last divisor is 37, so the HCF is 37. (Indeed, 1517 = 37 × 41 and 1961 = 37 × 53.)

For three numbers, find the HCF of any two, then the HCF of that answer with the third.

The difference shortcut

Any common factor of two numbers also divides their difference. So the HCF of 1517 and 1961 must divide 1961 − 1517 = 444, and 444 = $2^2 \times 3 \times 37$. Now you only need to test the factors of 444 that appear in the options. 1517 is odd, so 2 and 4 are out. Its digit sum is 14, so 3 is out. That leaves 37, and it works. In an MCQ this is often the fastest route of all.

LCM by common division

For the LCM of several numbers, keep dividing by a prime that goes into at least two of them, and bring the others down unchanged. Stop when no two numbers left share a factor. The LCM is the product of all the divisors and everything left at the bottom.

LCM of 12, 18 and 30:

• Divide by 2: 6, 9, 15
• Divide by 3: 2, 3, 5
• No two of 2, 3 and 5 share a factor, so the LCM is 2 × 3 × 2 × 3 × 5 = 180.

[SSC Tip] In an LCM question with options, test the options from the smallest. The first one that every given number divides into is the LCM. In an HCF question, test from the largest: the first one that divides every number is the HCF.

HCF and LCM of Fractions and Decimals

Fractions

Write each fraction in its lowest terms first. Then

$$\text{HCF} = \frac{\text{HCF of numerators}}{\text{LCM of denominators}} \qquad \text{LCM} = \frac{\text{LCM of numerators}}{\text{HCF of denominators}}$$

An easy way to remember which goes where: the HCF should come out small, so the big thing (an LCM) goes underneath. The LCM should come out big, so the small thing (an HCF) goes underneath.

For $\frac{2}{3}$, $\frac{8}{9}$ and $\frac{10}{27}$:

• HCF = HCF(2, 8, 10) over LCM(3, 9, 27) = $\frac{2}{27}$
• LCM = LCM(2, 8, 10) over HCF(3, 9, 27) = $\frac{40}{3}$

You can test the LCM: $\frac{40}{3}$ divided by each fraction gives 20, 15 and 36, all whole numbers, so it really is a common multiple.

Decimals

1. Give every number the same number of decimal places by adding zeros.
2. Drop the decimal point and find the HCF or LCM of the whole numbers.
3. Put back the same number of decimal places in the answer.

• LCM of 0.48, 0.72 and 1.08: work with 48, 72 and 108. Their LCM is 432, so the answer is 4.32. Their HCF is 12, so the HCF of the decimals is 0.12.
• HCF of 3.06 and 5.1: write 5.1 as 5.10, so work with 306 and 510. Their HCF is 102, giving 1.02.
• LCM of 1.5 and 2.4: work with 15 and 24. The LCM is 120, giving 12.0, which is just 12. The HCF is 3, giving 0.3.

The 3.06 and 5.1 case shows why step 1 matters. If you skip it and work with 306 and 51, you get 51 and then answer 0.51, which is wrong.

When one number is a multiple of the other

Spot it before doing any work. 142.5 × 2 = 285, so the LCM of 142.5 and 285 is 285 and the HCF is 142.5. SSC sets these with big-looking decimals to make you start long working that you don't need.

HCF × LCM = Product (of Two Numbers Only)

For any two numbers $a$ and $b$,

$$\text{HCF} \times \text{LCM} = a \times b$$

Why it works. For each prime, the HCF takes the smaller power and the LCM takes the bigger one. Between them they use both powers exactly once, which is just what the product does. With 72 = $2^3 \times 3^2$ and 60 = $2^2 \times 3 \times 5$, the HCF is $2^2 \times 3 = 12$ and the LCM is $2^3 \times 3^2 \times 5 = 360$, and 12 × 360 = 4320 = 72 × 60.

The three standard uses

• The other number. HCF 12, LCM 360 and one number 72. The other number is 12 × 360 ÷ 72 = 60.
• HCF from the product and LCM. The product of two numbers is 4800 and their LCM is 240. The HCF is 4800 ÷ 240 = 20. (60 and 80 fit.)
• LCM from the product and HCF. Same idea the other way: LCM = product ÷ HCF.

