6. Apti on Mathematical ability ( LCM , HCF )

 1. If the HCF of two numbers is 12 and their LCM is 360, find the numbers.

We know that the product of the HCF and LCM of two numbers is equal to the product of the two numbers.

So, for two numbers a and b with HCF = 12 and LCM = 360:

HCF × LCM = a × b
12 × 360 = a × b
4320 = a × b

Now, we need to find two numbers whose product is 4320 and HCF is 12. There can be multiple pairs of numbers that satisfy this condition, and one such pair is:

a = 120 and b = 36

Because 120 × 36 = 4320 and the HCF of 120 and 36 is 12.

If you want valid pairs, here are a few:

  • 12 and 360

  • 60 and 72

  • 36 and 120

2. Find the HCF of 36, 48, and 72.
  • Prime factorization of 36: 22 × 32
  • Prime factorization of 48: 24 × 3
  • Prime factorization of 72: 23 × 32

Common factors: 22 and 3 (take the minimum power)

So, the HCF of 36, 48, and 72 is 22 × 3 = 12.


3. What is the largest three-digit number that is exactly divisible by the HCF of 24 and 36?

he HCF of 24 and 36 is 12. To find the largest three-digit number that is exactly divisible by 12, we need to find the largest multiple of 12 that is less than 1000.

The largest multiple of 12 less than 1000 is 996 (83 × 12 = 996).

So, the largest three-digit number exactly divisible by the HCF of 24 and 36 is 996.


4. The sum of two numbers is 1001, and their HCF is 7. Find the numbers. answer: the nos can be 7x and 7y here 7 is HCF so, x and y are co- prime 7x +7y=1001 => (x+y)=143 hcf(x,y)=7 now find the pair that sum is 143 and the pairs are co-prime obviously (i.e their HCF is 1) such pair is (1, 142) the nos are 7*1=7 and 7*142=994


Note: product of two nos is equal to (LCM*HCF)

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