CLASS 7 · GANITA PRAKASH PART II · CHAPTER 3

Finding Common Ground

Understand prime factors, HCF and LCM, and use them to solve sharing, grouping and repeating-pattern problems.

90 worked answersAll 1 exercise setsAll 13 end-of-chapter questions

What you will learn

Find prime factorisations and list factors systematically.
Calculate HCF and LCM using prime powers and explain what each means.
Apply HCF and LCM to everyday situations and number patterns.

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90 worked answers · short steps first, explanations when you need them

3.1 The Greatest of All

Textbook pages 47–48

Question floor-tiles

A 12 ft by 16 ft room is tiled with equal square tiles of whole-number side length. What is the largest possible tile, and how many tiles cover the floor?

Steps only

HCF(12,16)=4.

12÷4=3 tiles across; 16÷4=4 tiles along.

Answer

Use 4 ft square tiles; 3×4=12 tiles.

Question fractional-tile

Would allowing a fractional tile side let the room use fewer square tiles?

Steps only

A square tile must still fit both 12 ft and 16 ft dimensions exactly.

Any side larger than 4 ft cannot divide 12 and 16 into whole tiles; a fractional side is not required to improve on the 4 ft tile.

Answer

No. The 4 ft tile already gives the fewest tiles in the stated whole-number setting; fractional subdivisions cannot produce a larger common side that tiles both dimensions without cutting.

Question rice-bags

Rice quantities are 84 kg and 108 kg. Bags must have equal whole-kilogram capacity, each contain rice from only one farm, and be as few as possible. Find the bag capacity and total bags.

Steps only

HCF(84,108)=12.

84÷12=7 bags; 108÷12=9 bags.

Answer

Each bag should hold 12 kg; 16 bags are needed altogether.

Question jump-a

Find the longest whole-number jump from 0 that lands on both 14 and 30.

Steps only

Find HCF(14,30)=2.

Both targets are multiples of 2 and no larger common factor divides both.

Answer

The longest jump is 2.

Question jump-b

Find the longest whole-number jump from 0 that lands on both 7 and 11.

Steps only

Find HCF(7,11)=1.

Both targets are multiples of 1 and no larger common factor divides both.

Answer

The longest jump is 1.

Question jump-c

Find the longest whole-number jump from 0 that lands on both 30 and 50.

Steps only

Find HCF(30,50)=10.

Both targets are multiples of 10 and no larger common factor divides both.

Answer

The longest jump is 10.

Question jump-d

Find the longest whole-number jump from 0 that lands on both 28 and 42.

Steps only

Find HCF(28,42)=14.

Both targets are multiples of 14 and no larger common factor divides both.

Answer

The longest jump is 14.

Prime factors and factors

Textbook pages 49–53

Question factor-105

Find the prime factorisation of 105.

Steps only

Divide by prime numbers until only primes remain.

Rebuild the product: 105=3×5×7.

Answer

The prime factorisation is 3×5×7.

Question factor-30

Find the prime factorisation of 30.

Steps only

Divide by prime numbers until only primes remain.

Rebuild the product: 30=2×3×5.

Answer

The prime factorisation is 2×3×5.

Question factor-1200

Find the prime factorisation of 1200.

Steps only

Divide by prime numbers until only primes remain.

Rebuild the product: 1200=2⁴×3×5².

Answer

The prime factorisation is 2⁴×3×5².

Question subparts-840

Using 840=2³×3×5×7, decide whether 28, 14, 8 and 27 are factors of 840.

Steps only

28=2²×7 uses available factors: yes.

14=2×7: yes. 8=2³: yes.

27=3³ needs three 3s, but 840 has one: no.

Answer

28, 14 and 8 are factors; 27 is not.

Question factors-225

Find all factors of 225.

Steps only

225=3²×5².

Choose 0, 1 or 2 copies of 3 and independently 0, 1 or 2 copies of 5.

Sort the products: 1,3,5,9,15,25,45,75,225.