It fails for three numbers

Take 2, 4 and 8. The HCF is 2 and the LCM is 8, so HCF × LCM = 16, but the product is 64. The formula is only for two numbers, so never use it for three. SSC sometimes puts the "product" answer in the options for a three-number question to catch exactly this.

An LCM that can't happen

The LCM is always a multiple of the HCF. If the HCF of two numbers is 16, the LCM could be 64, 96 or 240, but never 120, because 120 ÷ 16 is not a whole number. The question "Which of these cannot be the LCM?" is answered by picking the option that the HCF doesn't divide.

The same check works for "HCF and LCM of two numbers are given as …" questions. If the LCM is not a multiple of the HCF, no such pair of numbers exists.

[SSC Tip] After finding the other number, check it in five seconds. The HCF must divide it, and it must divide the LCM. For the 60 above: 60 ÷ 12 = 5 and 360 ÷ 60 = 6, so it passes.

Ratio, Sum and Co-prime Parts

If two numbers have HCF $h$, take $h$ out of both. The numbers become $ha$ and $hb$, where $a$ and $b$ are co-prime (if they still shared a factor, $h$ would not be the highest). Then

$$\text{LCM} = h \times a \times b \qquad \frac{\text{LCM}}{\text{HCF}} = a \times b \qquad \text{Product} = h^2 \times a \times b$$

Almost every "HCF with a twist" question uses these three lines.

Ratio and HCF. Two numbers are in the ratio 4 : 7 and their HCF is 9. The numbers are 36 and 63, and the LCM is 9 × 4 × 7 = 252. A ratio in lowest terms always has co-prime terms, so the multiplier is the HCF.

Ratio and LCM. Two numbers are in the ratio 3 : 5 and their LCM is 240. The numbers are $3x$ and $5x$, so the LCM is $15x = 240$, giving $x = 16$. The numbers are 48 and 80, and the HCF is 16. Be careful with three numbers: for the ratio 2 : 3 : 4 the LCM is $x$ × LCM(2, 3, 4) = $12x$, not $24x$, because 2 and 4 are not co-prime.

Sum and HCF. The sum of two numbers is 144 and their HCF is 12. Then $12a + 12b = 144$, so $a + b = 12$ with $a$ and $b$ co-prime. The pairs adding to 12 are (1, 11), (2, 10), (3, 9), (4, 8), (5, 7) and (6, 6). Only (1, 11) and (5, 7) are co-prime, so there are two pairs: 12 and 132, and 60 and 84.

Product and HCF. The product of two numbers is 7680 and their HCF is 16. Then $16^2 \times ab = 7680$, so $ab = 30$. The co-prime pairs with product 30 are (1, 30), (2, 15), (3, 10) and (5, 6). That gives four pairs: 16 and 480, 32 and 240, 48 and 160, 80 and 96.

HCF, LCM and difference. HCF 14, LCM 420 and difference 98. Here $ab$ = 420 ÷ 14 = 30 and $14(a - b) = 98$, so $a - b = 7$. From the pairs above only (10, 3) works, so the numbers are 140 and 42.

HCF and the other factors of the LCM. "The HCF of two numbers is 19 and the other two factors of their LCM are 7 and 11." The numbers are 19 × 7 = 133 and 19 × 11 = 209. The larger one is 209, and the LCM is 19 × 7 × 11 = 1463.

Greatest Number That Leaves Given Remainders

"Find the greatest number that divides 437 and 623 leaving remainders 5 and 11."

If the number leaves 5 from 437, then it divides 437 − 5 = 432 exactly. In the same way it divides 623 − 11 = 612 exactly. The greatest number that divides both is HCF(432, 612) = 36.

Rule: subtract each remainder from its own number, then take the HCF.

One check is worth doing: the answer must be bigger than every remainder. Here 36 is bigger than 11, so it holds. If the HCF came out smaller than one of the remainders, that remainder could not happen, so go back and read the question again.

