3^q is a factor of (300!/100!) where q is a positive integer. What is

Question

3^q is a factor of (300!/100!) where q is a positive integer. What is the highest possible value of q?

(A) 67 
(B) 89 
(C) 100 
(D) 114 
(E) 148

Lampard42 · Accepted Answer

No. of 3 in 100! is 100/3+100/9+100/27+100/81=33+11+3+1=48
No. of 3 in 300! is 300/3+300/9+300/27+300/81+300/243=100+33+11+3+1=148

So the 300!/100! prime factorized by 3 will be 3^148/3^48; Using exponent properties 3^100 IMO C

ScottTargetTestPrep · Answer

The number of factors of 3 in 300! is:

300/3 = 100

100/3 = 33

33/3 = 11

11/3 = 3

3/3 = 1

So there are 100 + 33 + 11 + 3 + 1 = 148 factors of 3 in 300!.

The number of factors of 3 in 100! is:

100/3 = 33

33/3 = 11

11/3 = 3

3/3 = 1

So there are 33 + 11 + 3 + 1 = 48 factors of 3in 300!.

Thus, there are 148 - 48 = 100 factors of 3 in (300!/100!), so the max value of q is 100.

Answer: C

CareerGeek · Answer

Definitely a >2 min question

300!/100! = 101*102*103* . . . . . . . . *298*299*300

Multiples of 3 from 101 to 300 are 
102, 105, 108 . . . . .300 (AP Series, Use last term to find number of multiples)
--> 300 = 102 + (x - 1)3
--> 198/3 = x - 1
--> x = 67

Multiples of 3^2 (9) from 101 to 300 are 
108, 117, 126 . . . . . 297 (AP Series, Use last term to find number of multiples)
--> 297 = 108 + (y - 1)9
--> 189/9 = y - 1
--> y = 22

Multiples of 3^3 (27) from 101 to 300 are 
108, 135, 162 . . . . . 297 (AP Series, Use last term to find number of multiples)
--> 297 = 108 + (z - 1)27
--> 189/27 = z - 1
--> z = 8

Multiples of 3^4 (81) from 101 to 300 are 
162, 243 = 2

Multiples of 3^5 (243) from 101 to 300 are 
243 = 1

Total factors = 67 + 22 + 8 + 2 + 1 = 100

So, 300!/100! = 3^100*k, where k is any non-multiple of 3

IMO Option C

Archit3110 · Answer

factors of for 300! ; 300!/3 + 300!/9+300!/27+300!/81+300!/243 100+33+11+3+1=148

factors of 3 for 100! ; 100!/3+ 100!/9+... = 48
so 3^148/3^48 ; 3^100
IMO C

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3^q is a factor of (300!/100!) where q is a positive integer. What is

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