Question

The sum of two numbers is 528 and their HCF is 33. The number of such pair is:
A) 2
B) 3
C) 4
D) 5

Answer 1

Hint: By reading the questions you get that the sum of the two numbers is 528 and the HCF is 33 which can be used in the sum of two different variables. secondly, the no having sum 33, should also have 1 only as an HCF

Complete step by step solution: We can stepwise get through the information provided as you can see that the first information is the sum of two numbers and second both have the highest common factor as 33.
From this we can conclude, both are divisible by 33 and we can assume both no as a multiple of 33.
So let’s Assume one no. as $33 \times x$ and the second number as $33 \times y$.
As we get back to our first information. Sum of both this number is equal to 524
Therefore,
$33x + 33y = 528$
$33\left( {x + y} \right) = 528$
$x + y = \dfrac{{528}}{{33}} = 16$
Now we need a pair of x and y which can have a sum equal to 16.
So pairs can be identified by assuming x as one variable then $16 - x$ as the second.
Therefore at
\[\begin{array}{l}
x = 1,\;\;y = 16 - x = 15\\
x = 2,\;\;y = 16 - 2 = 14\\
x = 3,\;\;y = 16 - 3 = 13\\
x = 4,\;\;y = 16 - 4 = 12\\
x = 5,\;\;y = 16 - 5 = 11\\
x = 6,\;\;y = 16 - 6 = 10\\
x = 7,\;\;y = 16 - 7 = 9\\
x = 8,\;\;y = 16 - 8 = 8
\end{array}\]
Further, for x it will repeat the same changing the values of x and y.

The 8 pairs (1, 15) (2, 14) (3, 13) (4, 12) (5, 11) (6, 10) (7, 9) and (8, 8) satisfies the condition appropriately, but the condition of being a highest common factor as 33 will be satisfied by few above pairs.

So now let’s eliminate the Pairs having HCF between them.
The Pairs which do not have HCF other than 1 are (1, 15) (3, 13) (5, 11) (7, 9).
Therefore, there are 4 Pairs.

Option C is the correct answer.

Note: As we proceed with the situation given in the question. Generally, we square or Square root both the side, which leads to the wrong solution turning into right for that particular stage. we need to check the final solution obtained with our initial conditions to get the final solution. Similarly, we got 8 pairs at some stage. but the condition of HCF reduced it to 4.