Same remainder, but the remainder is not given

"Find the greatest number that divides 158, 248 and 383 leaving the same remainder in each case."

If each number leaves the same remainder $r$, then the differences between the numbers leave no remainder at all. So the number divides 248 − 158 = 90, 383 − 248 = 135 and 383 − 158 = 225. HCF(90, 135, 225) = 45. The number is 45, and the common remainder is what 158 leaves when divided by 45, which is 23.

Rule: sort the numbers and take the HCF of the differences of neighbours. Two differences are enough, because the third one is just their sum.

SSC then usually asks for something about that number: the sum of its digits (4 + 5 = 9 here), the remainder itself (23), or the two added together. Read the last line of the question carefully before marking.

Same remainder, given

If the question says "leaving remainder 7 in each case", just subtract 7 from every number and take the HCF.

Exact division

With no remainders at all, it is the plain HCF. "The greatest number that divides 1517 and 1961 exactly" is 37.

[SSC Tip] Don't confuse "divides the numbers" with "is divisible by the numbers". A number that divides 432 and 612 is smaller than both, so it comes from the HCF. A number that 432 and 612 both divide into is bigger than both, so it comes from the LCM, which comes next.

Least Number That Leaves Given Remainders

Now the given numbers have to go into our answer, so it is LCM territory.

Same remainder $r$. The least number that leaves remainder 5 when divided by 12, 18 and 27: LCM(12, 18, 27) = 108, so the number is 108 + 5 = 113. Every such number has the form $108k + 5$. (The number 5 itself also leaves 5, but SSC questions mean a number bigger than the divisors, so take $k = 1$.)

Constant gap. Divided by 6, 8 and 10, a number leaves 4, 6 and 8. Each remainder is 2 short of its divisor, so the number plus 2 is divisible by all three. The number is LCM − 2 = 120 − 2 = 118.

With an extra condition. The least number that leaves remainder 4 when divided by 6, 9 and 15 and is also divisible by 7. These numbers are $90k + 4$: 94, 184, 274, 364, … The first multiple of 7 is 364 = 7 × 52. A faster test: 90 = 84 + 6, so 90 is −1 for 7, and $90k + 4$ leaves $4 - k$. That is 0 when $k = 4$, giving 364.

Least perfect square divisible by several numbers. The LCM of 6, 9 and 15 is $90 = 2 \times 3^2 \times 5$. A perfect square needs an even power of every prime, so multiply by 2 × 5. The answer is 900, which is $30^2$.

Largest and smallest n-digit numbers.

• Smallest 4-digit number divisible by 12, 15 and 20: the LCM is 60, and 1000 leaves 40, so add 60 − 40 = 20 to get 1020.
• Largest 4-digit number divisible by 12, 15 and 20: 9999 leaves 39, so subtract it to get 9960.
• Largest 4-digit number that leaves remainder 5 when divided by 8, 12 and 15: the LCM is 120, the largest 4-digit multiple is 9960, and adding 5 gives 9965. Always check that adding the remainder hasn't pushed you past 9999.

Least number to add. To make 2000 divisible by 12, 18 and 30: the LCM is 180, and 2000 leaves 20, so add 180 − 20 = 160.

How many in a range. Numbers between 1000 and 3000 that are divisible by 12, 18 and 30 are just the multiples of 180. They run from 1080 (180 × 6) to 2880 (180 × 16), so there are 16 − 6 + 1 = 11 of them.

Word Problems: Bells, Lights, Tracks, Tiles and Groups

First decide what the question wants: something that happens together again (LCM) or the biggest equal piece (HCF).

LCM problems

Bells and lights. Four bells ring every 6, 8, 12 and 15 minutes and ring together at 9:00 am. They next ring together after LCM(6, 8, 12, 15) = 120 minutes, at 11:00 am. In 6 hours (360 minutes) they ring together 360 ÷ 120 = 3 more times, or 4 times if you count the ring at 9:00. Always check whether the question counts the start.