Answer

The factors are 1, 3, 5, 9, 15, 25, 45, 75 and 225.

Question all-factors-90

List every factor of 90.

Steps only

Prime factorisation: 90=2×3²×5.

Form products using each prime no more often than it occurs; include 1 and 90.

Answer

1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90.

Question all-factors-105

List every factor of 105.

Steps only

Prime factorisation: 105=3×5×7.

Form products using each prime no more often than it occurs; include 1 and 105.

Answer

1, 3, 5, 7, 15, 21, 35, 105.

Question all-factors-132

List every factor of 132.

Steps only

Prime factorisation: 132=2²×3×11.

Form products using each prime no more often than it occurs; include 1 and 132.

Answer

1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66, 132.

Question all-factors-360

List every factor of 360.

Steps only

Prime factorisation: 360=2³×3²×5.

Form products using each prime no more often than it occurs; include 1 and 360.

Answer

1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60, 72, 90, 120, 180, 360.

Question all-factors-840

List every factor of 840.

Steps only

Prime factorisation: 840=2³×3×5×7.

Form products using each prime no more often than it occurs; include 1 and 840.

Answer

1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 14, 15, 20, 21, 24, 28, 30, 35, 40, 42, 56, 60, 70, 84, 105, 120, 140, 168, 210, 280, 420, 840.

Question conjecture-counterexample

Anshu claims that a larger number always has a longer prime factorisation. Is the claim true?

Steps only

Compare 96=2×2×2×2×2×3 (six prime factors) with 121=11×11 (two).

121 is larger than 96 but its prime factorisation is shorter.

Answer

No. 96 and 121 are a counterexample.

Find the HCF

Textbook pages 52–54

Question common-45-75

Find the common factors and HCF of 45 and 75.

Steps only

Prime factorise both numbers and keep the shared primes with the smaller exponent.

The common factors are 1, 3, 5, 15.

Answer

HCF(45,75)=15.

Question common-112-84

Find the common factors and HCF of 112 and 84.

Steps only

Prime factorise both numbers and keep the shared primes with the smaller exponent.

The common factors are 1, 2, 4, 7, 14, 28.

Answer

HCF(112,84)=28.

Question common-96-275

Find the common factors and HCF of 96 and 275.

Steps only

Prime factorise both numbers and keep the shared primes with the smaller exponent.

The common factors are 1.

Answer

HCF(96,275)=1.

Question common-50-60

Find the common factors and HCF of 50 and 60.

Steps only

Prime factorise both numbers and keep the shared primes with the smaller exponent.

The common factors are 1, 2, 5, 10.

Answer

HCF(50,60)=10.

Question common-140-275

Find the common factors and HCF of 140 and 275.

Steps only

Prime factorise both numbers and keep the shared primes with the smaller exponent.

The common factors are 1, 5.

Answer

HCF(140,275)=5.

Question common-77-725

Find the common factors and HCF of 77 and 725.

Steps only

Prime factorise both numbers and keep the shared primes with the smaller exponent.

The common factors are 1.

Answer

HCF(77,725)=1.

Question common-370-592

Find the common factors and HCF of 370 and 592.

Steps only

Prime factorise both numbers and keep the shared primes with the smaller exponent.

The common factors are 1, 2, 37, 74.

Answer

HCF(370,592)=74.

Question common-81-243

Find the common factors and HCF of 81 and 243.

Steps only

Prime factorise both numbers and keep the shared primes with the smaller exponent.

The common factors are 1, 3, 9, 27, 81.

Answer

HCF(81,243)=81.

Question hcf-24-180

Find the HCF of 24 and 180.

Steps only

Prime factorise: 24 and 180.

Keep each common prime to its smaller exponent.

Answer

HCF(24,180)=12.

Question hcf-42-75

Find the HCF of 42 and 75.

Steps only

Prime factorise: 42 and 75.

Keep each common prime to its smaller exponent.

Answer

HCF(42,75)=3.

Question hcf-240-378

Find the HCF of 240 and 378.

