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digits fun!

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Master | Next Rank: 500 Posts
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by Mclaughlin » Wed Jul 16, 2008 9:33 pm
How many 4-digit positive integers are multiples of each integer from 1 to 10?

A. 0
B. 1
c. 2
D. 3
E. 4

OA D

How do you do this? Where do you start? is this a LCM question? I have no idea!
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Re: digits fun!

by parallel_chase » Thu Jul 17, 2008 4:30 am
Mclaughlin wrote:How many 4-digit positive integers are multiples of each integer from 1 to 10?

A. 0
B. 1
c. 2
D. 3
E. 4

OA D

How do you do this? Where do you start? is this a LCM question? I have no idea!

Answer is D

2*3*2*5*7*2*3 = 2520

The question says how many 4 digit numbers are multiples of 2520

2520*1 = 2520
2520*2 = 5040
2520*3 = 7560

2520*4 = 10080


Therefore the answer is 3
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by floravaze » Thu Jul 17, 2008 5:16 am
Please can you explain your logic in the first step?i do not have a clue what it means.please!!!!
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by parallel_chase » Thu Jul 17, 2008 5:30 am
floravaze wrote:Please can you explain your logic in the first step?i do not have a clue what it means.please!!!!
The logic is we need a number which could be divisible by all the integers from 1 - 10

1 = 1
2 = 2
3 = 3
4 = 2*2
5 = 5
6 = 2*3
7 = 7
8 = 2*2*2
9 = 3*3
10= 2*5


Therefore that number will be

1*2*3*2(I)*5*(II)*7*2(III)*3(IV)*(V)

I - we have not multiplied by 4 since 4= 2*2 we have already multiplied 2 before so we just have to multiply one more 2 for it to be a multiple of 4 by 4)

II we have not multiplied by 6 since 6=3*2 we have already multiplied 2 and 3.

III we only have to multiply by 2 since 8 =2*2*2

IV we only have to multiply by 3 since 9 = 3*3

V we do not have to multiply by 10 since 10=5*2

I hope this clears everything.
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by floravaze » Thu Jul 17, 2008 5:42 am
Gee-i got it.Thank you so much.
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