CBSE Class 6 Mathematics Prime Time Worksheet Set 01

Long Answer Questions

Question 1. Answer the following about the numbers 36 and 48.
(i) Write all the factors of 36 and all the factors of 48.
(ii) Write all their common factors and name the greatest one.
(iii) Describe, step by step, the method you used to find the common factors of two numbers.
Answer:
(i) Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36.
Factors of 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48.
(ii) Numbers in both lists: 1, 2, 3, 4, 6 and 12. The greatest is 12; check: 36 = 12 × 3 and 48 = 12 × 4.
(iii) Step 1: Find all factors of the first number by checking which numbers divide it exactly. Factors come in pairs, such as 4 × 9 = 36, which speeds this up.
Step 2: Do the same for the second number.
Step 3: Compare the two lists and write every number found in both. These are the common factors.
Step 4: The largest number in this new list is the greatest common factor.
Since 1 is common to every pair, the list is never empty. Here it has six numbers, so 36 and 48 are not co-prime.
In simple words: List the factors of both numbers, keep the shared ones, and the biggest shared one is 12.

Teacher's Note:
1. Factor pairs prevent missing factors.
2. Always check the greatest common factor by dividing.
3. More than one common factor means the pair is not co-prime.

Question 2. Three important ideas of this chapter are prime numbers, composite numbers and co-prime numbers.
(i) Define each of the three ideas and give two examples of each.
(ii) Where does the number 1 fit among prime and composite numbers? Give a reason.
(iii) The numbers 2 and 3 are both prime. Are they also co-prime? Explain.
Answer:
(i) A prime number has exactly two factors, 1 and itself. Examples: 7 (factors 1, 7) and 13 (factors 1, 13).
A composite number has more than two factors. Examples: 15 (factors 1, 3, 5, 15) and 20 (factors 1, 2, 4, 5, 10, 20).
Two numbers are co-prime if their only common factor is 1. Examples: the pair 4 and 9, and the pair 10 and 21.
Prime and composite describe one number; co-prime describes a pair of numbers.
(ii) 1 has only one factor, itself. It does not have two factors, so it is not prime, and it does not have more than two, so it is not composite. So 1 is neither.
(iii) Yes. Factors of 2 are 1, 2 and factors of 3 are 1, 3. The only common one is 1, so 2 and 3 are co-prime. In fact any two different primes are co-prime.
In simple words: Primes have two factors, composites more, 1 is neither, and co-prime pairs share only 1.

Teacher's Note:
1. Give examples with their factors.
2. Keep the ‘one number’ and ‘pair’ ideas separate.
3. Any two different primes are co-prime.

Question 3. Two children start the factor tree of 180 in two different ways, as shown below.
(i) Complete both factor trees and write the prime factorisation of 180 that each one gives.
(ii) Compare the two answers. What do you notice about the primes used and how many times each one appears?
(iii) What does this tell you about the prime factorisation of a number?
Answer:
(i) Way 1: 180 = 2 × 90, then 90 = 2 × 45, 45 = 3 × 15 and 15 = 3 × 5. The leaves are 2, 2, 3, 3, 5, so 180 = 2 × 2 × 3 × 3 × 5.
Way 2: 180 = 4 × 45, then 4 = 2 × 2, 45 = 3 × 15 and 15 = 3 × 5. The leaves are again 2, 2, 3, 3, 5, so 180 = 2 × 2 × 3 × 3 × 5.

Multiplying back: 2 × 2 = 4, 4 × 3 = 12, 12 × 3 = 36, 36 × 5 = 180. Both trees are correct.
(ii) The same primes 2, 3 and 5 appear in both trees: 2 twice, 3 twice and 5 once. Only the order in which they were found is different.
(iii) A number can be split in many ways, but the final prime factorisation is always the same, apart from the order of the factors. So the prime factorisation of a number is unique. We usually write it in increasing order: 180 = 2 × 2 × 3 × 3 × 5.
In simple words: Different factor trees of 180 end with the same primes, so prime factorisation is unique.

Teacher's Note:
1. Keep splitting until every leaf is prime.
2. Compare the leaves, not the branches.
3. Uniqueness is the key idea here.

