Question

find the least number of 5 digits which is exactly divisible by 10,15, 20 and 25.

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Answer 1

In order for a number to be evenly divisible by 15, 20 and 25 it must be divisible by their least common multiple (LCM).

The prime factorizations of the given numbers are 3*5, 2*2*5 and 5*5, respectively. The LCM is 2*2*3*5*5 which equals 300.

Finally the smallest 5 digit number divisible by 300 is found by dividing the smallest 5 digit number (i.e., 10,000) by 300 and adding 300 minus the remainder. That is, 10,000/300 = 33R100. So the answer is 10,000 + (300–100) = 10,200.

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Answer 2

Decompose each into prime factors.

20 = 2*2*5

25 = 5*5

30 = 2*3*5

Lowest common multiple is then

2*2*3*5*5 = 300

Now you need the lowest number so that

300 * N >= 10000

then

N >= 10000/300

N > 33.3…

Being N an integer, N has to be 34.

300 * 34 = 10200

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