Question

7. Find the smallest number of 4 digits which when divided by 40, 50 and 60 leaves a remainder of 5 in each case.​

Answer 1

Answer:

1205 is the smallest 4 digit no divisible by 40 , 50 and 60 leaves remainder 5

Step-by-step explanation:

Given Numbers = 40 , 50 and 60

To find: Smallest 4 digit number when divided by given nos. leaves 5 as remainder.

First we find LCM of 40 , 50 and 60 as LCM is the smallest no which is divisible by given nos.

40 = 2 × 2 × 2 × 5

50 = 2 × 5 × 5

60 = 2 × 2 × 3 × 5

LCM = 2 × 5 × 2 × 2 × 5 × 3 = 600

Now we find multiples of 600 and select smallest 4 digit multiple of 600 as multiples of 600 also divisible by 40 , 50 and 60.

600 , 1200 , 1800 , ...

⇒ 1200 is the smallest 4 digit no divisible by given nos.

As its leave 5 as remainder we add 5 to 1200

⇒ 1200 + 5 = 1205

Therefore,  1205 is the smallest 4 digit no divisible by 40 , 50 and 60 leaves remainder 5