LCM and HCF Exercise 1

Least number of 5 digits is 10,000 L.C.M. of 12, 15 and 18 is 180.

On dividing 10000 by 180, the remainder is 100.

∴ Required number

= 10000 + (180 – 100) = 10080.

Greatest number of 4 digits is 9999. L.C.M. of 15, 25, 40 and 75 is 600.

On dividing 9999 by 600, the remainder is 399.

∴ Required number

= (9999 – 399) = 9600.

L.C.M. of 5, 6, 4 and 3 = 60.

On dividing 2497 by 60, the remainder is 37.

∴ Number to be added

= (60 – 37) = 23.

Required number

= HCF of 429 and 715 = 143

Required number

= HCF of (115 – 3), (149 –5) and (183 – 7)

= HCF of 112, 144 and 176 = 16

Required number

= 3000 – LCM of 7, 11, 13

= 3000 – 1001 = 1999

Let numbers be x and y.

Product of two numbers = their (LCM × HCF)

⇒ xy = 630 × 9

Also, x + y = 153 (given)

since x – y

⇒

H.C.F of co-prime numbers is 1.

So, L.C.M. = 117/1 = 117.

Required number

= LCM of ( 8, 11, 24 ) – 5

= 264 – 5 = 259

Product of numbers

= (LCM × HCF)

⇒ 480 × second number = 2400 × 16

⇒ second number = 80

Product of numbers = HCF × LCM

⇒ The other number

=

Let the numbers be 7x and 8x.

⇒ Their HCF = x

Now, LCM × HCF = Product of Numbers

i.e. 280 × x = 56x²

⇒ x = 5

Hence, the numbers are 35 and 40.

Let the numbers be x and 4x.

Then, 84 × 21 = x × 4x

⇒ 4x² = 1764

⇒ x² = 441

⇒ x = 21

⇒ 4x = 4 × 21 = 84

Thus the larger number = 84

LCM of 3 and 5 = 15

∴ 300/15 = 20 numbers