Question 4. The number 3640 is written on the blackboard.
(i) Use the divisibility tests of this chapter to check whether 3640 is divisible by 2, by 5 and by 10.
(ii) Use the tests to check whether it is divisible by 4 and by 8.
(iii) Write which of the numbers 2, 4, 5, 8 and 10 are factors of 3640, and explain in one line why each test looks only at the last digits.
Answer:
(i) By 2: the last digit 0 is even, so 3640 is divisible by 2.
By 5: the last digit is 0, so 3640 is divisible by 5.
By 10: it ends in 0, so 3640 is divisible by 10.
(ii) By 4: the last two digits form 40 = 4 × 10, so 3640 is divisible by 4.
By 8: the last three digits form 640 = 8 × 80, so 3640 is divisible by 8.
(iii) All five, 2, 4, 5, 8 and 10, are factors of 3640. Check: 3640 = 2 × 1820 = 4 × 910 = 5 × 728 = 8 × 455 = 10 × 364.
The tests look only at the last digits because 10 is a multiple of 2, 5 and 10, 100 is a multiple of 4, and 1000 is a multiple of 8; so the part in front of those digits is already a multiple, and only the last digits decide.
In simple words: 3640 passes every test, so 2, 4, 5, 8 and 10 all divide it.

Teacher's Note:
1. Name the test before using it.
2. For 8 always use three digits.
3. Show a check by division at the end.

Question 5. A class makes a chart of the numbers from 1 to 50 and uses the Sieve of Eratosthenes on it.
(i) Write the steps of the sieve in order, up to the point where the chart is finished.
(ii) List all the prime numbers from 1 to 50 and say how many there are.
(iii) Explain why crossing out the multiples of each circled number leaves only the primes behind.
Answer:
(i) Step 1: Cross out 1, as it is neither prime nor composite.
Step 2: Circle 2 and cross out every other multiple of 2: 4, 6, 8, 10, and so on.
Step 3: The next uncrossed number is 3. Circle it and cross out 6, 9, 12, and so on.
Step 4: The next uncrossed number is 5. Circle it and cross out 10, 15, 20, and so on.
Step 5: The next uncrossed number is 7. Circle it and cross out 14, 21, 28, 35, 42 and 49.
Step 6: Every number left is now circled, so the chart is finished.
(ii) 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47. There are 15 primes.
(iii) When a number gets circled, no smaller number except 1 divides it, so it has exactly two factors and is prime. Every composite number has a prime factor smaller than itself, so it is a multiple of some circled number and gets crossed out. So the crossed-out numbers are exactly the composites, and only the primes stay circled.
In simple words: Crossing out multiples removes every composite number and leaves the 15 primes up to 50.

Teacher's Note:
1. After 7 nothing new is crossed out up to 50, since 11 × 11 is more than 50.
2. Count the final list twice.
3. Every composite has a smaller prime factor.

Question 6. A museum keeps an old chest shut with two number locks. The chest can be opened only if some number bigger than 1 divides both lock numbers. The keeper wants a pair for which this is impossible, so that the chest stays safe.
(i) Using prime factorisation, test the pairs (14, 21), (16, 27) and (18, 30) and say which pair is safe.
(ii) What is the special name for such a safe pair of numbers?
(iii) Explain why two numbers have a common factor bigger than 1 exactly when they have a common prime factor.
Answer:
(i) 14 = 2 × 7 and 21 = 3 × 7 share the prime 7, so 7 divides both. Not safe.
16 = 2 × 2 × 2 × 2 and 27 = 3 × 3 × 3 share no prime. Safe.
18 = 2 × 3 × 3 and 30 = 2 × 3 × 5 share 2 and 3. Not safe.
So the keeper should use the pair 16 and 27.
(ii) Such a pair is called a pair of co-prime numbers. Here 16 and 27 are co-prime though neither is prime.
(iii) If two numbers have a common factor bigger than 1, that factor is either prime or can be broken into primes, and each of those primes divides both numbers; so a common prime factor exists. On the other hand, a common prime factor is itself a common factor bigger than 1. So checking the common prime factors settles the question completely.
In simple words: Only 16 and 27 share no prime, so they are co-prime and the chest stays safe.

Teacher's Note:
1. Test each pair separately with its factorisations.
2. Co-prime numbers need not be prime.
3. Argue both directions in part (iii).

