## Problems on H.C.F and L.C.M – Aptitude Questions and Answers

**Topics and Sub Topics in LCM and HCF are:** Factors, Lowest Common Multiples (LCM), Highest Common Factors, LCM of Fractions, HCF of Fraction, LCM of Decimals, HCF of Decimals, **LCM and HCF Formulas** and Practice Problems on LCM and HCF.

### H.C.F and L.C.M Problems With Solutions

**Factor:
**Factor of a given number is that number by which the given number can be divided completely.

**1. Write all the factors of :**

**(i) 15 (ii) 55 (iii) 48 (iv) 36 (v) 84**

Solution :

(i) Factors of 15 = 1, 3, 5 and 15

(ii) Factors of 55 = 1, 5, 11 and 55

(iii) Factors of 48 = 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48

(iv) Factors of 36 = 1, 2, 3, 4, 6, 9, 12, 18 and 36.

(v) Factors of 84 = 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42 and 84.

**Prime Numbers: **

A Natural number, which is divisible by 1 (one) and itself only is called a prime number.

**Prime Numbers between 1 to 100:**

**2. Write all prime numbers :**

**(i) less than 25**

**(ii) between 15 and 35**

**(iii) between 8 and 76**

Solution :

(i) 2, 3, 5, 7, 11, 13, 17, 19 and 23

(ii) 17, 19, 23, 29 and 31

(iii) 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71 and 73.

### Lowest Common Multiple – L.C.M:

**Common Multiple:** A number which is exactly divisible by all the given numbers is “Common Multiple”. Thus the common multiple of 3, 5, 6, and 8 is 120.

The **L.C.M.** of two or more given numbers is the lowest (smallest) number which is exactly **divisible by **each of the given numbers.

**LCM of two or more Numbers**

**Method of Prime Factors:**

Resolve the given numbers into their prime factors and then find the product of the highest power of all the factors that occur in the given numbers. This product will be the LCM.

**3. Find the LCM of 18, 24, 60 and 150.**

Solution :

Prime factors of 18 = 2 x 3 x 3 = 2^{1} x __3 ^{2}__

Prime factors of 24 = 2 x 2 x 2 x 3 =

__2__x 3

^{3}^{1}

Prime factors of 60 = 2 X 2 X 3 X 5 = 2

^{2}X 3

^{1}X 5

^{1}Prime factors of 150 = 2 X 3 X 5 X 5 = 2

^{1}X 3

^{1}X

__5__

^{2}Here, the prime factors that occur in the given numbers are 2, 3 and 5, and their highest powers are 2^{3}, 3^{2} and 5^{2} respectively.

Hence, the required LCM = 2^{3} x 3^{2} x 5^{2} = 1800

**4. L.C.M. of 18, 24 and 96**

Solution :

Prime factors of 18 = 2 x 3 x 3 =2^{1} x __3 ^{2}__

Prime factors of 24 = 2 x 2 x 2 x 3 =2

^{3}x 3

^{1}

Prime factors of 96 = 2 x 2 x 2 x 2 x 2 x 3 =

__2__x 3

^{5}^{1}

Here, the prime factors that occur in the given numbers are 2 and 3, and their highest powers are 2

^{3}, 3

^{2}and 5

^{2}respectively.

Hence, the required L.C.M = 2

^{5}× 3

^{2}

= 32 × 9 = 288

**Note:** The LCM of two numbers which are prime to each other is their product.

5. Find the LCM of 2^{3} × 3^{5} × 5^{2} × 7, 2^{0} × 3^{3} × 5^{2} × 7^{4} , 2^{3} × 3^{2} × 5^{2} × 7^{3}

Solution: 2^{3} × 3^{5} × 5^{2} × 7^{4
}Note: The highest Power should be the L.C.M

** Method of Division**

Write all the given numbers in a row. Divide them by any one of prime numbers 2, 3, 5, 7, 11 etc. which will divided at least two given numbers. Write down the quotients and the other undivided numbers in a row below the first. Repeat this process till you get a row of numbers, which are prime to one another. The product of all the divisors and the numbers in the last row is their LCM.

**6. What is the LCM of 12, 18, 32, and 40?
**At any stage of this method, we may cancel any number, which is a factor of other number in the same row.

Solution:

LCM = 2 × 3 × 2 × 2 × 3 × 4 × 5 = 1440.

**7. Find the LCM of 12, 15, 90, 108, 135, and 150.**

Solution:

∴ The required LCM =2 x 3 x 3 x 5 x 6 x 5 = 2700. In line (1), 12 and 15 are the factors of 108 and 90 respectively, therefore, 12 and 15 are struck- –off. Inline (2), 45 is a factor of 135, therefore 45 is struck off.

