CAT 2023 Slot 2 QA Question 6

"The number of positive integers less than 50, having exactly two distinct factors other than 1 and itself, is __."

Hmm, okay. So, we're looking for numbers less than 50 that have exactly two distinct factors besides 1 and themselves. First, I know that factors of a number are the numbers that divide it without leaving a remainder. For example, the factors of 6 are 1, 2, 3, and 6. So, excluding 1 and itself, 6 has two factors: 2 and 3. That seems to fit the description here.

Wait, so if a number has exactly two distinct factors other than 1 and itself, that means it has exactly four factors in total, right? Because 1, the number itself, and two others. So, the total number of factors is four.

in number theory, numbers with exactly four factors have a specific form. Let me recall. If a number has exactly four factors, it can be either the product of two distinct prime numbers or a cube of a prime number. Is that correct?

Let me verify. If a number is the product of two distinct primes, say p and q, then its factors are 1, p, q, and pq. That's four factors. So, that works. On the other hand, if a number is the cube of a prime, say p³, then its factors are 1, p, p², and p³. That's also four factors. So, both cases give exactly four factors.

Therefore, the numbers we're looking for are either the product of two distinct primes or the cube of a prime. Now, we need to find all such numbers less than 50.

Let me tackle each case separately.

We need to find all numbers less than 50 that are the product of two distinct primes. Let's list out the primes less than 50 first.

Primes less than 50 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47.

Now, we need to consider all possible pairs of these primes and compute their products, ensuring the product is less than 50.

Let me start with the smallest primes.

Next, starting with 3.

Now, starting with 5.

Starting with 7.

So, beyond this, any higher primes multiplied together will exceed 50. So, the products are:

6, 10, 14, 22, 26, 34, 38, 46, 15, 21, 33, 39, 35.

Let me count these: 6,10,14,22,26,34,38,46 (8 numbers), then 15,21,33,39,35 (5 numbers). So, total of 13 numbers.

Wait, let me list them all to make sure I didn't miss any:

From 2: 6,10,14,22,26,34,38,46 (8 numbers) From 3:15,21,33,39 (4 numbers) From 5:35 (1 number) Total: 8 + 4 + 1 = 13 numbers.

Wait, but earlier I thought 5*7=35 is included, which is correct. So, 13 numbers in this case.

Now, we need to find primes p such that p³ < 50.

Computing cubes of primes:

So, only 8 and 27 are cubes of primes less than 50.

So, that's two numbers: 8 and 27.

Now, we need to check if these numbers have exactly two distinct factors other than 1 and themselves.

For 8: factors are 1, 2, 4, 8. So, excluding 1 and 8, the factors are 2 and 4. That's two distinct factors. So, it fits.

For 27: factors are 1, 3, 9, 27. Excluding 1 and 27, the factors are 3 and 9. That's two distinct factors. So, it also fits.

Therefore, we have two more numbers: 8 and 27.

Now, combining both cases:

Total: 13 + 2 = 15 numbers.

Wait, but hold on. Let me make sure that none of the numbers in Case 1 are also in Case 2. For example, is 8 or 27 present in the list of products of two distinct primes? .

Looking back at the list from Case 1: 6,10,14,22,26,34,38,46,15,21,33,39,35. None of these are 8 or 27, so no overlap. So, total numbers are indeed 15.

Therefore, the number of positive integers less than 50 with exactly two distinct factors other than 1 and themselves is 15.

Let me just cross-verify this with another approach