The product of two integers p and q, where p > 60 and q > 60, is 7168 and their HCF is 16. The sum of these two integers is:
176
Understanding the Problem: Integers, HCF, and Product
We are given two integers, let's call them p and q. We know that both p and q are greater than 60. We are also given their product, which is 7168, and their Highest Common Factor (HCF), which is 16. Our goal is to find the sum of these two integers, p + q.Key Relationship: Product, HCF, and LCM
A fundamental property of two positive integers is that their product is equal to the product of their HCF and their Least Common Multiple (LCM). Let p and q be two integers. Product of p and q = \(p \times q\) HCF of p and q = HCF(p, q) LCM of p and q = LCM(p, q) The relationship is: \(p \times q = \text{HCF}(p, q) \times \text{LCM}(p, q)\)Representing Integers Using HCF
If the HCF of two integers p and q is 16, we can express these integers in terms of their HCF. Let \(p = 16a\) and \(q = 16b\), where 'a' and 'b' are two integers that are coprime. Coprime means that the HCF of 'a' and 'b' is 1, i.e., HCF(a, b) = 1. This is because if 'a' and 'b' had a common factor greater than 1, say 'k', then 16k would be a common factor of p and q, which would contradict 16 being the *highest* common factor.Using the Product to Find 'a' and 'b'
We know that the product of p and q is 7168. Substitute the expressions for p and q: \(p \times q = 7168\) \((16a) \times (16b) = 7168\) \(16 \times 16 \times a \times b = 7168\) \(256 \times a \times b = 7168\) Now, we can solve for the product \(a \times b\): \(a \times b = \frac{7168}{256}\) Let's perform the division: \(a \times b = 28\)Finding Coprime Factors 'a' and 'b' with Conditions
We need to find pairs of integers (a, b) such that their product \(a \times b = 28\) and they are coprime (HCF(a, b) = 1). We also have the conditions that p > 60 and q > 60. Since \(p = 16a\) and \(q = 16b\), the conditions \(p > 60\) and \(q > 60\) translate to: \(16a > 60 \implies a > \frac{60}{16} \implies a > 3.75\) \(16b > 60 \implies b > \frac{60}{16} \implies b > 3.75\) So, both 'a' and 'b' must be integers greater than 3.75. Let's list the pairs of factors for 28 and check the conditions:| Pair (a, b) where a × b = 28 | Is HCF(a, b) = 1? (Coprime) | Is a > 3.75? | Is b > 3.75? | Valid Pair? |
|---|---|---|---|---|
| (1, 28) | Yes (HCF=1) | No (1 < 3.75) | Yes (28 > 3.75) | No |
| (2, 14) | No (HCF=2) | No (2 < 3.75) | Yes (14 > 3.75) | No |
| (4, 7) | Yes (HCF=1) | Yes (4 > 3.75) | Yes (7 > 3.75) | Yes |
The only pair of coprime factors (a, b) for 28 where both a and b are greater than 3.75 is (4, 7).
Calculating the Integers and Their Sum
Using the valid pair (a, b) = (4, 7), we can find the integers p and q: \(p = 16a = 16 \times 4 = 64\) \(q = 16b = 16 \times 7 = 112\) Let's quickly verify the original conditions: - p > 60? 64 > 60 (Yes) - q > 60? 112 > 60 (Yes) - Product p * q = 64 * 112 = 7168 (Yes) - HCF(64, 112)? 64 = \(2^6\) 112 = \(2^4 \times 7\) HCF(64, 112) = \(2^4 = 16\) (Yes) All conditions are satisfied by the integers 64 and 112. Finally, we need to find the sum of these two integers: Sum = \(p + q = 64 + 112 = 176\) The sum of the two integers is 176.Revision Table: Integer Properties
This table summarizes the key information used in solving this problem related to integer properties.| Concept | Description | How it was used |
|---|---|---|
| HCF (Highest Common Factor) | The largest positive integer that divides two or more integers without leaving a remainder. | Used to express the unknown integers as multiples of the HCF (p=16a, q=16b). |
| Product of Two Integers | The result of multiplying the two integers. | Used to form an equation to find the product of the coprime parts (ab=28). |
| Relationship: Product = HCF × LCM | A fundamental property connecting the product, HCF, and LCM of two numbers. | Implied in the method; expressing numbers as HCF * coprime part relies on this relationship. |
| Coprime Numbers | Two integers are coprime (or relatively prime) if their HCF is 1. | The 'a' and 'b' in p=Ha and q=Hb must be coprime for H to be the *highest* common factor. Essential for identifying the correct pair (a, b). |
| Conditions on Integers | Specific requirements for the integers (e.g., p > 60, q > 60). | Used to filter the possible pairs of (a, b) derived from the product, ensuring the final numbers meet all problem constraints. |
Additional Information: Solving Similar Integer Problems
Solving problems involving the product and HCF of two integers often follows a standard pattern. Here are some additional points and strategies:- Let the two integers be p and q, and their HCF be H. Then \(p = Ha\) and \(q = Hb\), where HCF(a, b) = 1.
- The product \(p \times q = (Ha) \times (Hb) = H^2 \times a \times b\). This is a very useful formula derived from the product property.
- If you are given the product P and HCF H, you can find the product of the coprime parts 'a' and 'b' using \(ab = P / H^2\). In this problem, \(ab = 7168 / 16^2 = 7168 / 256 = 28\).
- Once you have \(ab\), list all pairs of factors (a, b) for that product.
- Check each pair to see if HCF(a, b) = 1. Discard pairs that are not coprime.
- Apply any other conditions given in the problem (like p > 60, q > 60) to the remaining coprime pairs. Remember to use \(p=Ha\) and \(q=Hb\) when checking these conditions.
- The pair that satisfies all conditions is the correct (a, b).
- Calculate the integers p and q using \(p = Ha\) and \(q = Hb\).
- Finally, calculate the required value, such as the sum (p+q), difference (p-q), or LCM (which would be \(H \times a \times b\)).
Similar Questions
If (x + k) is the HCF of x 2+ 5x + 6 and x 2+ 8x + 15, then what is the value of k?
A floor of a big hall has dimensions 30 m 60 cm and 23 m 40 cm. It is to be paved with square tiles of same size. What is the minimum number of tiles required ?
The LCM of two prime numbers p and q is 2231, where p > q. What is the value of p - q ?
Three runners are running in a circular track, and they complete one round in 20, 30 and 35 minutes respectively. When will they next meet at the starting point ?
What is the least perfect square which is divisible by 3, 4, 5, 6 and 7?
If (x - k) is the HCF of x 2+ ax + b and x 2+ cx + d, then what is the value of k?
HCF of two numbers is 12. Which one of the following can never be their LCM?
X, Y and Z start at some point and same time in the same direction to run around a circular stadium. X completes a round in 252 seconds, Y in 308 seconds and Z in 198 seconds. After what time will they meet again at the starting point?
What is the LCM of the polynomials x 3+ 3x 2+ 3x + 1, x 3+ 5x 2+ 5x + 4 and x 2+ 5x + 4?
HCF and LCM of two polynomials are (x + 3) and (x 3- 9x 2- x + 105). If one of the two polynomials is (x 2- 4x - 21), then the other is
Important Questions from LCM and HCF
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:
Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.
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?
What is the least number which when divided by 12,20 and 24 leaves in each case a remainder of 8?
Which of the following is a pair of co-primes?
