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tarek99
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rampuria
Here's a better approach...

Combination of 4 and 2 will give a product of eight. So if we find the max. no. of pairs of 4 and 2 we can find the max. of x.

Now the no. of 4's that we can extract out of 16! is less than the the no. of 2's. In other words 2's are available in plenty in comparison to 4's. So our answer gets further restricted to the max. no.s of of 4's.

Now the no. of 4's is given by the 16/4 = 4 and 16/ 4^2 =1

therefore 5

Agreed :)
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My approach was to figure out how many 8's I can make out of 16!

16 = 8 X 2 14 = 2 X 7 and so on for all even numbers.

8 can also be arrived at by 2 ^ 3 or 4 X 2 ( such numbers are formed from left out of numbers of 14 , 12 etc)

we get 5 eights and hence x=5
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rampuria
Here's a better approach...

Combination of 4 and 2 will give a product of eight. So if we find the max. no. of pairs of 4 and 2 we can find the max. of x.

Now the no. of 4's that we can extract out of 16! is less than the the no. of 2's. In other words 2's are available in plenty in comparison to 4's. So our answer gets further restricted to the max. no.s of of 4's.

Now the no. of 4's is given by the 16/4 = 4 and 16/ 4^2 =1

therefore 5

so according to your approach, how would you then solve this problem:



What is the greatest value of \(q\) such that \(9^q\) is a factor of 21! ?
(A) 1
(B) 3
(C) 4
(D) 5
(E) 6
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LiveStronger
C - 4

3^8 = 3^2q
q = 4

for whatever reason i m getting 3^9 = 3^2q
Pls help me see why ?

3 6 9 12 15 18 21 = 3^9
1+1+2+1+1+2+1=9



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