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Problem on HCF and LCM
  • Difficulty Level : Easy
  • Last Updated : 06 Jan, 2019

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= 22 x 3 x 11
216= 23x 33
HCF= 22 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= 22 * 32 * 7 = 252
To become perfect square all factors should be in power of 2.
So, multiply it by 7
LCM = 22 * 32 * 72 = 1764
And it is perfect square of 42.

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