# Problem on HCF and LCM

**Question 1: ** Find the HCF by long division method of two no’s the sequence of quotient from top to bottom is 9, 8, 5 and the last divisor is 16. Find the two no’s.

**Solution: ** Start with the divisor and last quotient.

Divisor x quotient + remainder = Dividend

16 x 5 + 0 = 80

80 x 8 + 16 = 656

**656** x 9 + 80 = **5984**

Hence, two numbers are** 656 and 5984.**

**Question 2: ** The LCM and HCF of two numbers is 210 and 5. Find the possible number of pairs.

**Solution: ** HCF = 5 so it should be multiple of both numbers.

So both numbers 5x : 5y

LCM = 5 * x * y = 210

x * y = 42

{1 x 42}, { 2 x 21}, {3 x 14}, { 6 x 7 } .

**Four pairs are possible**.

**Question 3: ** The sum of two numbers is 132 and their LCM is 216. Find both the numbers.

**Solution: **

**Note: ** HCF of Sum & LCM is also same as actual HCF of two numbers.

Factorize both 132 and 216 and find the HCF.

132= 2^{2} x 3 x 11

216= 2^{3}x 3^{3}

HCF= 2^{2} x 3 =12

Now, 12x + 12y = 132

x + y = 11

And 12 * x * y = 216

x * y = 18

Solve for x and y, we get y = 9 and x = 2. Hence both numbers are 12*2 = **24** and 12*9 = **108**

**Question 4: ** The LCM of two numbers is 15 times of HCF. The sum of HCF and LCM is 480. If both number are smaller than LCM. Find both the numbers.

**Solution: ** LCM = 15 * HCF

We know that

LCM + HCF = 480

16 * HCF = 480

HCF = 30

Then LCM = 450

LCM = 15 HCF

30 * x * y = 15 * 30

x * y = 15

Factors are {1 x 15} and { 3 x 5}

Both numbers less than LCM so take {3 x 5}

Hence numbers are 3 * 30 = **90** and 5 * 30 = **150**

**Question 5: ** Find the least perfect square number which when divided by 4, 6, 7, 9 gives remainder zero.

**Solution: ** Find the LCM for 4, 6, 7, 9

LCM= 2^{2} * 3^{2} * 7 = 252

To become perfect square all factors should be in power of 2.

So, multiply it by 7

LCM = 2^{2} * 3^{2} * 7^{2} = 1764

And it is perfect square of **42**.

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