# HCF and LCM Solved Examples

**Solved Examples on HCF and LCM**

**Question 1): Find the greatest number which on dividing 1657 and 2037 leaves remainder 6 and 5 respectively.**

Solution:

Required number = HCF of (1657-6) and (2037-5) = 127

Question 2): Find the largest number which divides 62, 132 and 237 to leave the same remainder in each case.

Solution:

Required number = HCF of (132-62), (237 - 132), and (237 -62) => 35

Question 3): Find the least number which when divided by 20, 25, 35 and 40 leaves remainder 14, 19, 29 and 34 respectively.

Solution:

Here, (20-14)=(25-19)=(35-29)=(40-34)=6

Requierd Number = (LCM of 20, 25, 35, 40) - 6 => 1394

Question 4): Find the least number which when divided by 5,6,7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder.

Solution: LCM of 5,6,7 and 8 is 840

required number is of the form 840k + 3

Least value of K for which (840k + 3) is divisible by 9 is k=2

required number = (840 x 2 +3) = 1683

Question 5): Three numbers which are co-prime to each other are such that the product of the first two is 551 and that of last two is 1073. the sum of the three numbers is:

Solution:

Given two products have middle number in common (Since they are co prime)

so, middle number = HCF of 551 and 1073 => 29

First number = 551/29 = 19;

Third number = 1073/29 = 37

required sum = 19 + 29 + 37 = 85

Question 6): The product of the HCF and LCM of two numbers is 24. The difference of two numbers is 2. Find the numbers.

Solution:

Let the numbers be x and x+2

So x * (x+2) = 24 => x+4

therefore numbers are 4 and 6

Question 7): Two numbers both greater than 29, have HCF 29 and LCM 4147. The sum of the numbers is?

Solution:

Product of numbers = 29 * 4147

Let the number be 29a and 29b, then 29a * 29b = 29 * 4147 => ab = 143

Now co-primes with product 143 are (1, 143) and (11, 13)

so the numbers are (29*1, 29*143) and (29*11, 29*13)

Since both numbers are greater than 29, the suitable pair is (319,377).

Question 8): The greatest number which can divide 1356, 1868, and 2764 leaving the same remainder 12 in each case, is:

Solution:

Required number = HCF of (1356 - 12), (1868-12) and (2764 - 12) = 64

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HCF and LCM Solved Examples
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