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If t is an integer, is 3^t a factor of 21! ?

(1) t is the product of two distinct single-digit prime numbers that are smaller than 7.
(2) 0 < t < 9


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The powers of 3 in 21! are 9
So, basically the question translates to:
\(\frac{3^9}{3^t}?\), if we can prove that t<=9, we should be good.

1)Case1: t = 2*3 = 6, this makes \(3^t\) a factor
Case2: t = 3*5 = 15, this doesn't make \(3^t\) a factor - Insufficient!
2)Giving the range for t, clearly the max possible value of t = 8. So, sufficient!
IMO, (B)
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The powers of 3 in 21!

21/3 + 21/9 + 21/ 27... = 7+ 2 +0 = 9

So up to 3^9 will be a factor of 21!

1) It could be under 9 but it could be over.

Not sufficient

2) 0<t<9
Sufficient

Answer B
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If t is an integer, is 3^t a factor of 21! ?

(1) t is the product of two distinct single-digit prime numbers that are smaller than 7.
t can take = 2*3 =6
2*5 = 10
3*5 = 15

We need check if 3^6 , 3^10 and 3^15 are factors of 21!
Roughly we can check for the number of times we have 3 is 21! which is 9.
So 3^6 is a factor but 3^10 is not a factor of 21!
Not sufficient.

(2) 0 < t < 9
Since t is a integer it can takes 2,3,4,5,6,7,8 only here
And we clearly saw 21! has 9 times 3
Hence sufficient.

IMO B
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If t is an integer, is 3^t a factor of 21! ?
t = [21/3] + [21/9] = 9, so question is t < =9 or not?

Stat1: t is the product of two distinct single-digit prime numbers that are smaller than 7.
if prime nos. are 2 and 3 then 2*3 =6, then t <=9, but if prime nos. are 2 and 5 then 2*5 =10, then t >9. So, Not sufficient.

Stat2: 0 < t < 9. It satisfies the condition. Sufficient.

So, I think B. :)
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