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Bunuel
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rkhosla
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35 ! = 1*2*3*4*5...*35
Count powers of 3 which comes out to 15
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Won't it be 10? C that is... since there are 11 multiples of 3 in the expansion of 35! (AP from 3 to 33 with c.d of 3). Thus cannot be 15 or 12, must be 10
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Won't it be 10? C that is... since there are 11 multiples of 3 in the expansion of 35! (AP from 3 to 33 with c.d of 3). Thus cannot be 15 or 12, must be 10

9 has two powers of 3
18 has two powers
27 has 3 powers

Total 7

Only three powers of 3 are accounted for by dividing as you did, so 4 more need to be added for a total of 15

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Here is my approach:
35!=1 * 2* 3* . . . 35!
I just looked at multiples of 3 since that is what I need to get a numberator portion that is a mutiple of 3
Hence need to find 3*6*9*12*15*18*21*24*27*30*33
Extract the 3s 3^15
Answer: A
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The question asks for the power of 3 in 35!
35/3 + 35/9 +35/27 = 11+3+1 = 15
A
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