a) Find the least number which when divided by 15, 18 and 20 leaves the remainder as 3 in each case. b)The HCF and LCM of two numbers are 99 and 2772 respectively. If one of the numbers is 297, find the other number.
Answers
Answer:
Step-by-step explanation:
Required number = (L.C.M of 15, 20, 48 and 36) + 9
= (2×2×3×5×4×3) = 720 + 9
= 729
(a) The least number which when divided by 15, 18 and 20 leaves the remainder as 3 in each case = 183
(b) The other number is 924
Given problems:
a) Find the least number which when divided by 15, 18 and 20 leaves the remainder as 3 in each case.
b)The HCF and LCM of two numbers are 99 and 2772 respectively. If one of the numbers is 297, find the other number
Solution:
In sum (a)
Given numbers are 15, 18, 20
To find the least number which when divided by 15, 18 and 20 leaves the remainder as 3 we need to find LCM(15, 18, 20)
To find LCM(15, 18, 20) write 15, 18, 20 as product of prime factors
=> 15 = 3 × 5
=> 18 = 2 × 3 × 3
=> 20 = 2 × 2 × 5
=> Lcm (15, 18, 20) = 2 × 2 × 3 × 3 × 5 = 180
Now add Reminder 3 to 180
The required number = 180 + 3 = 183
The least number which when divided by 15, 18 and 20 leaves the remainder as 3 in each case = 183
In sum (b)
Given HCF and LCM of two number are 99 and 2772
One of the number is 297 and let a be the other number
As we know Product of 2 number = product of LCM and HCF
⇒ 297 (a) = 99 (2772)
⇒ a = 99(2772) / 297 = 924
The other number is 924
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