Question 7. Study the numbers 216 and 90.
(i) Write the prime factorisation of each number.
(ii) Write their common prime factors.
(iii) Using the prime factorisations, write all the common factors of 216 and 90 and name the greatest one.
Answer:
(i) 216 = 8 × 27 = (2 × 2 × 2) × (3 × 3 × 3), so 216 = 2 × 2 × 2 × 3 × 3 × 3.
90 = 9 × 10 = (3 × 3) × (2 × 5), so 90 = 2 × 3 × 3 × 5.
(ii) The common prime factors are 2 and 3. 5 appears only in 90.
(iii) A common factor can use at most one 2 (90 has only one 2) and at most two 3s (90 has only two 3s).
With no 3: 1 and 2. With one 3: 3 and 6. With two 3s: 9 and 18.
So the common factors are 1, 2, 3, 6, 9 and 18, and the greatest is 18. Check: 216 = 18 × 12 and 90 = 18 × 5.
In simple words: 216 and 90 share one 2 and two 3s, so their common factors are 1, 2, 3, 6, 9, 18 and the greatest is 18.

Teacher's Note:
1. The smaller count of each prime limits the common factors.
2. Build the factors in an organised way.
3. Check the greatest one by division.

Question 8. Guna is exploring the divisibility test for 8.
(i) Find all the numbers between 2120 and 2140 that are divisible by 8.
(ii) Find all the numbers between 3120 and 3140 that are divisible by 8.
(iii) Write what you observe in the two answers, state the test for divisibility by 8 in your own words, and use it on 9416.
Answer:
(i) Counting in eights, the numbers are 2128 and 2136, since 2128 = 8 × 266 and 2136 = 8 × 267.
(ii) The numbers are 3128 and 3136, since 3128 = 8 × 391 and 3136 = 8 × 392.
(iii) In both ranges the answers end in the same three digits, 128 and 136; the thousands digit makes no difference. This is because 1000 = 8 × 125 is a multiple of 8, so any number of whole thousands is already divisible by 8.
Test: a number is divisible by 8 if the number formed by its last three digits is divisible by 8.
On 9416: the last three digits form 416 = 8 × 52, so 9416 is divisible by 8. Indeed 9416 = 8 × 1177.
In simple words: Only the last three digits decide divisibility by 8, because 1000 is a multiple of 8.

Teacher's Note:
1. ‘Between’ leaves out the end numbers 2120 and 3120.
2. Explain the test using 1000 = 8 × 125.
3. Apply the test and confirm by division.

Teacher's Note:
1. Start with a row or column that has only one possible split.
2. Always check all four products.
3. Uniqueness of prime factorisation removes guesswork.

Question 10. Answer the following, using prime factorisation wherever it helps.
(i) Find the smallest number that is a multiple of each of 2, 3, 4, 5 and 6, and explain how you found it.
(ii) Find the smallest number that is a multiple of 8, 9 and 10.
(iii) Is the number found in part (ii) divisible by 24? Justify your answer using prime factorisation, not long division.
Answer:
(i) List the multiples of the largest number, 6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60. The first one also divisible by 4 and by 5 is 60.
Check: 60 = 2 × 30 = 3 × 20 = 4 × 15 = 5 × 12 = 6 × 10. So the smallest such number is 60.
(ii) The number must contain the primes of 8 = 2 × 2 × 2, of 9 = 3 × 3 and of 10 = 2 × 5. Three 2s, two 3s and one 5 are enough for all three: 2 × 2 × 2 × 3 × 3 × 5 = 360.
Check: 360 = 8 × 45 = 9 × 40 = 10 × 36.
(iii) 360 = 2 × 2 × 2 × 3 × 3 × 5 and 24 = 2 × 2 × 2 × 3. Every prime of 24 (three 2s and one 3) is present in 360, so 360 is divisible by 24.
Grouping the factors shows it: 360 = (2 × 2 × 2 × 3) × (3 × 5) = 24 × 15.
In simple words: Collect just enough of each prime to cover every number; 360 contains all the primes of 24.

Teacher's Note:
1. Take each prime the greatest number of times it appears in any one number.
2. Check the answer by dividing by each number.
3. Grouping factors gives the quotient directly.

