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Find the smallest number of 4 digit which when divided by 40,50,and60 leaves remainder of 5 in each case

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Remark

First of all, you should notice that the number you want must end in 5 to get a remainder of 5. That's because your divisors all end in 0.

Step One

Factor the 3 divisors

40: 2 * 2 * 2 * 5

50: 2 * 5 * 5

60: 2 * 2 * 3 * 5

Step Two

Using these factors, find the LCM

There are 3 twos, 1 Three, and 2 fives.

The LCM is 8*3*25 = 600

Step Three

Find the first number over 1000 that is divisible by 600. That should be 1200 which is 600 * 2

So the answer to your question is 1205

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