Steps only

Prime factorise: 240 and 378.

Keep each common prime to its smaller exponent.

Answer

HCF(240,378)=6.

Question hcf-400-2500

Find the HCF of 400 and 2500.

Steps only

Prime factorise: 400 and 2500.

Keep each common prime to its smaller exponent.

Answer

HCF(400,2500)=100.

Question hcf-300-800

Find the HCF of 300 and 800.

Steps only

Prime factorise: 300 and 800.

Keep each common prime to its smaller exponent.

Answer

HCF(300,800)=100.

Question hcf-42-75-24

Find the HCF of 42, 75 and 24.

Steps only

42=2×3×7; 75=3×5²; 24=2³×3.

The only prime in all three factorizations is one copy of 3.

Answer

HCF(42,75,24)=3.

Question composite-factorisation

Given 72=6×12 and 144=8×18, does this show that 72 and 144 have no common factor except 1?

Steps only

No: a composite-number factorisation is not a prime factorisation.

144=2×72, so 72 is a common factor (and HCF(72,144)=72).

Answer

No. The claim is false; 72 itself divides both numbers.

3.2 Least, but not Last!

Textbook pages 55–57

Question toran

One person joins 6 cm strips and another joins 8 cm strips. What is the shortest length both can make?

Steps only

Find LCM(6,8).

Multiples first meet at 24: 6×4=24 and 8×3=24.

Answer

24 cm.

Question gajak

A sweet shop gives a treat every 7 days. Kabamai visits every 10 days. If both schedules align today, after how many days do they align again?

Steps only

Find LCM(7,10)=7×10 because 7 and 10 are coprime.

Answer

They align again after 70 days.

Question lcm-game-4-6

Find the first number called out for both 4 and 6 in the Idli-Vada game.

Steps only

Find the least shared multiple: LCM(4,6)=12.

Answer

The first shared call is at 12.

Question lcm-game-7-11

Find the first number called out for both 7 and 11 in the Idli-Vada game.

Steps only

Find the least shared multiple: LCM(7,11)=77.

Answer

The first shared call is at 77.

Question lcm-game-14-30

Find the first number called out for both 14 and 30 in the Idli-Vada game.

Steps only

Find the least shared multiple: LCM(14,30)=210.

Answer

The first shared call is at 210.

Question lcm-game-15-55

Find the first number called out for both 15 and 55 in the Idli-Vada game.

Steps only

Find the least shared multiple: LCM(15,55)=165.

Answer

The first shared call is at 165.

Question lcm-14-35

Find LCM(14,35). Is 2×3×5×7 also a common multiple?

Steps only

14=2×7 and 35=5×7.

The smallest product containing both sets is 2×5×7=70.

2×3×5×7=210 is divisible by both, but is larger than 70.

Answer

LCM(14,35)=70; 210 is a common multiple but not the least.

Question lcm-96-360

Find LCM(96,360).

Steps only

96=2⁵×3; 360=2³×3²×5.

Take maximum exponents: 2⁵×3²×5=32×9×5.

Answer

LCM(96,360)=1440.

Question lcm-30-72

Find the LCM of 30 and 72.

Steps only

Take the largest exponent of every prime appearing in either factorisation.

Answer

LCM(30,72)=360.

Question lcm-36-54

Find the LCM of 36 and 54.

Steps only

Take the largest exponent of every prime appearing in either factorisation.

Answer

LCM(36,54)=108.

Question lcm-222-370

Find the LCM of 222 and 370.

Steps only

Take the largest exponent of every prime appearing in either factorisation.

Answer

LCM(222,370)=1110.

Question lcm-105-195-65

Find the LCM of 105, 195 and 65.

Steps only

105=3×5×7; 195=3×5×13; 65=5×13.

Keep the greatest power of each prime appearing: 3×5×7×13=1,365.

Answer

LCM(105,195,65)=1,365.

