How many composite numbers are between 1 and 100?

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Answer 1 · 2017-05-27T13:53:10

$74$ $74$

Note that the following answer assumes that "between 1 and 100" is intended to include $1$ $1$ and $100$ $100$ , following common English usage.

Start with the numbers $1,2,3,...100$

$1$ is not composite, because it's a unit, so that leaves $99$ other numbers.
The numbers $4=2^2, 6, 8, ... , 100$ are all divisible by $2$ , so composite. There are $(100-4)/2+1 = 49$ of these, leaving $99-49 = 50$ other numbers.
The numbers $9=3^2, 15, 21,..., 99$ are all divisible by $3$ and not by $2$ . There are $(99-9)/6+1 = 16$ of these, leaving $50-16 = 34$ other numbers.
The numbers $25=5^2, 35, 55, 65, 85, 95$ are all divisible by $5$ and not by $2$ or $3$ . There are $6$ of these, leaving $34-6 = 28$ other numbers.
The numbers $49=7^2, 77, 91$ are divisible by $7$ and not by $2$ , $3$ or $5$ . There are $3$ of these, leaving $28-3 = 25$ other numbers.

These $25$ numbers must be prime, since $11^2 = 121 > 100$

So the total number of composite numbers is:

$49+16+6+3 = 74$

Answer 2 · 2017-05-27T22:52:12

There are $73$ composite numbers less than $100$

There are $100$ numbers from $1 " to " 100$

However, the question specifies numbers BETWEEN $1 and 100$ , so these two numbers are eliminated from the total.

So we are left with $98$ numbers to work with.

All the numbers between $1 and 100$ are either prime or composite.
(The only number that is neither is $1$ which has already been excluded)

There are $25$ prime numbers less than $100$ .

This is pretty easy to check by just counting them, but it is a small fact that is worth knowing.

Therefore, if $25$ of the $98$ numbers are prime, it means that all the rest are composite:

$98-25 =73$ composite numbers.