Question 1

Explore whether the LCM is a factor of the product in the following cases. If yes, identify the number that the LCM should be multiplied by to get the product. Do you see any pattern? Use these numbers:

(a) 45, 105

(b) 275, 352

(c) 222, 370

Accepted Answer

To explore whether the LCM is a factor of the product, we calculate the product of the two numbers and their LCM for each pair:

(a) 45 and 105
- Product = 45 × 105 = 4725
- Find HCF and LCM:
  * Prime factors of 45 = 3² × 5
  * Prime factors of 105 = 3 × 5 × 7
  * HCF = 3 × 5 = 15
  * LCM = 3² × 5 × 7 = 315
- Check if LCM divides product:
  * 4725 ÷ 315 = 15 (an integer)
- So, LCM × 15 = Product

(b) 275 and 352
- Product = 275 × 352 = 96800
- Prime factors:
  * 275 = 5² × 11
  * 352 = 2⁵ × 11
  * HCF = 11
  * LCM = 2⁵ × 5² × 11 = 8800
- Check:
  * 96800 ÷ 8800 = 11
- So, LCM × 11 = Product

(c) 222 and 370
- Product = 222 × 370 = 82140
- Prime factors:
  * 222 = 2 × 3 × 37
  * 370 = 2 × 5 × 37
  * HCF = 2 × 37 = 74
  * LCM = 2 × 3 × 5 × 37 = 1110
- Check:
  * 82140 ÷ 1110 = 74
- So, LCM × 74 = Product

In all cases, the number by which the LCM is multiplied to get the product is the HCF. — The product of two numbers equals the product of their HCF and LCM. This is because the prime factors common to both numbers (HCF) and the prime factors covering all primes in both numbers (LCM) multiply to give the product of the numbers. The calculations above demonstrate this property with examples.

Question 2

Do you see that, in each case, the number by which the LCM is multiplied to get the product is actually the HCF? Thus, our observations seem to suggest the following:

\mathrm{HCF} 	imes \mathrm{LCM} = 	ext{Product of the two numbers.}

Why does this happen? Can you give an explanation or proof?

Accepted Answer

This happens because the HCF contains the prime factors common to both numbers with the minimum powers, and the LCM contains all prime factors present in either number with the maximum powers. When multiplied, the common prime factors (HCF) and the remaining prime factors (LCM) together reconstruct the product of the two numbers.

Proof:
Let the prime factorization of two numbers be:

Number 1 = p1^a × p2^b × ...
Number 2 = p1^c × p2^d × ...

where p1, p2,... are primes and a,b,c,d,... are their powers.

Then,
HCF = p1^min(a,c) × p2^min(b,d) × ...
LCM = p1^max(a,c) × p2^max(b,d) × ...

Multiplying HCF and LCM:
= p1^{min(a,c)+max(a,c)} × p2^{min(b,d)+max(b,d)} × ...
= p1^{a+c} × p2^{b+d} × ...
= Number 1 × Number 2

Hence, HCF × LCM = Product of the two numbers. — The explanation uses prime factorization to show that the sum of minimum and maximum powers of each prime factor equals the sum of their powers in the two numbers, thus reconstructing the product.

Question 3

Explore whether this property holds when 3 numbers are considered.

Accepted Answer

For three numbers a, b, and c, the property HCF × LCM = product does not hold directly as it does for two numbers. Instead, the relationship is more complex.

For example, consider numbers 4, 6, and 8:
- Product = 4 × 6 × 8 = 192
- HCF(4,6,8) = 2
- LCM(4,6,8) = 24

Check HCF × LCM = 2 × 24 = 48, which is not equal to 192.

Thus, the simple product relation holds only for two numbers, not for three or more. However, there are formulas involving pairwise HCFs and LCMs for three numbers, but they are more complicated. — The property HCF × LCM = product is valid only for two numbers. For three numbers, the product of HCF and LCM is generally not equal to the product of the three numbers.

