# Tips Tricks and Shortcuts Of HCF and LCM

- Here, are some easy tips and tricks for you to solve HCF and LCM questions quickly, easily, and efficiently in competitive exams.

### HCF and LCM Tips and Tricks and Shortcuts

- The H.C.F of two or more numbers is smaller than or equal to the smallest number of given numbers
- The smallest number which is exactly divisible by a, b and c is L.C.M of a, b, c.
- The L.C.M of two or more numbers is greater than or equal to the greatest number of given numbers.
- The smallest number which when divided by a, b and c leaves a remainder R in each case.
**Required number = (L.C.M of a, b, c) + R** - The greatest number which divides a, b and c to leave the remainder R is
**H.C.F of****(a – R)**,**(b – R)**and**(c – R)** - The greatest number which divide x, y, z to leave remainders a, b, c is
**H.C.F of (x – a), (y – b)**and**(z – c)** - The smallest number which when divided by x, y and z leaves remainder of a, b, c (x – a), (y – b), (z – c) are multiples of M
**Required number = (L.C.M of x, y and z) – M**

**Type 1: Tips and Tricks and Shortcuts to find the greatest or smallest number **

**Question 1. Find the greatest 5 digit number divisible by 5, 15, 20, and 25**

**Options**

**A. 99900**

**B. 99000**

**C. 99990**

**D. 90990**

**Solution: **LCM of 5, 15, 20, and 25 is 300

The greatest 5 digit number is 99999

\frac{99999}{300} = 99

Therefore, required number 99999 – 99 = 99900

**Correct option: A**

**Type 2: Find the numbers, sum of numbers, product of numbers if**

**Their ratio and H.C.F. are given.****Product of H.C.F. and L.C.M are given**

**Question 2. The product of two numbers is 3888. If the H.C.F. of these numbers is 36, then the greater number is:**

**Options**

**A. 110**

**B. 108**

**C. 36**

**D. 120**

**Solution: **Let the two numbers be 36x and 36y

Now, 36x * 36y = 3888

xy = \frac{3888}{36 × 36}

xy = 3

Now, co-primes with product 3 are (1, 3).

Therefore, the required numbers are 36 * 1 = 36

36 * 3 = 108

Therefore the greatest number is 108

**Correct option: B**

**Type 3: Tips , Tricks and Shortcuts when sum of two numbers is given , LCM and HCF is given to find the sum of reciprocals.**

**Question: Sum of two numbers is 60 and the H.C.F. and L.C.M. of these numbers are 5 and 100 respectively, then the sum of the reciprocals of the numbers is equals to:**

**Options**

**A. \frac{3}{25}**

**B. \frac{11}{220}**

**C. \frac{21}{120}**

**D. \frac{11}{320}**

**Solution :** Let the numbers be a and b.

Now , given a+b = 60

a × b = HCF × LCM = 5 × 100

= 500

\frac{1}{a} +\frac{1}{b} = \frac{a+b}{a × b}

\frac{1}{a} +\frac{1}{b} = \frac{60}{500}

\frac{3}{25}

**Correct Option : A**

**Type 4: How to Solve** HCF, LCM Problems related to finding the biggest container to measure quantities

**Question : Suppose there are three different containers contain different quantities of a mixture of Sugar and rice whose measurements are 403 grams, 434 grams and 465 grams What biggest measure must be there to measure all the different quantities exactly?**

**Options :**

**A. 31 grams**

**B. 21 grams**

**C. 41 grams**

**D. 30 litres**

**Solution : **The HCF Of 403 , 434 and 465 is 31.

**Correct Option : A**

**Type 5 :Tips , tricks and Shortcuts of **HCF, LCM Problems related to Bell ring.

**Question: Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ?**

**Options :**

**A. 8**

**B. 16**

**C. 9**

**D. 10**

**Solution : **L.C.M. of 2, 4, 6, 8, 10, 12 is 120.

Hence, the bells will toll together after every 120 seconds(2 minutes).

Therefore, in 30 minutes ,number of times bells toll together is \frac{30}{2} + 1 = 16

**Correct Option B**

**Type 6 : Tips , tricks and Shortcuts of **HCF, LCM Problems related to Circle Based Runner Problem.

**Question: Two people P and Q start running towards a circular track of length 400 m in opposite directions with initial speeds of 10 m/s and 40 m/s respectively. Whenever they meet, P’s speed doubles and Q’s speed halves. After what time from the start will they meet for the third time?**

**Options**

**A. 30 seconds**

**B. 26 seconds**

**C. 10 seconds**

**D. 20 seconds**

**Solution :** Time taken to meet for the 1^{st} time= \frac{400}{40+10}=8 sec.

Now P’s speed = 20m/s and Q’s speed=20 m/s.

Time taken to meet for the 2^{nd} time= \frac{400}{20+20} = 10 sec.

Now P’s speed =40 m/sec and Q’s speed = 10 m/sec.

Time taken to meet for the 3^{rd} time= \frac{400}{10+40}=8 sec.

Therefore, Total time= (8+10+8) = 26 seconds.

**Correct Option B**

### Tips and Tricks and Shortcuts to find HCF easily

**Question. 3 Find HCF of 12 and 16.**

**Options**

**(A) 5**

**(B) 4**

**(C) 12**

**(D) 16**

**Solution **Find the difference between 12 and 16. The difference is 4. Now, check whether the numbers are divisible by the difference. 12 is divisible by 4 and 16 is divisible by 4.Hence, the HCF is 4.

**Correct Option B**

**Question. 4 Find HCF of 18 and 22.**

**Options**

**(A) 2**

**(B) 4**

**(C) 18**

**(D) 36**

**Solution **Find the difference between 18 and 22. The difference is 4. Now, check whether the numbers are divisible by the difference. Both 18 and 22 are not divisible by 4. So take the factors of the difference. The factors of 4 are 2*2*1. Now, check whether the numbers are divisible by the factors. 18 and 22 are divisible by factor 2.

Hence, the HCF is 2.

**Note: If there are more than two numbers, take the least difference.**

**Correct Option(A)**

### Tips and Tricks and Shortcuts to find LCM easily

**Question 5 Find LCM of 2,4,8,16.**

**Options**

**(A) 16**

**(B) 18**

**(C) 12**

**(D) 2**

**Solution **Factorize of above numbe

2 =2

8 = 2^{3}

16 = 2^{4}

Choose the largest number. In this example, the largest number is 16. Check whether 16 is divisible by all other remaining numbers. 16 is divisible by 2, 4, 8. Hence, the LCM is 16.

**Correct Option (A)**

### Tips and Tricks and Shortcuts to find LCM easily

**Question 6 Find the LCM of 2,3,7,21.**

**Options**

**(A) 21**

**(B) 44**

**(C) 36**

**(D) 42**

**Solution **Choose the largest number. The largest number is 21. Check whether 21 is divisible by all other remaining numbers. 21 is divisible by 3 and 7 but not by 2. So multiply 21 and 2. The result is 42. Now, check whether 42 is divisible by 2, 3, 7. Yes, 42 is divisible. Hence, the LCM is 42.

**Correct Option (D)**

**Read Also – How to Solve HCM and LCM Problems Quickly **

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