factor of 3675 = 3*5^2*7^2
now answer has to be one which is divisible by 3*5^2*7^2
A; 12! ; 12*11*10*9*8*7*6! is not completely divisibly by 3*5^2*7^2 as 7 gets left in dr.
B ; 14! ; is completely divisibly by 3*5^2*7^2
IMO B correct

3675 = \(3*5^2*7^2\)
--> The fact that 2 multiples of 7 are there implies that there should be at least multiples from 1 to 14
--> Least possible value of k = 14

IMO Option B

The largest power of a prime in x! (factorial) is the sum of quotients of x divided by powers of that prime less than x:
ie. 3's in 7! 7/3=2, so there are 3^2 in 7!
ie. 3's in 16! 16/3+16/9=5+1=6, so there are 3^6 in 16!

\(3675=5^27^23\)
\(f(k*)=5^27^23x\)
\(k≥14: 14/5=2…(5^2)…14/7=2…(7^2)…14/3+14/3^2=5…(3^5)\)
\(least(k)=14\)

Ans (B)

3675 can be factorized as 5^2*7^2*3
The highest prime factor is 7 and the power is 2. Hence from options we have to find the least factorial which has power of 7^2
option 1
The highest power of 7 in 12! is 1. hence mot the answer
option 2
The highest power of 7 in 14! is 2. Since the question asked least value of K , this must be the answer.
Hence B

NOTE : HOW TO FIND THE POWER OF PRIME FACTORS IN FACTORIAL ?
for example what is highest power of 2 in 14! ?
Step 1
14!/2 = 7
7/2 = 3
3/2 = 1
hence the highest power of 2 in 14 ! is 7+3+1 = 11

For any integer k greater than 1, the symbol k* denotes the product of all integers between 1 and k, inclusive. If k* is a multiple of 3,675, what is the least possible value of k?

(A) 12
(B) 14
(C) 15
(D) 21
(E) 25
Given,
k* is a multiple of 3,675.
Required: what is the least possible value of k?

Factors of 3,675 = 3*5^2*7^2.
Least possible value of k must be divisible by 3*5^2*7^2.
Option A: k = 12
k*= 12! = 12*11*10*9*8*7*6*5!, NOT

Option B: k= 14
K* = 14!= 14*13*12*10*9*8*7*6*5! must be divisible by 3*5^2*7^2.
Answer: B
Note: Option C, Option D and Option E are not least.

Breaking 3675 into prime factors we have:

3675 = 25 x 147 = 5^2 x 3 x 49 = 5^2 x 3^1 x 7^2

The factorization of 3675 indicates that we need two factors of 7 in our (factorial) answer. Thus, we see that k must be 14 because 14! is the smallest factorial that has two 7’s as factors (one from 7 and the second from 14). Notice that 14! is a multiple of 3675 since 14! also contains two 5’s as factors (one from 5 and the second from 10) and more than one factor of 3.

Answer: B
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k* = 3675x = 5^2*7^2*3x

(A) 12; Not feasible since 7^2 is needed
(B) 14; Feasible since 5^2, 7^2, 3 are available
(C) 15; Feasible since 5^2, 7^2, 3 are available
(D) 21; Feasible since 5^2, 7^2, 3 are available
(E) 25; Feasible since 5^2, 7^2, 3 are available

IMO B
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3675 =3*5*5*7*7
Since we want the least answer let us start from 12
In 12! We have
1*2*3*4*5*6*7*8*9*10*11*12 we have one 3 two 5s( 5&10) but only one 7... So it is not divisible by 3675
Next in B we have
1*2*3*4*5*6*7*8*9*10*11*12*13*14
Here we have a 3, two 5s (5&10) and two 7s(7&14)
So divisible by 3675.
Since the other choices are larger than 14, the answer is B

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