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Find the smallest number when divided by 30,32,48 and 24 leaves a remainder 5 in each case?
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Find the smallest number when divided by 30,32,48 and 24 leaves a rema...
Smallest number that leaves a remainder of 5 when divided by 30, 32, 48, and 24
To find the smallest number that leaves a remainder of 5 when divided by 30, 32, 48, and 24, we need to find the least common multiple (LCM) of these numbers and then add 5 to that LCM.

Finding the LCM of 30, 32, 48, and 24
- Prime factorization of the numbers:
- 30 = 2 * 3 * 5
- 32 = 2^5
- 48 = 2^4 * 3
- 24 = 2^3 * 3
- LCM will have all prime factors with the highest powers that appear in any of the numbers:
- LCM = 2^5 * 3 * 5 = 480

Calculating the smallest number
- Add 5 to the LCM:
- Smallest number = 480 + 5 = 485
Therefore, the smallest number that leaves a remainder of 5 when divided by 30, 32, 48, and 24 is 485.
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Find the smallest number when divided by 30,32,48 and 24 leaves a remainder 5 in each case?
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