3.3 Patterns, Properties, and a Pretty Procedure!

Textbook pages 58–62

Question factor-multiple

Describe pairs of positive whole numbers where the HCF equals one of the numbers.

Steps only

If a divides b, every factor of a is also a factor of b.

Thus the greatest common factor is a.

Answer

Exactly when one number divides the other (including equal numbers).

Question algebra-multiple

If one number is m and their HCF is m, what can the other number be?

Steps only

The HCF equals m exactly when m divides the other number.

Answer

The other number can be km for any positive whole number k.

Question algebra-7k

If one number is 7k, give a form for another number so their HCF is 7k.

Steps only

Choose the other number to be a whole-number multiple of 7k.

Answer

For example, 7kr for a positive integer r.

Question hcf-consecutive-even

What is the HCF of two consecutive even numbers?

Steps only

Write them as 2n and 2n+2.

Their HCF is 2×HCF(n,n+1), and consecutive integers have HCF 1.

Answer

Their HCF is 2.

Question hcf-consecutive-odd

What can the HCF of two consecutive odd numbers be?

Steps only

Their difference is 2.

They are both odd, so their common factor cannot be 2.

Answer

Their HCF is 1.

Question hcf-two-even

What can be said about the HCF of two even numbers?

Steps only

Both are divisible by 2.

Answer

Their HCF is at least 2; its exact value depends on the numbers.

Question hcf-consecutive

What is the HCF of two consecutive whole numbers?

Steps only

Any common factor divides their difference.

The difference is 1.

Answer

Their HCF is 1.

Question hcf-coprime

What is the HCF of two co-prime numbers?

Steps only

Co-prime means their only common positive factor is 1.

Answer

Their HCF is 1.

Question lcm-factor-multiple

Describe pairs whose LCM equals one of the numbers.

Steps only

If a divides b, b is a common multiple of both.

Every common multiple must be a multiple of b, so b is the least one.

Answer

When one number is a multiple of the other; the LCM is the larger number.

Question lcm-multiples-3

What can be said about the LCM of two multiples of 3?

Steps only

Both numbers are divisible by 3, so their LCM is also divisible by 3.

Answer

It is a multiple of 3; its exact value depends on the pair.

Question lcm-consecutive-even

Find a general expression for the LCM of 2n and 2n+2.

Steps only

Their HCF is 2.

For two positive integers, HCF×LCM=product.

LCM=(2n)(2n+2)÷2.

Answer

The LCM is 2n(n+1).

Question lcm-consecutive

What is the LCM of consecutive positive integers n and n+1?

Steps only

Their HCF is 1.

LCM=product÷HCF.

Answer

LCM=n(n+1).

Question lcm-coprime

Find the LCM of co-prime numbers a and b.

Steps only

HCF(a,b)=1.

HCF×LCM=a×b.

Answer

LCM(a,b)=ab.

Question shared-multiplier-180-270

Find the HCF of 180 and 270.

Steps only

Factor out the common multiplier.

HCF(180,270) is 90.

Answer

The HCF is 90.

Question shared-multiplier-380-210

Find the HCF of 380 and 210.

Steps only

Factor out the common multiplier.

HCF(380,210) is 70.

Answer

The HCF is 70.

Question shared-multiplier-192-240

Find the HCF of 192 and 240.

Steps only

Factor out the common multiplier.

HCF(192,240) is 48.

Answer

The HCF is 48.

Question shared-multiplier-65-100

Find the HCF of 65 and 100.

Steps only

Factor out the common multiplier.

HCF(65,100) is 25.

Answer

The HCF is 25.

Question procedure-84-180

Find HCF and LCM of 84 and 180.

Steps only

Use prime factorisations or divide both by common factors.

HCF=12; verify HCF×LCM=15120.

Answer

HCF=12 and LCM=1260.

Question procedure-300-150

Find HCF and LCM of 300 and 150.

Steps only

Use prime factorisations or divide both by common factors.

HCF=150; verify HCF×LCM=45000.

Answer

HCF=150 and LCM=300.