In line (5), 3 is a factor of 6, therefore 3 is struck off.

**8. Find-the L.C.M of 5, 6, 8, 9, and 10?**

Solution:

LCM = 2 × 3 × 3 × 4 × 5 = 360.

**9. Find the least square number which is exactly divisible by 3, 4, 5, and 6?**

Solution:

L.C.M = 2 x 2 x 3 x 5 = 2^{2} x 3 x 5

Least Square Number = (2^{2} x 3 x 5) X (3 x 5)

(Multiply by 3 x 5 to get a square Number)

= 2^{2} X 3^{2} X 5^{2}

= 900

**10. The smallest number which is divisible by 12, 15, and 20 and is a perfect square is?**

Solution:

∴ L.C.M of 12, 15, and 20 is 5 X 3 X 4 = 60.

Required number = (60 X 60) = 3600

**11. 4 bells ring at 10 min, 12 min, 15 min and 20 min respectively, beginning together, after what interval of time do they ring together is?**

Solution:

LCM (10, 12, 15, 20)

(∵ 10 is a factor of 20, so 10 is not considered in the division method while finding LCM)

5 x 3 x 4 = 60 min (or) 1 hr.

∵ After 1 hr. they will ring together.

### LCM of Decimals:

First make (if necessary) the same number of decimal places in all the given numbers; then find their LCM as if they were integers, and mark in the result as many decimal places as there are in each of the numbers.

**12. Find the LCM of 0.6, 9.6 and 0.36.**

Solution:

The given numbers are equivalent to 0.60, 9.60 and 0.36

Now, find the LCM of 60, 960 and 36. Which is equal to 2880.

∴ The required LCM = 28.80.

**13. Find the H.C.F. and the L.C.M. of 0.48, 0.72 and 0.108**

Solution:

Given, 0.48, 0.72 and 0.108

Converting each of the following decimals into like decimals we get;

0.480, 0.720 and 0.108

Now, expressing each of the numbers without the decimals as the product of primes we get

480 = 2 × 2 × 2 × 2 × 2 × 3 × 5 = 2^{5} × 3 × 5

720 = 2 × 2 × 2 × 2 × 3 × 3 × 5 = 2^{4} × 3^{2} × 5

108 = 2 × 2 × 3 × 3 × 3 = 2^{2} × 3^{3
}L.C.M. of 480, 720 and 108 = 2^{5} × 3^{3} × 5 = 4320

Therefore, L.C.M. of 0.48, 0.72, 0.108 = 4.32 (taking 3 decimal places)

### LCM of Fractions:

**To find the least number which is exactly divisible by x, y and z.**

Required number = LCM of x, y and z.

14. Find the least number which is exactly divisible by 8, 12, 15 and 21.

Solution:

By the above rule, we have,

The required least number = LCM of 8, 12, 15 and 21 = 840

∴ Ans = 840

**To find the least number which when divided by x, y and z leaves the remainders a, b and c respectively. It is always observed that, (x – a) = (y – b) = (z- c) = K (say)
**

**Required number = (LCM of x, y and z) – K**

**15. What is the least number which, when divided by 52, leaves 33 as the remainder, and when divided by 78 leaves 59, and when divided by 117 leaves 98 as the respective remainders.**

Solution:

Since (52 – 33) = 19, (78 – 59) = 19, (117 -98) = 19

We see that the remainder in each case is less than the divisor by 19. Hence, if 19 is added to the required number, it becomes exactly divisible by 52, 78 and 117. Therefore, the required number is 19 less than the LCM of 52, 78 and 117.

The LCM of 52, 78 and 117 = 468

The required number = 468 -19 = 449

**To find the least number which, when divided by x, y and z leaves the same remainder r in each case.
Required number = (LCM of x, y and z) + r**

**16. Find the least number which, upon being divided by 2, 3, 4, 5 and 6 leaves in each case a remainder of 1. **

Solution:

By the above rule, we have,

Required least number = (LCM of 2, 3, 4, 5 and 6) + 1 = 60 + 1 = 61

**To find the n-digit greatest number which, when divided by x, y and z **

**(i) Leaves no remainder (i.e. exactly divisible)**

**Following step wise methods are adopted.**

Step I: LCM of x, y and z = L

Step II: L) n-digit greatest number (

Remainder (R)

Step III: Required number = n – digit smallest number + (L – R)

