factorial ds

If t is an integer,3^t is a factor of 21!? The highest power of 3 in 21! is 8(3^8) How ? one each from the numbers in bold.

21! =1*2*3*4*5*6*7*8*9*10*11*12*13*14*15*16*17*18*19*20*21

The question can be rephrased to If t is an integer, Is 0<t<9?

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

t can be 2*3 or 2*5 or 3*5.
If t = 6, Is 0<t<9 Yes !
If t = 10, Is Is 0<t<9 No!
Hence Insufficient!

(2) 0 < t < 9

Sufficient!

IMO B

i got value of t between 0 and 10. so can you explain how you come to the value of 3^8 as for me it is 3^9?

Need to figure out what the question is asking, which is "how many threes are in 21!"?

so, you can list out like this

21 - (3)
18 - (3,3)
15 - (3)
12 - (3)
9 - (3,3)
6 - (3)
3 - (3)

By this, we know we have a total of nine 3s

Now, with that said, we need to see if either can fulfill this.

A- No, since 3*5 would go out of scope

B - Yes, anything less than or equal to 9 works.

ANSWER:
B

How did you come up with the following, can you pls explain

Need to figure out what the question is asking, which is "how many threes are in 21!"?

so, you can list out like this

21 - (3)
18 - (3,3)
15 - (3)
12 - (3)
9 - (3,3)
6 - (3)
3 - (3)

As a short cut you could also do this as shown,

3 | 21
3 | 7 - 0
3 | 2 - 1

Its similar to division but keep dividing the number until it becomes smaller than the divisor (neglect the remainder) So you have 7+2 = 9

iwillsurvive101 wrote:How did you come up with the following, can you pls explain

Need to figure out what the question is asking, which is "how many threes are in 21!"?

so, you can list out like this

21 - (3)
18 - (3,3)
15 - (3)
12 - (3)
9 - (3,3)
6 - (3)
3 - (3)