Case Study Based Questions

Question 11. Chairs for the Annual Day
The students of Class 6 are setting up the school hall for Annual Day. They have 48 chairs and want to place all of them in equal rows, so that every row has the same number of chairs and no chair is left over. Meera suggests 6 rows of 8 chairs. Ravi says 7 rows would look neater, but when the class tries it, some chairs are left over. Their teacher tells them that a seating plan works only when the number of rows is a factor of 48. The class then decides to list every possible plan before choosing one.
(i) Write all the factors of 48.
(ii) Explain why rows of 7 chairs cannot be made.
(iii) If the chairs are placed in 8 equal rows, how many chairs are there in each row?
(iv) How many different equal-row plans are possible in all? Is 48 a prime or a composite number?
Answer:
(i) The factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48.
(ii) 7 is not a factor of 48: 48 = 7 × 6 + 6, so 6 chairs would be left over.
(iii) 48 ÷ 8 = 6 chairs in each row.
(iv) Each factor of 48 gives one plan, and 48 has ten factors, so ten plans are possible. Since 48 has more than two factors, it is composite.
In simple words: Equal rows work only for factors of 48; it has ten factors, so ten plans.

Teacher's Note:
1. Use factor pairs: 1 × 48, 2 × 24, 3 × 16, 4 × 12, 6 × 8.
2. A remainder means the plan fails.

Question 12. Two Bells
In Shanti’s village, the temple bell rings every 4 minutes and the school bell rings every 6 minutes. One morning both bells ring together exactly at 7:00 am. Shanti sits by the window with a notebook and writes down the minute at which each bell rings after 7:00 am. She soon sees that at some minutes only one bell rings, and at a few minutes the two bells ring together. Her brother says that the minutes when they ring together are the common multiples of 4 and 6.
(i) Write the first four multiples of 4 and the first four multiples of 6 after 7:00 am.
(ii) At which minute after 7:00 am do the two bells ring together for the first time?
(iii) Write all the minutes within one hour at which both bells ring together.
(iv) How many times in that one hour do the two bells ring together?
Answer:
(i) Temple bell (multiples of 4): 4th, 8th, 12th and 16th minute. School bell (multiples of 6): 6th, 12th, 18th and 24th minute.
(ii) The first common minute is 12, so they ring together again at 7:12 am.
(iii) Common multiples of 4 and 6 up to 60: 12, 24, 36, 48, 60, that is, at 7:12, 7:24, 7:36, 7:48 and 8:00 am.
(iv) They ring together 5 times in that hour (not counting 7:00 am itself).
In simple words: The bells meet every 12 minutes, so 5 times in the hour after 7:00 am.

Teacher's Note:
1. All common multiples are multiples of the first one, 12.
2. Say clearly whether the start time is counted.

Question 13. The Idli-Vada Game with 4 and 6
A Class 6 group plays the idli-vada game with the numbers 4 and 6, counting from 1 up to 60. A player says ‘idli’ at a multiple of 4, ‘vada’ at a multiple of 6, and ‘idli-vada’ when the number is a multiple of both. Asha is surprised that the first ‘idli-vada’ comes well before 24, which is the product of 4 and 6. Her teacher asks the class to look carefully at the factors of 4 and 6 before explaining why.
(i) At which number is ‘idli-vada’ said for the first time?
(ii) Write all the numbers up to 60 at which ‘idli-vada’ is said.
(iii) At the number 30, what should the player say? Give a reason.
(iv) Why is the first common multiple of 4 and 6 smaller than their product?
Answer:
(i) The first common multiple of 4 and 6 is 12.
(ii) 12, 24, 36, 48 and 60, the common multiples of 4 and 6 up to 60.
(iii) ‘Vada’ only: 30 = 6 × 5 is a multiple of 6, but 30 = 4 × 7 + 2 is not a multiple of 4.
(iv) 4 and 6 are not co-prime; they share the factor 2. For such a pair the first common multiple comes before the product, so it is 12 and not 24.
In simple words: 4 and 6 share the factor 2, so their multiples meet at 12, before 24.

Teacher's Note:
1. Check each number against both 4 and 6.
2. Link the early meeting to the shared factor.

Question 14. The Gardener’s Rows
A gardener has 36 rose saplings and 24 marigold saplings. He wants to plant them in rows so that every row has the same number of saplings and each row has only one kind of sapling. He first tries 5 saplings in a row, but saplings are left over in both beds. His helper suggests that the number in each row must be a factor of 36 as well as a factor of 24.
(i) Write all the factors of 24.
(ii) Write all the common factors of 36 and 24.
(iii) What is the largest number of saplings he can put in one row?
(iv) With that largest number, how many rows will he get in all?
Answer:
(i) Factors of 24: 1, 2, 3, 4, 6, 8, 12 and 24.
(ii) Factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36, so the common factors are 1, 2, 3, 4, 6 and 12.
(iii) The largest common factor, 12 saplings per row.
(iv) Roses: 36 ÷ 12 = 3 rows; marigolds: 24 ÷ 12 = 2 rows; in all 3 + 2 = 5 rows.
In simple words: 12 is the biggest number that divides both 36 and 24, giving 5 rows.