Watch the units. Lights that flash every 20, 30 and 45 seconds flash together every 180 seconds, which is 3 minutes, so 20 times in the hour after they start.

Traffic lights. Lights change every 36, 54 and 90 seconds and change together at 7:00:00 am. The LCM is 540 seconds = 9 minutes, so they next change together at 7:09:00 am.

Runners on a circular track. A, B and C take 36, 48 and 60 seconds for one round and start together. They are all back at the start together after LCM = 720 seconds = 12 minutes. By then A has run 20 rounds, B 15 and C 12.

HCF problems

Square tiles. A floor is 6 m 24 cm by 4 m 32 cm. Convert to centimetres first: 624 and 432. The largest square tile has side HCF(624, 432) = 48 cm, and the number of tiles is (624 ÷ 48) × (432 ÷ 48) = 13 × 9 = 117.

Longest tape or scale. To measure 3 m 85 cm, 7 m 70 cm and 12 m 95 cm exactly, the longest tape is HCF(385, 770, 1295) = 35 cm.

Equal rows or groups. 391 boys and 323 girls stand in rows of equal size, each row only boys or only girls. The biggest row has HCF(391, 323) = 17 students, and the number of rows is 23 + 19 = 42. The fewest rows always come from the biggest row size.

Cutting rods. Rods of 84 cm, 140 cm and 196 cm are cut into equal pieces, as long as possible. Each piece is 28 cm, and there are 3 + 5 + 7 = 15 pieces.

Tins and tanks. 120 L, 168 L and 216 L of three oils go into identical tins, without mixing. The fewest tins come from the biggest tin, HCF = 24 L, so the number of tins is 5 + 7 + 9 = 21.

HCF or LCM? Spotting the Type, and Two Shortcuts

The words alone can mislead you. "Least number of tiles" is an HCF question, because the fewest tiles come from the biggest tile. "Greatest 4-digit number divisible by 12, 15 and 20" uses the LCM, because the numbers go into the answer. So ask yourself one thing: is the answer smaller than the given numbers (it goes into them, so HCF) or bigger than them (they go into it, so LCM)?

Checking options without working

• An HCF option must divide every given number. Throw out any option that doesn't.
• An LCM option must be divisible by every given number. Throw out any option that isn't.
• An LCM option must also be a multiple of the HCF, if the HCF is given.

Shortcut 1: scaling

Multiplying both numbers by the same $k$ multiplies the HCF and the LCM by $k$. The HCF of 18 and 24 is 6, so the HCF of 180 and 240 is 60 and the HCF of 1.8 and 2.4 is 0.6. Their LCM is 72, so the LCM of 180 and 240 is 720. Also, the HCF of $a$ and $b$ is the same as the HCF of $a$ and $b - a$, which is the difference shortcut from earlier.

Shortcut 2: numbers of the form $2^m - 1$ [Tier 2]

For any whole number base $a$ of 2 or more,

$$\text{HCF}(a^m - 1,\ a^n - 1) = a^{\text{HCF}(m, n)} - 1$$

So the HCF of $2^{100} - 1$ and $2^{120} - 1$ is $2^{20} - 1$, because the HCF of 100 and 120 is 20. In the same way, the HCF of $3^{12} - 1$ and $3^{18} - 1$ is $3^6 - 1 = 728$.

This rule is only for the "minus 1" form. Numbers like $2^m + 1$ behave differently, so don't stretch it to them.

Solved Examples

Question 1

Find the HCF and LCM of 84, 126 and 210.

Answer. I break them into primes: 84 = $2^2 \times 3 \times 7$, 126 = $2 \times 3^2 \times 7$ and 210 = 2 × 3 × 5 × 7.

For the HCF I take what all three share, with the smallest powers: 2 × 3 × 7 = 42.

For the LCM I take every prime with its biggest power: $2^2 \times 3^2 \times 5 \times 7$ = 4 × 9 × 35 = 1260.

Question 2

Find the HCF of 1591 and 2279.

