The product of two numbers is 2028 and their HCF is 13. How many pairs of such numbers are possible?Options :A.1B.2C.3D.4

The solution to this problem involves understanding the relationship between the Highest Common Factor (HCF) and the product of two numbers.

Step 1: We know that the product of two numbers is equal to the product of their HCF and their Lowest Common Multiple (LCM).

Step 2: In this case, we know that the product of the two numbers is 2028 and their HCF is 13. Therefore, their LCM is 2028/13 = 156.

Step 3: Now, we need to find the number of pairs of numbers that have an LCM of 156. The pairs of numbers that have an LCM of 156 are (1,156), (2,78), (3,52), (4,39), (6,26), (12,13).

Step 4: However, we need to consider only those pairs where the HCF is 13. The HCF of (1,156), (2,78), (3,52), (4,39), (6,26) is not 13. The HCF of (12,13) is 13.

Therefore, there is only 1 pair of such numbers possible. So, the answer is A.1.

The solution to this problem involves understanding the relationship between the Highest Common Factor (HCF) and the product of two numbers.

Step 1: We know that the product of two numbers is equal to the product of their HCF and their Lowest Common Multiple (LCM).

Step 2: In this case, we are given that the product of the two numbers is 2028 and their HCF is 13. Therefore, their LCM is 2028/13 = 156.

Step 3: Now, we need to find the number of pairs of numbers that have 156 as their LCM. The pairs of numbers are the factors of 156.

Step 4: The factors of 156 are 1, 2, 3, 4, 6, 12, 13, 26, 39, 52, 78, 156.

Step 5: We can pair these factors in the following way: (1,156), (2,78), (3,52), (4,39), (6,26), (12,13).

Step 6: However, since the HCF of the two numbers is 13, we can only consider the pairs where 13 is a factor. These pairs are (1,156), (3,52), (12,13).

Step 7: Therefore, there are 3 pairs of such numbers.

So, the answer is C.3.

The product of two numbers is given as 2028 and their HCF is 13.

Step 1: First, we need to find the other factor when the numbers are divided by their HCF. So, we divide 2028 by 13. The result is 156.

Step 2: Now, we need to find the number of pairs of factors of 156. The pairs of factors of 156 are (1, 156), (2, 78), (3, 52), (4, 39), (6, 26), and (12, 13).

Step 3: However, since the HCF of the two numbers is 13, we need to find the pairs of factors where both numbers are co-prime (i.e., their HCF is 1). The pairs (1, 156), (2, 78), and (4, 39) are not co-prime pairs, so we discard them.

Step 4: The remaining pairs are (3, 52), (6, 26), and (12, 13). These are co-prime pairs, so they are the possible pairs of numbers.

So, the answer is C. 3. There are 3 pairs of such numbers possible.

The first step is to understand the problem. We know that the product of two numbers is 2028 and their highest common factor (HCF) is 13.

The second step is to divide the product of the numbers by their HCF. So, 2028 divided by 13 equals 156.

The third step is to find all the pairs of factors of 156. The pairs of factors are (1, 156), (2, 78), (3, 52), (4, 39), (6, 26), (12, 13).

The fourth step is to check which pairs have an HCF of 13. The pairs (1, 156), (2, 78), (3, 52), (4, 39), (6, 26) do not have an HCF of 13. Only the pair (12, 13) has an HCF of 13.

So, there is only 1 pair of such numbers possible. Therefore, the answer is A.1.

The product of two numbers is given as 2028 and their HCF is 13.

Step 1: First, we need to find the other factor when the numbers are divided by their HCF. So, we divide 2028 by 13. The result is 156.

Step 2: Now, we need to find the number of pairs of factors of 156. The pairs of factors of 156 are (1, 156), (2, 78), (3, 52), (4, 39), (6, 26), and (12, 13).

Step 3: However, since the numbers must have an HCF of 13, we can only consider the pairs where both numbers are divisible by 13. These pairs are (1, 156), (3, 52), and (12, 13).

Step 4: Therefore, there are 3 pairs of such numbers possible.

So, the answer is C. 3.

The product of two numbers is 2028 and their highest common factor (HCF) is 13. This means that the two numbers are both multiples of 13.

Let's divide 2028 by 13 to simplify the problem. 2028 ÷ 13 = 156.

So, the two numbers we are looking for are factors of 156 that multiply together to give 156.

The factors of 156 are 1, 2, 3, 4, 6, 12, 13, 26, 39, 52, 78, and 156.

We are looking for pairs of factors that multiply together to give 156. These pairs are (1, 156), (2, 78), (3, 52), (4, 39), (6, 26), and (12, 13).

However, since the two numbers must also be multiples of 13, we can eliminate the pairs that do not include 13. This leaves us with the pairs (12, 13) and (13, 156).

So, there are 2 pairs of such numbers possible.

Therefore, the answer is B. 2.