Question 4

In the two rows below, colours repeat as shown. When will the blue stars meet next?

Accepted Answer

To find when the blue stars meet next, we need to find the least common multiple (LCM) of the intervals at which the blue stars appear in each row.

Assuming the blue stars appear every m positions in the first row and every n positions in the second row, the blue stars will meet after LCM(m, n) positions.

Without exact intervals given in the figure, the general approach is:
- Identify the repeating intervals for blue stars in both rows.
- Calculate LCM of these intervals.
- The blue stars will meet after that many positions.

Hence, the answer depends on the intervals shown in the figure. — The problem involves finding the LCM of the repeating intervals of blue stars in two rows to determine when they coincide again.

Question 5

(a) Is 5 × 7 × 11 × 11 a multiple of 5 × 7 × 7 × 11 × 2?
(b) Is 5 × 7 × 11 × 11 a factor of 5 × 7 × 7 × 11 × 2?

Accepted Answer

(a) Check if 5 × 7 × 11 × 11 is a multiple of 5 × 7 × 7 × 11 × 2.

Calculate both:
- Left: 5 × 7 × 11 × 11 = 5 × 7 × 121 = 5 × 847 = 4235
- Right: 5 × 7 × 7 × 11 × 2 = 5 × 49 × 11 × 2 = 5 × 49 × 22 = 5 × 1078 = 5390

Is 4235 a multiple of 5390? No, because 4235 < 5390.

So, (a) No.

(b) Is 5 × 7 × 11 × 11 a factor of 5 × 7 × 7 × 11 × 2?

Check if 5390 is divisible by 4235:
- 5390 ÷ 4235 ≈ 1.27 (not an integer)

So, (b) No.

Therefore, neither is the first number a multiple nor a factor of the second. — By calculating the numerical values and checking divisibility, we conclude neither multiple nor factor relation holds.

Question 6

Find the HCF and LCM of the following (state your answers in the form of prime factorisations):

(a) 3 × 3 × 5 × 7 × 7 and 12 × 7 × 11

(b) 45 and 36

Accepted Answer

(a) Numbers:
- First number: 3 × 3 × 5 × 7 × 7 = 3² × 5 × 7²
- Second number: 12 × 7 × 11 = (2² × 3) × 7 × 11 = 2² × 3 × 7 × 11

HCF:
- Common primes with minimum powers:
  * 3: min(2,1) = 1
  * 5: min(1,0) = 0
  * 7: min(2,1) = 1
  * 2: min(0,2) = 0
  * 11: min(0,1) = 0
- So, HCF = 3¹ × 7¹ = 3 × 7

LCM:
- Primes with maximum powers:
  * 2: max(0,2) = 2
  * 3: max(2,1) = 2
  * 5: max(1,0) = 1
  * 7: max(2,1) = 2
  * 11: max(0,1) = 1
- So, LCM = 2² × 3² × 5 × 7² × 11

(b) Numbers: 45 and 36
- 45 = 3² × 5
- 36 = 2² × 3²

HCF:
- Common primes with minimum powers:
  * 2: min(0,2) = 0
  * 3: min(2,2) = 2
  * 5: min(1,0) = 0
- HCF = 3² = 9

LCM:
- Primes with maximum powers:
  * 2: max(0,2) = 2
  * 3: max(2,2) = 2
  * 5: max(1,0) = 1
- LCM = 2² × 3² × 5 = 4 × 9 × 5 = 180 — HCF is found by taking the minimum powers of common prime factors; LCM is found by taking the maximum powers of all prime factors present.

Question 7

Find two numbers whose HCF is 1 and LCM is 66.

Accepted Answer

Since HCF is 1, the two numbers are co-prime (no common prime factors).

