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Question

The sum of two positive numbers is 240 and their HCF is 15. Find the number of pairs of numbers satisfying the given condition.

This question was previously asked in
SSC CGL 2019 (Tier 2) GS Finance & Economics Previous Year Paper (17-Nov-2020)
The correct answer is

4

Understanding the Problem of Number Pairs and HCF

The question asks us to find the number of unique pairs of positive integers that satisfy two conditions: their sum is 240, and their Highest Common Factor (HCF) is 15.

Let the two positive numbers be \(a\) and \(b\). We are given:

  • \(a + b = 240\)
  • HCF\((a, b) = 15\)

When the HCF of two numbers \(a\) and \(b\) is 15, it means that both \(a\) and \(b\) are multiples of 15. We can write \(a\) and \(b\) in terms of their HCF and two other numbers.

Let \(a = 15x\) and \(b = 15y\), where \(x\) and \(y\) are positive integers.

An important property when expressing numbers this way is that the numbers \(x\) and \(y\) must be coprime. Coprime means that their Highest Common Factor is 1, i.e., HCF\((x, y) = 1\) or gcd\((x, y) = 1\). This ensures that 15 is indeed the highest common factor of \(a\) and \(b\).

Setting up the Equation with HCF

Now substitute \(a = 15x\) and \(b = 15y\) into the sum equation:

\(15x + 15y = 240\)

We can factor out 15 from the left side:

\(15(x + y) = 240\)

To find the sum of \(x\) and \(y\), divide both sides by 15:

\(x + y = \frac{240}{15}\)

\(x + y = 16\)

Finding Coprime Pairs (x, y) that Sum to 16

We now need to find pairs of positive integers \((x, y)\) such that their sum is 16 and they are coprime (gcd\((x, y) = 1\)). Since \(x\) and \(y\) are interchangeable for the pair \(\{a, b\}\), we can list pairs \((x, y)\) where \(x \le y\) to avoid counting the same pair of numbers \(\{a, b\}\) twice, but we must ensure we check the coprime condition for each pair.

Let's list the pairs of positive integers \((x, y)\) whose sum is 16:

  • (1, 15)
  • (2, 14)
  • (3, 13)
  • (4, 12)
  • (5, 11)
  • (6, 10)
  • (7, 9)
  • (8, 8)

Checking the Coprime Condition (gcd(x, y) = 1)

Now we check the HCF (or gcd) for each pair \((x, y)\) to see which ones are coprime:

Pair (x, y) Sum (x + y) gcd(x, y) Coprime?
(1, 15) 16 gcd(1, 15) = 1 Yes
(2, 14) 16 gcd(2, 14) = 2 No
(3, 13) 16 gcd(3, 13) = 1 Yes
(4, 12) 16 gcd(4, 12) = 4 No
(5, 11) 16 gcd(5, 11) = 1 Yes
(6, 10) 16 gcd(6, 10) = 2 No
(7, 9) 16 gcd(7, 9) = 1 Yes
(8, 8) 16 gcd(8, 8) = 8 No

The pairs \((x, y)\) that are coprime are (1, 15), (3, 13), (5, 11), and (7, 9).

Each of these coprime pairs corresponds to a unique pair of positive numbers \((a, b)\) with HCF 15 and sum 240:

  • For (x, y) = (1, 15): \(a = 15 \times 1 = 15\), \(b = 15 \times 15 = 225\). Pair: (15, 225). Sum = 15 + 225 = 240. HCF(15, 225) = 15.
  • For (x, y) = (3, 13): \(a = 15 \times 3 = 45\), \(b = 15 \times 13 = 195\). Pair: (45, 195). Sum = 45 + 195 = 240. HCF(45, 195) = 15.
  • For (x, y) = (5, 11): \(a = 15 \times 5 = 75\), \(b = 15 \times 11 = 165\). Pair: (75, 165). Sum = 75 + 165 = 240. HCF(75, 165) = 15.
  • For (x, y) = (7, 9): \(a = 15 \times 7 = 105\), \(b = 15 \times 9 = 135\). Pair: (105, 135). Sum = 105 + 135 = 240. HCF(105, 135) = 15.

We found 4 such coprime pairs \((x, y)\). Each coprime pair corresponds to a distinct pair of numbers \(\{a, b\}\).

Therefore, there are 4 pairs of numbers satisfying the given conditions.

Conclusion on the Number of Pairs

Based on our analysis, there are 4 pairs of positive numbers whose sum is 240 and whose HCF is 15.

Revision Table: Key Concepts for Number Problems

Concept Definition/Property Application in this Problem
HCF (Highest Common Factor) The largest positive integer that divides each of the integers. Used to express numbers as \(a=HCF \times x\) and \(b=HCF \times y\).
Coprime Numbers Two integers are coprime (or relatively prime) if their HCF is 1. \(x\) and \(y\) must be coprime for HCF\((15x, 15y)\) to be 15.
Sum of Numbers The result of adding numbers. Used to form the equation \(15x + 15y = 240\).

Additional Information: General Approach for HCF and Sum/Product Problems

Problems involving the sum or product of two numbers and their HCF can often be solved using the approach demonstrated above.

General steps:

  1. Represent the two numbers as \(a = \text{HCF} \times x\) and \(b = \text{HCF} \times y\), where \(x\) and \(y\) are positive integers and gcd\((x, y) = 1\).
  2. Use the given condition (sum or product) to form an equation in terms of \(x\) and \(y\).
    • If sum \(S\) is given: \(\text{HCF}(x + y) = S \implies x + y = S / \text{HCF}\).
    • If product \(P\) is given: \((\text{HCF} \times x)(\text{HCF} \times y) = P \implies \text{HCF}^2 \times xy = P \implies xy = P / \text{HCF}^2\).
  3. Find pairs of positive integers \((x, y)\) that satisfy the equation from step 2 AND the coprime condition gcd\((x, y) = 1\).
  4. Each valid coprime pair \((x, y)\) corresponds to a unique pair of the original numbers \((a, b)\). Count the number of such coprime pairs.

This systematic approach helps ensure you find all possible pairs and correctly apply the properties of HCF and coprime numbers.

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Similar Questions

  1. A and B are two prime numbers such that A > B and their LCM is 209. The value of B 2 –  A is:

  2. Choose the option in which the numbers are in correct ascending order.

  3. The HCF of two numbers is 17 and the other two factors of their LCM are 11 and 19. The smaller of the two numbers is:

  4. The HCF of two numbers is 8 and their LCM is 2520. If one of the numbers is 56, then the other number is:

  5. The LCM of 1.2 and 2.7 is:

  6. Find the HCF of 4.08 and 6.63.

  7. The HCF of three numbers 98, 175 and 210 will be:

  8. Determine the LCM of two numbers if their HCF is 9 and their ratio is 14 : 19.

  9. Find the HCF of 60, 148 and 382.

  10. If the highest common factor (HCF) of x and y is 15, then the HCF of 36x2 - 81y2 and 81x2 - 9y2 is divisible by ______.


Important Questions from LCM and HCF

  1. The HCF and LCM of two numbers are 12 and 72, respectively. If the ratio of the two numbers is 2 ∶ 3, then the larger of the two numbers is:

  2. Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.

  3. Joseph visits the club on every 5 th day, Harsh visits on every 24 th day, while Sumit visits on every 9 th day. If all three of them met at the club on a Sunday, then on which day will all three of them meet again?

  4. What is the least number which when divided by 12,20 and 24 leaves in each case a remainder of 8?

  5. Which of the following is a pair of co-primes?

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