Question procedure-630-770

Find HCF and LCM of 630 and 770.

Steps only

Use prime factorisations or divide both by common factors.

HCF=70; verify HCF×LCM=485100.

Answer

HCF=70 and LCM=6930.

Question procedure-90-150

Find HCF and LCM of 90 and 150.

Steps only

Use prime factorisations or divide both by common factors.

HCF=30; verify HCF×LCM=13500.

Answer

HCF=30 and LCM=450.

Question procedure-84-132

Find HCF and LCM of 84 and 132.

Steps only

Use prime factorisations or divide both by common factors.

HCF=12; verify HCF×LCM=11088.

Answer

HCF=12 and LCM=924.

Question lcm-not-greater-product

Explain why the LCM of two positive whole numbers cannot be greater than their product.

Steps only

The product a×b is divisible by a and by b.

So it is a common multiple.

The least common multiple cannot exceed any common multiple.

Answer

LCM(a,b)≤a×b.

Question product-lcm-105-95

For 105 and 95, find their HCF and LCM and check the product relation.

Steps only

HCF=5 and LCM=1995.

5×1995=9975; 105×95=9975.

Answer

HCF×LCM=the product 9975.

Question product-lcm-45-105

For 45 and 105, find their HCF and LCM and check the product relation.

Steps only

HCF=15 and LCM=315.

15×315=4725; 45×105=4725.

Answer

HCF×LCM=the product 4725.

Question product-lcm-275-352

For 275 and 352, find their HCF and LCM and check the product relation.

Steps only

HCF=11 and LCM=8800.

11×8800=96800; 275×352=96800.

Answer

HCF×LCM=the product 96800.

Question product-lcm-222-370

For 222 and 370, find their HCF and LCM and check the product relation.

Steps only

HCF=74 and LCM=1110.

74×1110=82140; 222×370=82140.

Answer

HCF×LCM=the product 82140.

Figure it Out — end-of-chapter questions

Textbook pages 63–64

Question repeat-stars

Two rows show repeating colour patterns. Find the first position at which the blue stars meet again.

Steps only

In the displayed rows, blue repeats every 6 stars in the first row and every 5 in the second.

The first row blue star is at position 4; the second row blue star is at position 4.

Next shared position is 4+LCM(6,5)=34.

Answer

The blue stars next meet at position 34.

Question compare-prime-products-a

Is 5×7×11×11 a multiple of 5×7×7×11×2?

Steps only

The first product has no factor 2 and only one 7.

The second product requires both 2 and two 7s.

Answer

No. The first product is not divisible by the second.

Question compare-prime-products-b

Is 5×7×11×11 a factor of 5×7×7×11×2?

Steps only

Cancel the shared 5, 7 and 11.

The remaining comparison would require 11 to divide 14, which it does not.

Answer

No. The first product is not a factor of the second.

Question hcf-lcm-factorised-a

Find HCF and LCM of 3×3×5×7×7 and 12×7×11.

Steps only

First number=3²×5×7²; second=2²×3×7×11.

HCF uses 3×7; LCM uses 2²×3²×5×7²×11.

Answer

HCF=3×7=21; LCM=2²×3²×5×7²×11.

Question hcf-lcm-45-36

Find HCF and LCM of 45 and 36.

Steps only

45=3²×5; 36=2²×3².

HCF=3²=9; LCM=2²×3²×5=180.

Answer

HCF=9; LCM=180.

Question coprime-lcm-66

Give a pair of numbers with HCF 1 and LCM 66.

Steps only

66=2×3×11.

Choose factors with no shared prime, for example 6=2×3 and 11.

Answer

One answer is 6 and 11.

Question 5

A cowherd has fewer than 200 cows. They can be divided equally into groups of 3, 5 or 7, and one cow remains when divided into groups of 2. How many cows are there?

Steps only

LCM(3,5,7)=105, so the positive possibilities below 200 are 105 only.

105÷2 leaves remainder 1.