LCM = 66 = 2 × 3 × 11

Two numbers whose product is 66 and HCF is 1 could be:
- 6 and 11 (6 = 2 × 3, 11 = 11)
- 2 and 33 (33 = 3 × 11)

Check HCF:
- HCF(6,11) = 1
- HCF(2,33) = 1

Thus, two such numbers are 6 and 11 or 2 and 33. — Two numbers with HCF 1 and LCM 66 must multiply to 66 and share no common prime factors.

Question 8

A cowherd took all his cows to graze in the fields. The cows came to a crossing with 3 gates. An equal number of cows passed through each gate. Later at another crossing with 5 gates again an equal number of cows passed through each gate. The same happened at the third crossing with 7 gates. If the cowherd had less than 200 cows, how many cows did he have? (Based on the folklore mathematics from Karnataka.)

Accepted Answer

The number of cows is divisible by 3, 5, and 7 (since equal number passed through each gate).

Find LCM of 3, 5, and 7:
- LCM = 3 × 5 × 7 = 105

The number of cows is a multiple of 105 and less than 200.

Multiples of 105 less than 200 are 105 only.

Therefore, the cowherd had 105 cows. — The number must be divisible by 3, 5, and 7 and less than 200, so the smallest such number is 105.

Question 9

The length, width, and height of a box are 12 cm, 18 cm, and 36 cm respectively. Which of the following sized cubes can be packed in this box without leaving gaps?

(a) 9 cm

(b) 6 cm

(c) 4 cm

(d) 3 cm

(e) 2 cm

Accepted Answer

To pack cubes without gaps, the cube's side length must exactly divide the length, width, and height of the box.

Check each option:

(a) 9 cm:
- 12 ÷ 9 = 1.33 (not integer)
- Not suitable.

(b) 6 cm:
- 12 ÷ 6 = 2
- 18 ÷ 6 = 3
- 36 ÷ 6 = 6
- All integers, suitable.

(c) 4 cm:
- 12 ÷ 4 = 3
- 18 ÷ 4 = 4.5 (not integer)
- Not suitable.

(d) 3 cm:
- 12 ÷ 3 = 4
- 18 ÷ 3 = 6
- 36 ÷ 3 = 12
- All integers, suitable.

(e) 2 cm:
- 12 ÷ 2 = 6
- 18 ÷ 2 = 9
- 36 ÷ 2 = 18
- All integers, suitable.

Therefore, cubes of size 6 cm, 3 cm, and 2 cm can be packed without gaps. — Cube side length must divide all dimensions of the box exactly to avoid gaps.

Question 10

Among the numbers below, which is the largest number that perfectly divides both 306 and 36?

(a) 36

(b) 612

(c) 18

(d) 3

(e) 2

(f) 360

Accepted Answer

We need to find the highest common factor (HCF) of 306 and 36.

Prime factorization:
- 306 = 2 × 3 × 3 × 17 = 2 × 3² × 17
- 36 = 2² × 3²

HCF is product of common primes with minimum powers:
- 2: min(1,2) = 1
- 3: min(2,2) = 2
- 17: min(1,0) = 0

HCF = 2¹ × 3² = 2 × 9 = 18

Among the options, 18 is present (option c).

Hence, the largest number that divides both is 18. — By prime factorization, HCF is 18, which is option (c).

Question 11

Find the smallest number that is divisible by 3, 4, 5 and 7, but leaves a remainder of 10 when divided by 11.

Accepted Answer

Step 1: Find LCM of 3, 4, 5, and 7.
- 3 = 3
- 4 = 2²
- 5 = 5
- 7 = 7

LCM = 2² × 3 × 5 × 7 = 4 × 3 × 5 × 7 = 420

Step 2: Find the smallest number N such that:
- N is divisible by 420
- When divided by 11, remainder is 10

Let N = 420k

We want:
420k ≡ 10 (mod 11)

Calculate 420 mod 11:
- 11 × 38 = 418
- 420 - 418 = 2

So, 420k mod 11 = 2k mod 11 ≡ 10

Find k such that 2k ≡ 10 (mod 11)

Try k = 5:
2 × 5 = 10 ≡ 10 (mod 11)

So, k = 5

Therefore, N = 420 × 5 = 2100

Answer: 2100 — The number must be a multiple of LCM(3,4,5,7) and satisfy the modular condition with 11. Using modular arithmetic, k=5 satisfies the condition.

