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Question

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

The correct answer is

350

Understanding Prime Numbers and LCM

The question asks us to find the value of \(A^2 - B\), where \(A\) and \(B\) are prime numbers, \(A > B\), and their Least Common Multiple (LCM) is 209.

Let's first understand what prime numbers are. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. Examples include 2, 3, 5, 7, 11, 13, 17, 19, etc.

The LCM of two numbers is the smallest positive integer that is a multiple of both numbers.

LCM of Two Prime Numbers

A key property of two distinct prime numbers is that their only common factor is 1. This means their Highest Common Factor (HCF) is 1.

For any two numbers, say X and Y, the product of their LCM and HCF is equal to the product of the numbers themselves:

\(\text{LCM}(X, Y) \times \text{HCF}(X, Y) = X \times Y\)

Since \(A\) and \(B\) are prime numbers and \(A > B\), they must be distinct. Thus, their HCF is 1.

\(\text{HCF}(A, B) = 1\)

Using the relationship above, we get:

\(\text{LCM}(A, B) \times 1 = A \times B\)

So, the LCM of two distinct prime numbers is simply their product.

Finding the Prime Numbers A and B

We are given that \(\text{LCM}(A, B) = 209\). Since \(A\) and \(B\) are distinct prime numbers, we know that \(A \times B = \text{LCM}(A, B)\).

\(A \times B = 209\)

To find \(A\) and \(B\), we need to find the prime factors of 209. We can test small prime numbers:

  • 209 is not divisible by 2 (it's odd).
  • The sum of digits is \(2+0+9 = 11\), which is not divisible by 3, so 209 is not divisible by 3.
  • 209 does not end in 0 or 5, so it's not divisible by 5.
  • \(209 \div 7\). \(7 \times 20 = 140\), \(209 - 140 = 69\). 69 is not a multiple of 7. \(7 \times 9 = 63\), \(7 \times 10 = 70\). So, 209 is not divisible by 7.
  • \(209 \div 11\). \(11 \times 10 = 110\), \(209 - 110 = 99\). \(11 \times 9 = 99\). So, \(11 \times 10 + 11 \times 9 = 11 \times (10+9) = 11 \times 19\).

So, the prime factors of 209 are 11 and 19. Both 11 and 19 are prime numbers.

We have \(A \times B = 11 \times 19\).

We are given the condition that \(A > B\). Since 19 is greater than 11, we must have:

\(A = 19\)

\(B = 11\)

Let's verify: Are A and B prime numbers? Yes, 19 and 11 are prime. Is \(A > B\)? Yes, \(19 > 11\). Is their LCM 209? Yes, \(\text{LCM}(19, 11) = 19 \times 11 = 209\). The values fit all the conditions.

Calculating the Value of A² - B

Now that we have \(A = 19\) and \(B = 11\), we can calculate the value of \(A^2 - B\).

\(A^2 - B = 19^2 - 11\)

First, calculate \(19^2\):

\(19^2 = 19 \times 19\)

1 9
19 1 × 1 = 1 1 × 9 = 9
9 × 1 = 9 9 × 9 = 81
Multiply and Add diagonally:
1
9 + 9 = 18 (carry 1)
81 (carry 8 from 81, add 8 to 18+1=19 --> 19, carry 1 from 19, add 1 to 1 --> 2)
Or standard multiplication:
19
x 19
---
171 (19 × 9)
190 (19 × 10)
---
361

So, \(19^2 = 361\).

Now substitute this value back into the expression \(A^2 - B\):

\(A^2 - B = 361 - 11\)

\(361 - 11 = 350\)

The value of \(A^2 - B\) is 350.

Conclusion

We found that the two prime numbers A and B, with \(A > B\) and LCM 209, are \(A=19\) and \(B=11\). We then calculated \(A^2 - B\) as \(19^2 - 11 = 361 - 11 = 350\).

Revision Table: Key Steps

Step Action Result
1 Understand properties of prime numbers and LCM for primes. For distinct primes A, B, LCM(A, B) = A × B.
2 Set up equation using given LCM. A × B = 209.
3 Find prime factors of 209. 209 = 11 × 19.
4 Assign values to A and B based on A > B. A = 19, B = 11.
5 Calculate \(A^2\). \(19^2 = 361\).
6 Calculate \(A^2 - B\). \(361 - 11 = 350\).

Additional Information: Properties of Prime Numbers and LCM

  • Prime Numbers: Numbers greater than 1 divisible only by 1 and themselves (e.g., 2, 3, 5, 7, 11, 13, 17, 19, ...). The number 1 is not a prime number. 2 is the only even prime number.
  • Composite Numbers: Natural numbers greater than 1 that are not prime (e.g., 4, 6, 8, 9, 10, 12, ...).
  • Prime Factorization: Expressing a composite number as a product of its prime factors. For example, \(12 = 2^2 \times 3\). Finding prime factors is crucial for calculating LCM and HCF.
  • HCF (Highest Common Factor): The largest positive integer that divides both numbers without leaving a remainder. For example, HCF(12, 18) = 6.
  • LCM (Least Common Multiple): The smallest positive integer that is a multiple of both numbers. For example, LCM(12, 18) = 36.
  • Relationship between LCM and HCF: For any two positive integers a and b, \(\text{LCM}(a, b) \times \text{HCF}(a, b) = a \times b\). This property is particularly simple for distinct prime numbers where HCF is always 1.
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Important Questions from LCM and HCF

  1. Six bells begin to toll together and toll, respectively, at intervals of 3, 4, 6, 7, 8 and 12 seconds. After how many seconds, will they toll together again?

  2. Find the least number which when divided by 12, 18, 24 and 30 leaves 4 as remainder in each case, but when divided by 7 leaves no remainder.

  3. Calculate the HCF of \(\frac{12}{5}\) \(\frac{14}{15}\)  and  \(\frac{16}{17}\) .

  4. Three numbers are in the proportion of 3 : 8 : 15 and their LCM is 8280. What is their HCF?

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

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