Answer

The cowherd has 105 cows.

Question prime-factorisation-table

For 90, 240, 441 and 1,001, write the prime factorisation and list all positive factors.

Steps only

90=2×3²×5: 12 factors.

240=2⁴×3×5: 20 factors.

441=3²×7²: 9 factors.

1,001=7×11×13: 8 factors.

Answer

90: 1,2,3,5,6,9,10,15,18,30,45,90. 240: 1,2,3,4,5,6,8,10,12,15,16,20,24,30,40,48,60,80,120,240. 441: 1,3,7,9,21,49,63,147,441. 1,001: 1,7,11,13,77,91,143,1,001.

Question largest-cube-side

A 12 cm by 18 cm by 36 cm box is filled with equal cubes without gaps. Find the largest possible cube side and number of cubes.

Steps only

HCF(12,18,36)=6 cm.

12÷6=2, 18÷6=3, 36÷6=6.

2×3×6=36 cubes.

Answer

Use 6 cm cubes; 36 cubes fill the box.

Question largest-divisor

Which is the greatest common factor of 306 and 36 among 36, 612, 18, 3, 2 and 360?

Steps only

306=18×17; 36=18×2.

No common factor can exceed 36, and the common prime factors give 18.

Answer

18.

Question smallest-multiple

Find the smallest number divisible by 1, 2, 3, 4, 5, 6, 8, 9 and 10.

Steps only

The largest required prime powers are 2³, 3² and 5.

8×9×5=360.

Answer

360.

Question children-game

A group divides equally into teams of 6 or 9; when divided into teams of 10, some children remain. Which listed totals 72, 90, 45, 3, 36 or none are possible?

Steps only

The total must be a multiple of LCM(6,9)=18.

From the list, 72 and 90 are multiples of 18.

Neither 72 nor 90 is divisible by 10.

Answer

72 and 90 are possible.

Question equal-to-lcm

Can the LCM of two distinct prime numbers be less than either prime, between them, greater than both, less than their product, or greater than their product?

Steps only

Distinct primes have no common factor other than 1.

Their LCM is their product.

Answer

It is greater than both primes and equal to their product.

Question dog-rabbit

A dog is 150 ft behind a rabbit. In each leap the dog covers 9 ft and the rabbit 7 ft. How many leaps until the dog catches the rabbit?

Steps only

The gap closes by 9−7=2 ft each leap.

150÷2=75.

Answer

75 leaps.

Question smallest-multiple-remainder-11

Find the least positive number divisible by 3, 4, 5 and 7 that leaves remainder 10 when divided by 11.

Steps only

The number must be 420k, since LCM(3,4,5,7)=420.

420 leaves remainder 2 modulo 11, so 420k leaves remainder 2k.

2k≡10 (mod 11); the least positive k is 5.

Answer

The least number is 2,100.

Question three-number-hcf-lcm

Test whether HCF×LCM equals the product for 6, 10 and 15.

Steps only

HCF(6,10,15)=1.

LCM(6,10,15)=30.

HCF×LCM=30, while 6×10×15=900.

Answer

The two-number relation does not extend as HCF×LCM=product to three numbers.

Question 13

Add 8/15, 1/36, 7/63, 11/63 and 1/21. How can the sum be found efficiently?

Steps only

LCM(15,36,63,21)=1260.

Rewrite as 672/1260+35/1260+140/1260+220/1260+60/1260.

Add the numerators: 1127.

Answer

The sum is 1127/1260.

Mystery Colours — prime-factor pattern

Textbook pages 65

Question decode-number-designs

The chapter shows a circle design for every number from 1 to 100. Explain the rule and give the prime factors represented for 12, 30 and 49.

Steps only

Each colour marks one prime divisor: 2, 3, 5 or 7.

12=2²×3, so its design must show two 2-marks and one 3-mark.

30=2×3×5; 49=7².

Answer

12 has prime factors 2, 2 and 3; 30 has 2, 3 and 5; 49 has 7 and 7.