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sriramsundaram91
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How many integers less than 1000 have no factors (other than 1) in com 24 Apr 2026, 13:09
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This is such a good problem,
Here is how I solved it.
Basically even numbers have a common factor with 1000 so immediately 500 of them are gone (500 even integers are there from 1-1000).
Therefore the Odd numbers = 500,
In odd numbers, only numbers ending in 5 have a common divisor, numbers ending in 5 are odd multiples of 5
there are 200 multiples of 5 till 1000, 100 even and 100 odd, remove 100 odd multiples from the previously obtained 500
You get 400.
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How many integers less than 1000 have no factors (other than 1) in com 18 May 2026, 13:51
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I tried this way.
all positive numbers less than 1000 which have no factors in common to 1000 other than 1=?
since 1000=2^3*5^3
first we must discard all even no.'s less than 1000 since have 2 as common factor=500 -1(for 1000)=499
then discard all odd multiples of 5 less than 1000,since even multiples are included in 499=> we have total of 200 multiples of 5 of which 100 should be odd ones i.e. 100 (no need to subtract 1 from this as 1000 not part of this)
so total discarded=499+100=599
hence required value=999-599=400
A
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How many integers less than 1000 have no factors (other than 1) in com 22 May 2026, 02:19
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Bunuel


First of all it should be "how many positive integers less than 1000 have no factors (other than 1) in common with 1000", as if we consider negative integers answers will be: infinitely many.

\(1000=2^3*5 ^3\) so basically we are asked to calculate the # of positive integrs less than 1000, which are not multiples of 2 or/and 5.

Multiples of 2 in the range 0-1000, not inclusive - \(\frac{998-2}{2}+1=499\);
Multiples of 5 in the range 0-1000, not inclusive - \(\frac{995-5}{5}+1=199\);
Multiples of both 2 and 5, so multiples of 10 - \(\frac{990-10}{10}+1=99\).

Total # of positive integers less than 1000 is 999, so # integers which are not factors of 2 or 5 equals to \(999-(499+199-99)=400\).

Answer: A.
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This may sound a very very stupid question, but I want to ask, how do you figure out multiples of 2 and 5 in a given range? Like you have mentioned:

Multiples of 2 in the range 0-1000, not inclusive - \(\frac{998-2}{2}+1=499\);

Is there any formula for this? Or is it just some logic that I am unable to figure out?
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How many integers less than 1000 have no factors (other than 1) in com 22 May 2026, 02:32
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freebunny

This may sound a very very stupid question, but I want to ask, how do you figure out multiples of 2 and 5 in a given range? Like you have mentioned:

Multiples of 2 in the range 0-1000, not inclusive - \(\frac{998-2}{2}+1=499\);

Is there any formula for this? Or is it just some logic that I am unable to figure out?
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\(Number \ of \ multiples \ of \ x \ in \ the \ range = \)

\(=\frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\).

Example 1: how many multiples of 5 are there between -7 and 35, not inclusive?

Last multiple of 5 IN the range is 30;
First multiple of 5 IN the range is -5;

\(\frac{30-(-5)}{5}+1=8\).

Example 2: How many multiples of 4 are there between 12 and 96, inclusive?
\(\frac{96-12}{4}+1=22\).

Example 3:
How many multiples of 7 are there between -28 and -1, not inclusive?
Last multiple of 7 IN the range is -7;
First multiple of 7 IN the range is -21;

\(\frac{-7-(-21)}{7}+1=3\).
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How many integers less than 1000 have no factors (other than 1) in com 18 Jun 2026, 12:04
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How about using Euler's totient formula? we can solve such problems quickly.

f(n)=n*(1-(1/p1))*(1-(1/p2))*.......* (1-(1/pn)) where p1 to pn are the prime factors of n.

1000= 2^3*5^3
f(1000)= 1000*(1-(1/2))*(1-(1/5))
=1000*1/2*4/5=400
Bunuel


First of all it should be "how many positive integers less than 1000 have no factors (other than 1) in common with 1000", as if we consider negative integers answers will be: infinitely many.

\(1000=2^3*5 ^3\) so basically we are asked to calculate the # of positive integrs less than 1000, which are not multiples of 2 or/and 5.

Multiples of 2 in the range 0-1000, not inclusive - \(\frac{998-2}{2}+1=499\);
Multiples of 5 in the range 0-1000, not inclusive - \(\frac{995-5}{5}+1=199\);
Multiples of both 2 and 5, so multiples of 10 - \(\frac{990-10}{10}+1=99\).

Total # of positive integers less than 1000 is 999, so # integers which are not factors of 2 or 5 equals to \(999-(499+199-99)=400\).

Answer: A.
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