Answer. I use the division method:

• 2279 = 1 × 1591 + 688
• 1591 = 2 × 688 + 215
• 688 = 3 × 215 + 43
• 215 = 5 × 43 + 0

The last divisor is 43, so the HCF is 43. Check: 1591 = 43 × 37 and 2279 = 43 × 53.

Question 3

Find the HCF and LCM of $\frac{4}{5}$, $\frac{6}{25}$ and $\frac{8}{15}$.

Answer. All three are already in lowest terms. HCF = HCF(4, 6, 8) over LCM(5, 25, 15) = $\frac{2}{75}$. LCM = LCM(4, 6, 8) over HCF(5, 25, 15) = $\frac{24}{5}$. So the HCF is 2/75 and the LCM is 24/5.

Question 4

Find the HCF and LCM of 2.52, 0.84 and 1.4.

Answer. I give them all two decimal places: 2.52, 0.84 and 1.40. Without the point they are 252, 84 and 140.

252 = $2^2 \times 3^2 \times 7$, 84 = $2^2 \times 3 \times 7$ and 140 = $2^2 \times 5 \times 7$. So the HCF is $2^2 \times 7 = 28$ and the LCM is $2^2 \times 3^2 \times 5 \times 7 = 1260$.

Putting the two decimal places back, the HCF is 0.28 and the LCM is 12.60, which is 12.6.

Question 5

Find the LCM and HCF of 112.4 and 337.2.

Answer. Before any working I check whether one is a multiple of the other: 112.4 × 3 = 337.2. It is. So the LCM is the bigger number, 337.2, and the HCF is the smaller one, 112.4.

Question 6

The HCF of two numbers is 18 and their LCM is 540. If one number is 108, find the other.

Answer. HCF × LCM = product of the two numbers, so the other number is 18 × 540 ÷ 108 = 9720 ÷ 108 = 90. Check: 108 and 90 do have HCF 18 and LCM 540.

Question 7

The product of two numbers is 6912 and their LCM is 288. Find their HCF.

Answer. HCF = product ÷ LCM = 6912 ÷ 288 = 24. The numbers 72 and 96 fit: their HCF is 24 and their LCM is 288.

Question 8

The HCF of two numbers is 14. Which of these can NOT be their LCM? (a) 98 (b) 126 (c) 168 (d) 189

Answer. The LCM has to be a multiple of the HCF. 98 = 14 × 7, 126 = 14 × 9 and 168 = 14 × 12 all work. 189 is odd, so 14 can't go into it. The answer is (d) 189.

Question 9

Two numbers are in the ratio 5 : 8 and their HCF is 7. Find their LCM.

Answer. The numbers are 7 × 5 = 35 and 7 × 8 = 56. Since 5 and 8 are co-prime, the LCM is 7 × 5 × 8 = 280.

Question 10

The sum of two numbers is 216 and their HCF is 12. How many such pairs of numbers are there?

Answer. I write the numbers as $12a$ and $12b$ with $a$ and $b$ co-prime. Then $12(a + b) = 216$, so $a + b = 18$.

The pairs adding to 18 are (1, 17), (2, 16), (3, 15), (4, 14), (5, 13), (6, 12), (7, 11), (8, 10) and (9, 9). The co-prime ones are (1, 17), (5, 13) and (7, 11). So there are 3 pairs: 12 and 204, 60 and 156, 84 and 132.

Question 11

The HCF of two numbers is 13, and the other two factors of their LCM are 11 and 15. Find the larger number.

Answer. The numbers are 13 × 11 = 143 and 13 × 15 = 195. The larger one is 195. (Their LCM is 13 × 11 × 15 = 2145.)

Question 12

Find the greatest number that divides 701 and 1109 leaving remainders 5 and 7 respectively.

Answer. I take the remainders away first: 701 − 5 = 696 and 1109 − 7 = 1102. The number must divide both exactly, so it is their HCF.

• 1102 = 1 × 696 + 406
• 696 = 1 × 406 + 290
• 406 = 1 × 290 + 116
• 290 = 2 × 116 + 58
• 116 = 2 × 58 + 0

The HCF is 58, and it is bigger than both 5 and 7, so those remainders are possible.

