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Find the smallest number which when divided by 25, 40 and 60 leaves remainder 7 in each case?
  • a)
    607                             
  • b)
    608          
  • c)
    609                             
  • d)
    610
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Find the smallest number which when divided by 25, 40 and 60 leaves re...

5×4×5×2×3=600
Required numbers is 600+7=607
This question is part of UPSC exam. View all Class 5 courses
Most Upvoted Answer
Find the smallest number which when divided by 25, 40 and 60 leaves re...
To find the smallest number that leaves a remainder of 7 when divided by 25, 40, and 60, we can use the concept of the least common multiple (LCM).

Finding the LCM:
1. Begin by finding the prime factorization of each number:
- 25 = 5 * 5
- 40 = 2 * 2 * 2 * 5
- 60 = 2 * 2 * 3 * 5

2. Identify the highest power of each prime factor present in any of the numbers:
- 2^3, 3^1, 5^2

3. Multiply these prime factors with their highest power to find the LCM:
- LCM = 2^3 * 3^1 * 5^2 = 8 * 3 * 25 = 600

Finding the smallest number:
1. Since we need to find the smallest number that leaves a remainder of 7 when divided by 25, 40, and 60, we can add the remainder (7) to the LCM (600) to get the desired number:
- Smallest number = LCM + remainder = 600 + 7 = 607

Therefore, the smallest number that leaves a remainder of 7 when divided by 25, 40, and 60 is 607.
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Find the smallest number which when divided by 25, 40 and 60 leaves remainder 7 in each case?a)607 b)608c)609 d)610Correct answer is option 'A'. Can you explain this answer?
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