Teacher's Note:
1. The row size must divide both counts.
2. Add the rows of both kinds at the end.

Question 15. Two Dripping Taps
In the school washroom, one tap drips every 9 seconds and another drips every 12 seconds. The drips of the two taps happen together at the moment the caretaker starts his stopwatch. The caretaker notices that most of the time the drips come one after another, but now and then the two taps drip at exactly the same instant. He wants to know how often that happens so that he can place a single bucket under both taps at that moment.
(i) Write the first four multiples of 9 and the first four multiples of 12.
(ii) After how many seconds do the two taps drip together for the first time?
(iii) In 3 minutes, how many times do the taps drip together?
(iv) Are 9 and 12 co-prime? Give a reason.
Answer:
(i) Multiples of 9: 9, 18, 27, 36. Multiples of 12: 12, 24, 36, 48.
(ii) The first number in both lists is 36, so after 36 seconds.
(iii) 3 minutes = 180 seconds and 180 ÷ 36 = 5, so they drip together 5 times (at 36, 72, 108, 144 and 180 seconds, not counting the start).
(iv) No. Factors of 9 are 1, 3, 9 and factors of 12 are 1, 2, 3, 4, 6, 12, so 3 is a common factor other than 1.
In simple words: The taps meet every 36 seconds, 5 times in 3 minutes; 9 and 12 share 3.

Teacher's Note:
1. Change minutes to seconds first.
2. Here 9 × 12 = 108 is not the first meeting.

Question 16. The Sieve Chart
Class 6 makes a big chart of the numbers from 1 to 60. Following the Sieve of Eratosthenes, the children first cross out 1. Then they circle 2 and cross out every other multiple of 2, circle 3 and cross out every other multiple of 3, and carry on in the same way with the next uncrossed number each time. At the end, the circled numbers are pinned up on the class board under the title “Building blocks of numbers”.
(i) Why is 1 crossed out at the very beginning?
(ii) After the multiples of 2, 3, 5 and 7 have been crossed out, what kind of numbers are still left uncrossed?
(iii) How many prime numbers are there from 1 to 60?
(iv) Write the only pair of twin primes that lies between 40 and 60.
Answer:
(i) 1 has only one factor, itself, so it is neither prime nor composite.
(ii) Only prime numbers are left, because every composite number up to 60 is a multiple of 2, 3, 5 or 7 and has been crossed out.
(iii) 17 primes: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53 and 59.
(iv) 41 and 43; both are prime and they differ by 2.
In simple words: The sieve leaves the 17 primes up to 60; 41 and 43 are the twin primes between 40 and 60.

Teacher's Note:
1. Every composite up to 60 has a prime factor 7 or less.
2. 59 and 61 are twins, but 61 is beyond 60.

Question 17. Tiling a Verandah
A verandah is a rectangle of length 120 cm and breadth 90 cm. A tile-layer wants to cover it completely with identical square tiles, without cutting a single tile. He explains to the family that the side of the tile must be a factor of 120 and also a factor of 90, otherwise the last row or the last column will not fit.
(i) Write all the common factors of 120 and 90.
(ii) Can square tiles of side 8 cm be used? Give a reason.
(iii) What is the largest square tile that can be used?
(iv) How many tiles of that largest size will be needed?
Answer:
(i) Factors of 120: 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120. Factors of 90: 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90.
Common factors: 1, 2, 3, 5, 6, 10, 15 and 30.
(ii) No. 8 divides 120 but not 90 (90 = 8 × 11 + 2), so the tiles would not fit along the breadth.
(iii) The largest common factor is 30, so the largest tile has side 30 cm.
(iv) Along the length 120 ÷ 30 = 4 tiles, along the breadth 90 ÷ 30 = 3 tiles, so 4 × 3 = 12 tiles.
In simple words: The tile side must divide both 120 and 90; the biggest is 30 cm, needing 12 tiles.

Teacher's Note:
1. The side must fit both length and breadth exactly.
2. Count tiles as (tiles along length) × (tiles along breadth).