Question 13

N is the greatest number that divides 1023, 1527 and 2283 leaving the same remainder each time. Find the sum of the digits of N, and the remainder.

Answer. Same remainder, so I use the differences: 1527 − 1023 = 504 and 2283 − 1527 = 756. For their HCF: 756 − 504 = 252, and 504 = 2 × 252, so the HCF is 252. So N = 252, and its digit sum is 2 + 5 + 2 = 9.

The remainder is what 1023 leaves: 1023 − 4 × 252 = 1023 − 1008 = 15.

Question 14

Find the least number which leaves remainder 7 when divided by 15, 20, 24 and 32.

Answer. First the LCM: 15 = 3 × 5, 20 = $2^2 \times 5$, 24 = $2^3 \times 3$ and 32 = $2^5$. Taking the biggest powers, LCM = $2^5 \times 3 \times 5 = 480$. Adding the remainder gives 487.

Question 15

Find the least number which, when divided by 9, 12 and 15, leaves remainders 7, 10 and 13 respectively.

Answer. Each remainder is 2 less than its divisor: 9 − 7 = 12 − 10 = 15 − 13 = 2. So the number plus 2 is divisible by all three. LCM(9, 12, 15) = 180, so the number is 180 − 2 = 178.

Question 16

Find the least number which leaves remainder 3 when divided by 5, 6 and 8, and is divisible by 11.

Answer. LCM(5, 6, 8) = 120, so the number is $120k + 3$. Now 120 = 121 − 1, so 120 is −1 for 11, and $120k + 3$ leaves $3 - k$. That is 0 when $k = 3$. So the number is 360 + 3 = 363, and 363 = 11 × 33. (Taking $k = 0$ gives 3, which is not divisible by 11.)

Question 17

Find the largest 4-digit number which leaves remainder 4 when divided by 6, 10 and 14.

Answer. LCM(6, 10, 14) = 210. 9999 ÷ 210 gives 47 and leaves 129, so the largest 4-digit multiple of 210 is 9999 − 129 = 9870. Adding the remainder gives 9874, which is still a 4-digit number.

Question 18

Four bells ring at intervals of 9, 12, 15 and 18 minutes. They ring together at 8:00 am. How many times will they ring together from 8:00 am to 8:00 pm, counting both these times?

Answer. LCM(9, 12, 15, 18) = 180 minutes, which is 3 hours. So they ring together at 8 am, 11 am, 2 pm, 5 pm and 8 pm. That is 5 times.

Question 19

A, B and C run round a circular track, taking 30, 45 and 54 seconds per round. They start together from the same point. After how long will they all be together at the starting point again?

Answer. I need LCM(30, 45, 54). 30 = 2 × 3 × 5, 45 = $3^2 \times 5$ and 54 = $2 \times 3^3$, so the LCM is $2 \times 3^3 \times 5 = 270$ seconds, which is 4 minutes 30 seconds. By then A has run 9 rounds, B 6 and C 5.

Question 20

A floor measures 5 m 44 cm by 3 m 74 cm. It is to be covered with square tiles, all the same size, as large as possible. Find the side of each tile and the number of tiles.

Answer. In centimetres the floor is 544 by 374. For the HCF: 544 = 1 × 374 + 170, 374 = 2 × 170 + 34 and 170 = 5 × 34. So each tile is 34 cm on a side. The number of tiles is (544 ÷ 34) × (374 ÷ 34) = 16 × 11 = 176.

Question 21

A shop has 357 litres, 408 litres and 459 litres of three different oils. It wants to fill them into identical tins without mixing, with every tin full. What is the least number of tins needed?

Answer. The fewest tins come from the biggest tin, so the tin size is HCF(357, 408, 459). 357 = 3 × 7 × 17, 408 = $2^3 \times 3 \times 17$ and 459 = $3^3 \times 17$, so the HCF is 3 × 17 = 51 litres. The number of tins is 357 ÷ 51 + 408 ÷ 51 + 459 ÷ 51 = 7 + 8 + 9 = 24.