Find the least possible numbers with which 270! should be multiplied so that it can be exactly divisible by 54^48?
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Answer:
3^10
Step-by-step explanation:
firstly, let's find the power of 54 that divides 270! completely
Since, 54=2*3*3*3--->We require three 3s because the restriction on the power is based on the number of 3s and not on the number of 2s.
Total number of 3s = [270/3]+[270/9]+[270/27]+[270/81]+[270/243]=
90+30+10+3+1= 134 threes in 270!
Therefore, the power of 54 that divides 270! completely is [134/3]=44 and remainder is 2 [i.e. two 3s are extra]
270!/(54)^(44) leaves no remainder.
Secondly, according to the question the power is 48, there the extra power is 4.------>(28*3*3*3)^4
therefore the power of 3 that needs to be multiplied is 3^12
but since we already have two extra 3s, we need to multiply only 3^10
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