Question 18. Markers on a Race Track
A 120 m straight track is being made ready for a school race. Distance markers are painted every 8 m from the starting line, and water points are placed every 10 m from the same starting line. The organiser wants to know the places where a marker and a water point fall at the very same spot, so that he can put a flag there.
(i) At which distance from the start do a marker and a water point first come together?
(ii) Write all the distances on the 120 m track where a flag will be needed.
(iii) Are 8 and 10 co-prime? Give a reason.
(iv) If the markers had been painted every 7 m instead, at what distance would a marker and a water point first come together?
Answer:
(i) Markers: 8, 16, 24, 32, 40 m; water points: 10, 20, 30, 40 m. They first meet at 40 m.
(ii) Common multiples of 8 and 10 up to 120: 40 m, 80 m and 120 m.
(iii) No. Both 8 and 10 have 2 as a factor, so they have a common factor other than 1.
(iv) 7 and 10 are co-prime, so their first common multiple is their product: 7 × 10 = 70 m.
In simple words: Markers and water points meet every 40 m; with 7 m markers they would first meet at 70 m.

Teacher's Note:
1. Common multiples give the meeting points.
2. For co-prime numbers, multiply to get the first meeting.

Question 19. Sharing the Mela Collection
A school mela collects Rs 3540 on one day. The organisers want to share the whole amount equally, with no money left over, among the different clubs of the school. Four clubs, five clubs, eight clubs and ten clubs all ask for a share, and the treasurer decides to settle the question with the divisibility tests instead of long division.
(i) Is 3540 divisible by 10? Give the test you used.
(ii) Is 3540 divisible by 4? Give the test you used.
(iii) Is 3540 divisible by 8? Show the working of the test.
(iv) Among 2, 4, 5, 8 and 10, write which numbers of clubs can share the money equally, and explain why the answer for 4 and the answer for 8 are different.
Answer:
(i) Yes. A number is divisible by 10 when its last digit is 0; 3540 ends in 0. (3540 = 10 × 354.)
(ii) Yes. For 4 only the last two digits matter; they form 40 = 4 × 10. (3540 = 4 × 885.)
(iii) No. For 8 the last three digits matter; they form 540 = 8 × 67 + 4, which is not a multiple of 8.
(iv) The money can be shared equally among 2, 4, 5 or 10 clubs, but not among 8 clubs. The test for 4 needs only the last two digits, 40, to be a multiple of 4, which they are; the test for 8 needs the last three digits, 540, to be a multiple of 8, which they are not.
In prime factors, 3540 = 2 × 2 × 3 × 5 × 59 has only two 2s, while 8 needs three.
In simple words: 3540 passes the tests for 2, 4, 5 and 10 but fails the test for 8.

Teacher's Note:
1. Divisible by 4 does not mean divisible by 8.
2. Show the remainder 4 for the test of 8.

Question 20. Two Numbered Lockers
In a school store there are two lockers, one numbered 126 and the other numbered 275. The storekeeper has a rule for her keys. A single key opens both lockers only if some number bigger than 1 divides both locker numbers. She also uses small labels: a label numbered 14 may be put on a locker only if the locker number is divisible by 14, and a label numbered 55 only if the locker number is divisible by 55. She checks everything by prime factorisation.
(i) Write the prime factorisation of 126 and of 275.
(ii) Will one key open both lockers? Name the relation between 126 and 275.
(iii) Can the label 14 be put on locker 126? Justify using prime factorisation.
(iv) Can the label 55 be put on locker 275? Justify using prime factorisation.
Answer:
(i) 126 = 2 × 63 = 2 × 3 × 3 × 7 and 275 = 5 × 55 = 5 × 5 × 11.
(ii) No. The prime factors of 126 are 2, 3, 7 and those of 275 are 5, 11. None is common, so the only common factor is 1. The numbers 126 and 275 are co-prime.
(iii) Yes. 14 = 2 × 7, and 126 contains one 2 and one 7. So 126 is divisible by 14: 126 = 14 × 9.
(iv) Yes. 55 = 5 × 11, and 275 contains two 5s and one 11, so it includes one 5 and one 11. So 275 is divisible by 55: 275 = 55 × 5.
In simple words: 126 and 275 share no prime, so they are co-prime; 14 divides 126 and 55 divides 275.

Teacher's Note:
1. Write factorisations neatly before answering the parts.
2. The left-over primes give the quotient.