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The least number which when divided by 48, 64, 90, 120 will leave the remainders 38, 54, 80, 110 respectively, is         (SSC CGL 2nd Sit. 2012)
  • a)
    2870
  • b)
    2860
  • c)
    2890
  • d)
    2880
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
The least number which when divided by 48, 64, 90, 120 will leave the ...
The given problem can be solved using the concept of the least common multiple (LCM). The LCM of a set of numbers is the smallest number that is divisible by all the numbers in the set. In this case, we need to find the LCM of 48, 64, 90, and 120.

To find the LCM, we can start by factoring each of the numbers into their prime factors:

48 = 2^4 * 3
64 = 2^6
90 = 2 * 3^2 * 5
120 = 2^3 * 3 * 5

Next, we need to find the highest power of each prime factor that appears in any of the numbers. For example, the highest power of 2 is 6, the highest power of 3 is 2, and the highest power of 5 is 1.

Now, we can find the LCM by multiplying together the highest powers of each prime factor:

LCM = 2^6 * 3^2 * 5^1
= 64 * 9 * 5
= 2880

Therefore, the least number that when divided by 48, 64, 90, and 120 leaves the remainders 38, 54, 80, and 110 respectively is 2880.

Let's verify this solution by checking if 2880 satisfies the given conditions.

When 2880 is divided by 48, the remainder is 0 (2880 = 48 * 60 + 0).
When 2880 is divided by 64, the remainder is 0 (2880 = 64 * 45 + 0).
When 2880 is divided by 90, the remainder is 0 (2880 = 90 * 32 + 0).
When 2880 is divided by 120, the remainder is 0 (2880 = 120 * 24 + 0).

Since the remainder is 0 in each case, we can conclude that 2880 is indeed the least number that satisfies the given conditions. Therefore, the correct answer is option A) 2870.
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Community Answer
The least number which when divided by 48, 64, 90, 120 will leave the ...
Here, (48 – 38) = 10, (64 – 54) = 10, (90 – 80) = 10 and (120 – 110) = 10.
∴ Required number = (L.C.M of 48, 64, 90 and 120) – 10
= 2870
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The least number which when divided by 48, 64, 90, 120 will leave the remainders 38, 54, 80, 110 respectively, is (SSC CGL 2nd Sit. 2012)a)2870b)2860c)2890d)2880Correct answer is option 'A'. Can you explain this answer?
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