**17. Find the greatest number of four digits which, when divided by 12, 15, 20 and 3 5 leaves no remainder. **

Solution:

Using the above method, we get,

Step I: LCM of 12, 15, 20 and 35 = 420

Step II:

Step III: The required number = 9999 – 339 = 9660

**(ii) Leaves remainder K in each case**

**Following step wise method is adopted.**

Step I: LCM of x, y and z = L

Step II: L) n-digit greatest number (

Remainder (R)

Step III: Required number = (n-digit greatest number – R) + K

**To find the n – digit smallest number which, when divided by x, y and z.**

**(i) Leaves no remainder (i.e. exactly divisible)**

**Following steps are followed.**

Step I: LCM of x, y and z = L

Step II: L) n-digit smallest number (

Remainder (R)

Step III: The required number = n-digit smallest number + (L – R)

**18. Find the 4-digit smallest number which when divided by 12, 15 20 and 35 leaves no remainder.**

Solution

Using the above method

Step I: LCM of 12, 15, 20 and 35 =420

Step II:

Step III: The required number = 1000 + (420 – 160) = 1260

### Highest Common Factor – HCF

“Factors” are numbers we can multiply together to get another number.

When we find the factors of two or more numbers, and then find some factors are the same (“common”), then they are the “common factors”.

**Highest Common Factor (HCF):** The greatest number that divides all the given numbers exactly is “HCF”. Thus 8 is the HCF of 24, 48 and 64.

**HCF of Two are More Numbers**

**1. Method of Prime Factors:**

Break the given numbers into prime factors and then find the product of all the prime factors common to all the numbers. The product will be the required HCF.

**19. Find the HCF of 42 and 70.**

Solution:

42 = __2 __x 3 × __7__

70 = __2__ x 5 × __7__

HCF = 2 × 7 = 14

**20. Find the HCF of 1365, 1560 and 1755.**

Solution:

1365 = __3__ × __5__ × 7 × __13__

1560 = 2 × 2 × 2 × __3__ × __5__ × __13__

1755 = 3 × 3 × __3__ × __5__ × __13__

HCF = 3 × 5 × 13 = 195

**(b) HCF of more than two numbers**

Find the HCF of any two of the numbers and then find the HCF of this HCF and the third number and so on. The last HCF will be the required HCF.

**Method II: Method of Division**

**(a) HCF of two numbers:**

Divide the greater number by the smaller number, divide the divisor by the remainder, and divide the remainder by the next remainder, and so on until no remainder is left. The last divisor is the required HCF.

**21. Find the HCF of 42 and 70 by the method of division.**

Solution:

**22. Find the HCF of 1365, 1560 and 1755.**

**Ans:**

**To find the HCF of two or more concrete quantities**

First, the quantities should be reduced to the same unit.

**23. Find the greatest weight which can be contained exactly in 1 kg 235 gm and 3 kg 430 gm**

Solution:

1 kg 235 gm = 1235 gm

3 kg 430 gm = 3430 gm

The greatest weight required is the HCF of 1235 and 3430, which will be found to be 5 gm.

### HCF of decimals:

First make (if necessary) the same number of decimal places in all the given numbers, then find their HCF as if they are integers and marks off in the result as many decimal places as there are in each of the numbers.

**24. Find the HCF of 16.5, 0.45 and 15.**

Solution:

The given numbers are equivalent to 16.50, 0.45 and 15.00

Step I: First we find the HCF of 1650, 45 and 1500. Which comes to 15.

Step II: The required HCF = 0.15.

**25. Find the HCF of 1.7, 0.51, and 0.153.**

Solution:

Step I: First we find the HCF of 1700, 510 and 153. Which comes to 17.

Step II: The required HCF = 0.017.

### HCF of Fractions

**Question 26:**

**HCF of Numbers × LCM of Numbers = Product of Numbers**

**Question 27.**

**To find the greatest number that will exactly divide x, y and z. **

Required number = HCF of x, y and z

**28. What is the greatest number that will exactly divide 1365, 1560 and 1755?**

Solution:

Applying the above rule, the required greatest number = HCF of 1365, 1560 and 1755 = 195

**To find the greatest number that will divide x, y and z leaving remainders a, b and c respectively.**

Required number = HCF of (x – a), (y – b) & (z – c)

**29. What is the greatest number that will divide 38, 45, and 52 and leave as remainders 2, 3 and 4 respectively?**

Solution:

Applying the above rule, we have,

The required greatest number = HCF of (38 – 2), (45 – 3) and (52-4) or 36, 42 and 48 = 6

∴ Ans = 6

**There are n numbers. If the HCF of each pair is x and the LCM of all the n numbers is y, then the product of n numbers is given by ** or Product of ‘n’ numbers = (HCF of each pair)^{n-1} × (LCM of n numbers).

