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Common Integers !!

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Source: — Problem Solving |
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by gmatmachoman » Tue May 25, 2010 12:11 pm
niksdoon wrote:Hi, can anyone pls help me get this solve?

How many integers less than 1000 have no factors(other than 1) in common with 1000 ?

a. 400
b. 399
c. 410
d. 420
1000 = 2^3 * 5 ^3

We can use Venn Diagram to solve this!!

n (AuB) = n (A) + n(B) - n(AnB)

n(A) : Total numbers divisible by 2 : 1000/2 = 500

n(B) : Total Numbers divisible by 5 : 1000/5 = 200

n (AnB) : Overlapping/ Toatl numbers that divisible by both 2 & 5 : 1000/10 =100

Apply these values in formula stated

500 + 200 -100

600.

Since the stem asks How many integers less than 1000 have no factors(other than 1) in common with 1000 ? we have to find : Total - n (AuB) -1

1000 -600-1 = 399

IMO B
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by albatross86 » Tue May 25, 2010 12:12 pm
1000 = 2 * 500 = 2 * 2 * 250 = 2 * 2 * 2 * 125 = 2^3 *5^3

Any number with prime factors of 2 and/or 5 would share factors with 1000

The number of integers less than or equal to 1000 and divisible by 2 = 1000/2=500.
The number of integers less than or equal to 1000 and divisible by 5 = 1000/5=200.
The number of integers less than or equal to 1000 and divisible by 2 AND 5 = The number of integers less than or equal to 1000 and divisible by 10 = 1000/10=100.

Hence N=500+200-100=600, so the answer you seek is 1000-600=400.
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by niksdoon » Wed May 26, 2010 5:23 am
Hi albatross86

Don't you think that we need to exclude 1 from list of integers? Question asked to consider only those int which have no factors(other than 1) in common to 1000.

I agree with gmatmachoman...

Thanks,
niks
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by albatross86 » Wed May 26, 2010 5:27 am
Aah true. Well spotted!
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