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The least possible number with which 270! should be multiplied so that it can be exactly divisible by (54)48 is?
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The least possible number with which 270! should be multiplied so that...



Calculating the Least Possible Number for Divisibility

To find the least possible number with which 270! should be multiplied so that it can be exactly divisible by (54)48, we need to consider the highest power of each prime factor in the divisor.

Prime Factorization of (54)48

  • (54)48 = 2^48 * 3^48


Prime Factorization of 270!

  • 270! = 2^x * 3^y * ...


Determining the Least Possible Number

In order for 270! to be divisible by (54)48, we need to ensure that the power of 2 in the factorization of 270! is at least 48, and the power of 3 is at least 48.

Let's find the powers of 2 and 3 in 270!:

  • Power of 2 in 270! = floor(270/2) + floor(270/4) + floor(270/8) + ... = 135 + 67 + 33 + ... = 135 + 67 + 33 + 16 + 8 + 4 + 2 + 1 = 266
  • Power of 3 in 270! = floor(270/3) + floor(270/9) + floor(270/27) = 90 + 30 + 10 = 130


Since the powers of both 2 and 3 in 270! are already greater than 48, we don't need to multiply 270! by any additional number to make it divisible by (54)48.

Conclusion

The least possible number with which 270! should be multiplied so that it can be exactly divisible by (54)48 is 1.


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The least possible number with which 270! should be multiplied so that it can be exactly divisible by (54)48 is?
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