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

Question

For any integer n greater than 1, n* denotes the product of all the integers from 1 to n, inclusive. How many multiples of 4 are there between 4* and 5*, inclusive?

A. 5
B. 6
C. 20
D. 24
E. 25

Could please someone explain me the logic of the solution for this task since the official explanation isn't clear for me

Bunuel · Answer

"n* denotes the product of all the integers from 1 to n, inclusive" means that n*=1*2*3*...*n=n!. Which means that, 4*=4!=24 and 5*=5!=120. So, the question basically asks "how many multiples of 4 are there between 24 and 120, inclusive" # of multiples of 4 between 24 and 120, inclusive is (last-first)/multiple+1=(120-24)/4+1=25 (check this: how-many-multiples-of-4-are-there-between-12-and-94862.html). Answer: E. Hope it helps.

galiya · Answer

I'm familiar w this formula, but i need to grasp it
subtracting 24 from 120 we eliminate some multiples of four between 24 and 120: 100, 104, ...112...

Bunuel · Answer

You can derive it from arithmetic progression formula.

Multiples of 4 represent evenly spaced set (aka arithmetic progression).

Now, if we have  n  terms in arithmetic progression, the first term is  a_1  and the common difference of successive members is  d , then the  n_{th}  term of the sequence is given by:  a_ n=a_1+d(n-1)  -->  n=\frac{a_ n-a_1}{d}+1 .

You can see that it's basically the same formula as in my previous post.

aalba005 · Answer

The more I browse the more tricks I learn, this question would have taken me a good 3 minutes. Thanks for the good explanation.

stunn3r · Answer

here, 4* = 4! and 5* = 5!

question is asking no. of terms in an AP whose first term is 24(4!), last term is 120(5!) and common difference is 4.

last term OR nth term = first term +(no. of tems -1)(common difference)
Tn = a + (n-1)d

120 = 24 + (n-1)4
96/4 = n-1

n = 25 .....

eddy8700 · Answer

4*=4!=24
5*=5!=120

multiples of b/w 24 and 120= (120-24)/4+1=24+1=25
answer E

ScottTargetTestPrep · Answer

We are given that for any integer n greater than 1, n* denotes the product of all the integers from 1 to n inclusive. Thus the notation n* is same as the notation n!.

Thus, 4* = 4! = 24 and 5* = 5! = 120.

We need to determine how many multiples of 4 are between 24 and 120.

(largest multiple of 4 in the set - smallest multiple of 4 in the set)/4 + 1

(120 - 24)/4 + 1 = 96/4 + 1 = 25

Answer: E

Aeyvei · Answer

An alternative approach, which is practically an application of the same principles as @Bunuel's solution but just reduces little arithmetic.

\frac{(5!-4!)}{4} + 1
\frac{2*3*4(5-1)}{4} + 1
= 25

Aeyvei · Answer

An alternative approach, which is practically an application of the same principles as @Bunuel's solution but just reduces little arithmetic and will be suitable for larger numbers.

(5!-4!)/4 + 1
= /4 + 1
= 25

BrushMyQuant · Answer

Given that n∗ denotes the product of all the integers from 1 to n, inclusive and we need to find How many multiples of 4 are there between 4* and 5*, inclusive

4* = Product of all the integers from 1 to 4, inclusive = 1*2*3*4 = 4*1*2*3 = 4*6
5* = Product of all the integers from 1 to 5, inclusive = 1*2*3*4*5 = 4*1*2*3*5 = 4*30

So, number of multiples of 4 from 4* to 5* inclusive = Number of multiples of 4 from 4*6 to 4*30 inclusive = (30-6)+1 = 24 + 1 = 25

So,  Answer will be E .
Hope it helps!

Watch the following video to learn the Basics of Functions and Custom Characters

https://www.youtube.com/watch?v=b7JoMX0gk8M

DrHuber · Answer

In this question, the symbol * is interchangeable with the
concept of "factorial", usually abbreviated with "!". Thus, 4* = 4x3x2x1 =
24, and 5* = 5 x 4 x 3 x 2 x 1 = 120. The question boils down to: how many
multiples of 4 are there between 24 and 120, inclusive. For that, find the difference: 120-24 = 96. Divide by 4: 96/4 = 24. 
Finally, add 1, because the endpoints are both multiples of 4, and the question uses the
word "inclusive": 24 + 1 = 25, choice (E).

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

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