**30. There are 4 numbers. The HCF of each pair is 3 and the LCM of all the 4 numbers is 116. What is the product of 4 numbers?**

Solution:

Applying the above rule, we have,

The required answer = (3)^{4-1} x 116 = 3132

### Practice Problems on LCM and HCF

**1. Find the HCF of 144 and 192.**

**2. Find the greatest weight which can be contained exactly in 6 kg 7 hg 4 dag 3g and 9 kg 9 dag 7 g.**

**a) 11 g ****b) 27 g ****c) 12 g ****d) 17g**

**3. Find the HCF of 405.9 and 219.**

**4. Find the LCM of 40, 36 and 126.**

**5. Find the LCM of 3, 1.2 and 0.06.**

**6. Find the HCF of 3/4, 5/6 and 6/7.**

**7. The LCM of two numbers is 64699, their GCM (or HCF) is 97 and one of the numbers is 2231. Find the other.**

**a) 2183 ****b) 2813 ****c) 2831 ****d) 2381.**

**8. What is the greatest number that will exactly divide 96, 528 and 792?**

**a) 12 ****b) 48 ****c) 36 ****d) 24**

**9. Find the greatest number that will divide 728 and 900, leaving the remainders 8 and 4 respectively.**

**a) 16 ****b) 15 ****c) 14 ****d) 24**

**10. Find the least number which is exactly divisible by 72, 90 and 120.**

**a) 260 ****b) 630 ****c) 360 ****d) 620**

**11. Find the greatest number of six digits which on being divided by 6, 7, 8, 9 and 10 leaves 4, 5, 6, 7 and 8 as remainder respectively.**

**a) 997920 ****b) 997918 ****c) 998918 ****d) 999918**

**12. What is the least number, which when divided by 98 and 105 has in each case 10 as remainder?**

**a) 1840 ****b) 1400 ****c) 1460 ****d) 1480**

**13. Find the greatest number which will divide 260,720 and 1410 so as to leave the remainder 7 in each case.**

**a) 33 ****b) 43 ****c) 32 ****d) 23**

**14. Find the greatest number of three digits which, when divided by 3, 4 and 5 leaves no remainder.**

**a) 960 ****b) 860 ****c) 690 ****d) 680**

**15. Find the smallest 3-digit number, such that they are exactly divisible by 3, 4 and 5.**

**a) 105 ****b) 120 ****c) 115 ****d) 130**

**16. Find the least number which being divided by 2, 3, 4, 5, 6, leaves in each case a remainder 1, but when divided by 7 leaves no remainder.**

**a) 301 ****b) 201 ****c) 302 ****d) 310**

**17. There are 4 numbers. The HCF of each pair is 7 and the LCM of all the 4 numbers is 1470. What is the product of 4 numbers?**

**a) 504210 ****b) 502410 ****c) 504120 ****d) Can’t be determined.**

**18. The LCM of a number is 10 and the 1st number is 12, 2nd number is 4 then what is HCF?**

**19. Find the greatest number of 5 digits which is exactly divisible by 8, 9, 15, 21 is?**

**20. The greatest number of 4 digits which when divided by 2, 3, 4, 5, 6, and 7 is ?**

**21. Find the least no. of 5 digits which is exactly divisible by 4, 12, 15 and 18 is?**

**22. Find the least number of 6 digits which is exactly divisible by 30, 36, 48, 72 is?**

**Answers**

1. 48

2. a; Hint: 6 kg 7 hg 4 dag 3 g = 6743 g

9 kg 9 dag 7 g = 9097 g.

3. 0.3

4. 2520

5. 600

6. 40

7. b

8. d

9. a

10. c

11. b; Hint: The LCM of 6,7,8,9 and 10 = 2520

The greatest number of six digits is 999999.

Dividing 999999 by 2520 we get 2079 as remainder.

Hence the number divisible by 2520 is 999999 – 2079 or 997920.

Since 6 – 4 = 2, 7 – 5 = 2, 8 – 6 = 2, 9 – 7 = 2, 10 – 8 = 2, the remainder in each case is less than the divisor by 2.

∴ the required number = 997920 – 2 = 997918.

12. d

13. d

14. a

15. b

16. a

17. a

18. 4.8

19. 98280.

20. 9660.

21. 10080.

22. 100080

Hello says

It’s a very helpful please make same thing of other chapters