Triple Prime Product

We would start with the smallest prime numbers to find our product to be less than $100$ . To get the highest prime number that can be a part of the triplets, we would check it by trial and error as to which number gives product less than $100$ on multiplication by $2$ and $3$ , as they are the smallest primes. By doing this, we get $13$ as that number, because $2\times 3\times 13 = 78$ which is less than $100$ . Note that all the primes less than or equal to $13$ are $2, 3, 5, 7, 11$ and $13$ .

To get all the numbers less than $100$ which are products of $3$ distinct primes, we must do case work.

Firstly, we will consider the smallest prime numbers, that are $2$ and $3$ . Then will consider their products with other primes, from which we get that all the products, that are: $2\times 3\times 5 = 30$ $2\times 3\times 7 = 42$ $2 \times3 \times 11 = 66$ and $2 \times3 \times 13 = 78$ are less than $100$ . Note that $2 \times 3 \times 17 = 102$ is not wanted, as it's product is exceeding $100$ . Hence, till here, we get $4$ desired values.

Now we consider the next prime number with $2$ , that is, $5$ . Now, considering the product of $2$ and $5$ with other prime numbers, we get only one desired value, that is $2 \times 5 \times 7 = 70$ , because it is less than $100$ . Note that we are not doing $2 \times 5 \times 3 = 30$ again here, because we have already considered it in previous case. Also, we can't consider $2 \times 5 \times 11 = 110$ or any bigger prime number, as it will end us up with a number which is greater than $100$ . Thus from here, we get only $1$ desired value.

Now considering the next prime number with $2$ , that is $7$ , we notice that $2 \times 7 \times 11 = 154$ or any bigger prime number can't be considered, as it too ends up with a value which is greater than $100$ .

Thus, this enables us to stop here, and sum up all the desired values, which yields us $4 + 1 = \fbox{5}$ . Thus, there are $\fbox{5}$ numbers less than $100$ which are the product of $3$ distinct primes.

To get all possible numbers with the given combination ,

We do casework on the numbers , for which product is less than 100 by assuming other two numbers .

i.e from the smallest possible products to the maximum possible products by taking two prime numbers at a time and eliminating the ones with product greater than $100$ .

$$

$\begin{array}{c|c|c} & Product & Possible-numbers & Product & Product<100\\ \hline Case-1 & 2 \cdot 3 & 5 & 30 & Yes \\ & & 7 & 42 & Yes \\ & & 11 & 66 & Yes \\ & & 13 & 78 & Yes\\ & & 17 & 102 & No \\ & & & \\ Case-2 & 2 \cdot 5 & 7 & 70 & Yes \\ & & 11 & 110 & No \\ & & & \\ Case-3 & 3 \cdot 5 & 7 & 105 & No \\ \end{array}$

$$

Since , no further values or cases are possible ,

Thus , the numbers less than 100 and product of 3 distinct primes will be :

$30 , 42 , 66 , 70 , 78$

Hence, there are $\boxed{5}$ such numbers .

Cheers !

Hope that helps .

If any queries, please comment below .

You could use this method:

$2 * 3 * 5=30$

$2 * 3 * 7=42$

$2 * 3 * 11=66$

$2 * 5 * 7=70$

$2 * 3 * 13=78$

There are 5 numbers all up.

Multiply the 3 smallest primes until the product exceeds 99. We have: 2 3 5 = 30, 2 3 7 = 42, 2 3 11 = 66, 2 3 13 = 78, 2 5 7 = 70. Ed Gray

see minimum prime ni=2 now we can take 3=6 so trird possible prime no 5 ,7 ,11, 13.........now 2,5,7 so total 5 possibilities

30,42,66,70, and 78

First take the two smallest primes 2 and 3. Find x such that 2 × 3 × x ≤ 100 ≤ 2 × 3 × (x+1). So, x = 16. Prime numbers less than 16 are 2, 3, 5, 7, 11, 13 2 × 3 × 13 = 78 < 100 2 × 3 × 11 = 66 < 100 2 × 3 × 7 = 42 < 100 2 × 3 × 5 = 30 < 100 30, 42, 66, 78 are the product of exactly 3 prime numbers.

Now, take the primes 2 and 5. 2 × 5 × 7 = 70 < 100 < 110 = 2 × 5 × 11 70 is the product of exactly 3 prime numbers.

Now, 2 × 7 × 11 = 154 > 100 3 × 5 × 7 = 105 > 100 There is no other such numbers less than 100 which are the product of exactly 3 prime numbers.

So, 30, 42, 66, 70 and 78 are the only numbers less than 100 which are the product of exactly 3 prime numbers.

possible values are 2x3x..here we can fill with either by 5,7,11,13and 17..by which the product is less than 100