Question 12

Children are playing 'Fire in the Mountain'. When the number 6 was called out, no one got out. When the number 9 was called out, no one got out. But when the number 10 was called out, some people got out. How many children could have been playing initially?

(a) 72

(b) 90

(c) 45

(d) 3

(e) 36

(f) None of these

Accepted Answer

The game 'Fire in the Mountain' involves children getting out if their number divides the called number.

Given:
- No one got out at 6 → No child's number divides 6
- No one got out at 9 → No child's number divides 9
- Some got out at 10 → At least one child's number divides 10

So, the number of children is such that:
- It is not divisible by 2 or 3 (since 6 and 9 are multiples of 2 and 3)
- It divides 10 (which is 2 × 5)

Check options:
(a) 72: divisible by 2 and 3 → No
(b) 90: divisible by 2 and 3 → No
(c) 45: divisible by 3 → No
(d) 3: divides 6 and 9 → No
(e) 36: divisible by 2 and 3 → No

None satisfy the condition except (f) None of these.

Hence, answer is (f) None of these. — Children's numbers must not divide 6 or 9 but must divide 10. None of the options satisfy this, so 'None of these' is correct.

Question 13

Tick the correct statement(s). The LCM of two different prime numbers (m, n) can be:

(a) Less than both numbers

(b) In between the two numbers

(c) Greater than both numbers

(d) Less than m × n

(e) Greater than m × n

Accepted Answer

For two different prime numbers m and n:
- Their LCM is m × n because primes have no common factors.

Check options:
(a) Less than both numbers → No, LCM = m × n which is greater.
(b) In between the two numbers → No, LCM is product, which is greater.
(c) Greater than both numbers → Yes.
(d) Less than m × n → No, equal to m × n.
(e) Greater than m × n → No.

Correct statements: (c) — LCM of two distinct primes is their product, which is greater than both numbers.

Question 14

A dog is chasing a rabbit that has a head start of 150 feet. It jumps 9 feet every time the rabbit jumps 7 feet. In how many leaps does the dog catch up with the rabbit?

Accepted Answer

Let the number of leaps be n.

Distance covered by dog in n leaps = 9n feet
Distance covered by rabbit in n leaps = 7n feet

Dog catches rabbit when:
Distance by dog = Distance by rabbit + 150

So,
9n = 7n + 150

9n - 7n = 150
2n = 150
n = 75

Answer: The dog catches the rabbit in 75 leaps. — Set up equation equating dog's distance to rabbit's distance plus head start, solve for n.

Question 15

What is the smallest number that is a multiple of 1, 2, 3, 4, 5, 6, 8, 9, 10? Do you remember the answer from Grade 6, Chapter 5?

Accepted Answer

Find LCM of 1, 2, 3, 4, 5, 6, 8, 9, 10.

Prime factors:
- 1 (no prime factors)
- 2 = 2
- 3 = 3
- 4 = 2²
- 5 = 5
- 6 = 2 × 3
- 8 = 2³
- 9 = 3²
- 10 = 2 × 5

LCM takes highest powers:
- 2³ (from 8)
- 3² (from 9)
- 5 (from 5 or 10)

LCM = 2³ × 3² × 5 = 8 × 9 × 5 = 360

Answer: 360 — LCM is found by taking highest powers of all primes appearing in the numbers.

Finding Common

Finding Common — Study Notes

Introduction

Multiples and Common Multiples

Finding the Least Common Multiple (LCM)

Practice Questions — Finding Common

All 7 Chapters in Ganita Prakash-II