For any integer k greater than 1, the symbol k* denotes the product of

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

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

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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K* denotes ---> K!


Concept: for K! to be a Multiple of 3,675 ---> K! must be divisible by Each of the Factors of 3,675

in other words, if K! is divisible by 3,675's Prime Factorization, then K! will be divisible by Each of the Factors of 3,675 ----> and will thus be Divisible by 3,675 itself


3,675 = 25 * 147 = (5)(5) * (7) (21) = (3) * (5)^2 * (7)^2


K! is the Multiplication of: (1) (2) (3) (4) ...... up through *(K)

what is the minimum Factorial that will include 2 Prime Factors of 7 in its Prime Factorization?

(*note* if the Factorial includes two 7's in its Prime Factorization, it will surely contain two 5's and one 3)

we need at least 2 Multiples of 7 in the Multiplication

14! = 1 * 2 * 3 .... * 7 ..... * 14


since 7 and 14 are the first two multiples of 7 ----> 14! is the minimum value that is divisible by 3,675

essentially asking which factorial satisfies the 3,675 multiple condition
factors of 3,675 are 5^2* 7^2 * 3
14! has 2-5s and 2-7s clearly satisfying the condition
Therefore IMO B

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