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## Difficult Math Problem #109 - Number Theory

This topic has 2 member replies
800guy Master | Next Rank: 500 Posts
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#### Difficult Math Problem #109 - Number Theory

Wed Mar 14, 2007 8:42 am
How many integers less than 1000 have no factors (other than 1) in common with 1000?

(1) 400
(2) 410
(3) 411
(4) 412
(5) None of the above

from diff math doc. oa coming after some people respond with explanations

gabriel Legendary Member
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Wed Mar 14, 2007 9:14 am
800guy wrote:
How many integers less than 1000 have no factors (other than 1) in common with 1000?

(1) 400
(2) 410
(3) 411
(4) 412
(5) None of the above

from diff math doc. oa coming after some people respond with explanations
So we need numbers that have factors other than 2 or 5..... that wuld be equal to.. total- ( numbers with 2 or 5 as a factor )

Now, the number of integers that have 2 as a factor and less than 1000 is 499.....

now while counting factors of 5 we do not have to include numbers like 10,20,30... bcoz they are already included in the list of integers with 2 as a factor... so we only consider the odd multiples of 5... ie 5,15,25... which wuld be equal to 100.. therefore tha answer is 999-( 499+100) = 400.. therefore the answer is A....

800guy Master | Next Rank: 500 Posts
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Fri Mar 16, 2007 8:59 am
here's the oa:

1000 - multiples of 2 and/or 5

multiples of 2 = 500 (all even #)
multiples of 5 = (995 -5)/10 + 1 [ Using AP formula]
= 100

Answer = 1000 - (500 + 100)
= 400

You cannot calculate for all multiples of 5 because you have already removed all even integers (including 10, 20, and 30). The difference in the AP series should be 10 instead of 5 because you're looking for the integers that have 5 as a unit's digit. Therefore we divide by 10 and not 5.

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