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Question

Rama made necklaces with either 24 beads, 36 beads or 40 beads, such that no bead was left over. What is the least number of beads Rama could have?

This question was previously asked in
RRB Group D 2024 Question Paper PDF (21-Dec-2025) (Shift 2)
The correct answer is

360

To solve the problem of finding the least number of beads Rama could have, let's determine the least common multiple (LCM) of the bead counts for the necklaces she makes: 24, 36, and 40.

The LCM is the smallest number that is a multiple of each of these numbers. To calculate the LCM, we will use the prime factorization method:

  1. Prime factorization of 24: 24 = 2^3 \times 3^1
  2. Prime factorization of 36: 36 = 2^2 \times 3^2
  3. Prime factorization of 40: 40 = 2^3 \times 5^1

Next, we take the highest power of each prime number present in the factorization:

  • 2^3 (from 24 and 40)
  • 3^2 (from 36)
  • 5^1 (from 40)

The LCM is thus calculated as follows:

\[ \text{LCM} = 2^3 \times 3^2 \times 5^1 = 8 \times 9 \times 5 = 360 \]

So, the least number of beads Rama could have, ensuring no beads are left over in any type of necklace, is 360.

Let's verify the answer by dividing 360 by each number:

  • 360 \div 24 = 15 (Whole number)
  • 360 \div 36 = 10 (Whole number)
  • 360 \div 40 = 9 (Whole number)

Since 360 can be divided evenly by 24, 36, and 40, the answer is confirmed. Therefore, 360 is the correct answer.

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Similar Questions

  1. The ratio of two numbers is \(3:4\) and their HCF is 7. Their LCM is:

  2. Find the LCM of the numbers 780, 2080 and 2600.

  3. The sum of two numbers is 88 and their L.C.M. is 870. The two numbers are:


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

  3. 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.

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

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

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