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Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  Let f(x,y) be defined as the remainder when (x–y)! is divided by x. If

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Intern  Joined: 08 May 2014
Posts: 10
Concentration: General Management, Operations
Let f(x,y) be defined as the remainder when (x–y)! is divided by x. If  [#permalink]

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Question Stats: 64% (01:50) correct 36% (01:42) wrong based on 236 sessions

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Let f(x,y) be defined as the remainder when (x–y)! is divided by x. If x=25, what is the maximum value of y for which f(x,y)=0

A. 5
B. 10
C. 15
D. 20
E. 25

Originally posted by prakash85 on 06 Nov 2014, 03:55.
Last edited by Bunuel on 06 Nov 2014, 05:26, edited 1 time in total.
Renamed the topic and edited the question.
Senior Manager  Joined: 13 Jun 2013
Posts: 266
Re: Let f(x,y) be defined as the remainder when (x–y)! is divided by x. If  [#permalink]

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prakash85 wrote:
Let f(x,y) be defined as the remainder when (x–y)! is divided by x. If x=25, what is the maximum value of y for which f(x,y)=0

1. 5
2. 10
3. 15
4.20
5. 25

we have to find the value of y for which (25-y)!/ 25 yields zero remainder

if y=20, then (25-20)!=5!, which clearly is not divisible by 5^2
thus our answer will be less than 20

let's try 15 (25-15)!=10!, which is clearly divisible by 5^2.

hence C
Intern  Joined: 15 Sep 2014
Posts: 5
Re: Let f(x,y) be defined as the remainder when (x–y)! is divided by x. If  [#permalink]

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1/25 no reminder

So simple

(X-y)!/x = no reminder
If x = 25
Then (25-x)!/25 = no reminder
Let y = 25
Then 0!/25 , 0! = 1
So 1/25 which have 0 reminder
Math Expert V
Joined: 02 Sep 2009
Posts: 58465
Re: Let f(x,y) be defined as the remainder when (x–y)! is divided by x. If  [#permalink]

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Bhanupriyachauhan wrote:
1/25 no reminder

So simple

(X-y)!/x = no reminder
If x = 25
Then (25-x)!/25 = no reminder
Let y = 25
Then 0!/25 , 0! = 1
So 1/25 which have 0 reminder

That's not true. 1 divided by 25 yields the remainder of 1, not 0.
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Intern  Joined: 15 Sep 2014
Posts: 5
Re: Let f(x,y) be defined as the remainder when (x–y)! is divided by x. If  [#permalink]

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Totally agree.
It was my mistake.
Sorry

Posted from my mobile device
Manager  Joined: 21 Jul 2014
Posts: 119
Re: Let f(x,y) be defined as the remainder when (x–y)! is divided by x. If  [#permalink]

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manpreetsingh86 wrote:
prakash85 wrote:
Let f(x,y) be defined as the remainder when (x–y)! is divided by x. If x=25, what is the maximum value of y for which f(x,y)=0

1. 5
2. 10
3. 15
4.20
5. 25

we have to find the value of y for which (25-y)!/ 25 yields zero remainder

if y=20, then (25-20)!=5!, which clearly is not divisible by 5^2
thus our answer will be less than 20

let's try 15 (25-15)!=10!, which is clearly divisible by 5^2.

hence C

It's important to note why 10! is clearly divisible by 5^2, while 5! is not.

It is possible to find the prime factors and their powers of a factorial by counting the amount of times a prime factor p is contained within some number n for n!.

This is most easily done by taking the floor value, or rounding to the next highest whole integer, when dividing out n by p to see how many numbers between p and n have a factor of p.
One must also remember to be mindful of the fact that there most likely exist numbers with multiple factors of p, in which case you must keep dividing by p until floor(p) = 0.

In this case, 5! clearly has no factors of 5^2 because floor(5/5) (or n/p, where n is taken from 5! and p is taken from 5^2) = 1, and floor(1/5) = 0. This means that the power of 5 in the prime factorization of 5! is 1, so it does not contain 5^2.

10!, however, clearly has a factor of 5^2 because floor(10/5) = 2, so the prime factorization of 10! contains 5^2. Continuing this factorization further purely to see how this method works, the prime factorization contains 8 2's (floor(10/2) = 5, floor(5/2) = 2, floor(2/2) = 1, 5+2+1 = 8).

I hope this method can help to quickly identify prime factors of factorials!
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Joined: 23 Jan 2013
Posts: 525
Schools: Cambridge'16
Re: Let f(x,y) be defined as the remainder when (x–y)! is divided by x. If  [#permalink]

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The question is in finding y to be (25-y)! multiple of 25. That means we should have at least 5*5 in (25-y)!

Go directily answer choices and start with 15, meaning we have 10!, having 5*5. But we should get maximum for y, so go 20, meaning we have 5!, not having 5*5.

C
Intern  Joined: 22 Dec 2014
Posts: 33
Let f(x,y) be defined as the remainder when (x–y)! is divided by x. If  [#permalink]

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prakash85 wrote:
Let f(x,y) be defined as the remainder when (x–y)! is divided by x. If x=25, what is the maximum value of y for which f(x,y)=0

A. 5
B. 10
C. 15
D. 20
E. 25

(x-y)! = x*n + f(x,y)
f(x,y) = 0; x=25 --> (25-y)! = 25*n --> (25-y)! is divisible by 5*5. Among options, D and E make (25-y)! not divisible by 5*5: eliminate
Find Max y = find Min (25-y) --> among A, B, C, C offers the min (25-y) and max y. --> C
Manager  S
Joined: 21 Jul 2017
Posts: 187
Location: India
Concentration: Social Entrepreneurship, Leadership
GMAT 1: 660 Q47 V34 GPA: 4
WE: Project Management (Education)
Re: Let f(x,y) be defined as the remainder when (x–y)! is divided by x. If  [#permalink]

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Manager  B
Joined: 30 Apr 2017
Posts: 58
Re: Let f(x,y) be defined as the remainder when (x–y)! is divided by x. If  [#permalink]

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rever08 wrote:

I have the same question, 0/25.... reminder will be 0..
is it wrong?
Math Expert V
Joined: 02 Sep 2009
Posts: 58465
Re: Let f(x,y) be defined as the remainder when (x–y)! is divided by x. If  [#permalink]

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soodia wrote:
rever08 wrote:
Let f(x,y) be defined as the remainder when (x–y)! is divided by x. If x=25, what is the maximum value of y for which f(x,y)=0

A. 5
B. 10
C. 15
D. 20
E. 25

I have the same question, 0/25.... reminder will be 0..
is it wrong?

f(x,y) is the remainder when (x–y)! is divided by x;

If x = y = 25, then f(x,y) is the remainder when (25 - 25)! = 0! = 1 divided by 25, so f(x,y) = f(25, 25) = 1, NOT 0.
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Posts: 9705
Location: Pune, India
Re: Let f(x,y) be defined as the remainder when (x–y)! is divided by x. If  [#permalink]

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prakash85 wrote:
Let f(x,y) be defined as the remainder when (x–y)! is divided by x. If x=25, what is the maximum value of y for which f(x,y)=0

A. 5
B. 10
C. 15
D. 20
E. 25

f(x, y) = Remainder of $$\frac{(x - y)!}{x}$$

For 0 remainder, (x - y)! should have x (=25) as a factor. So (25 - y)! should have at least two 5s.

Note that the first factorial to have two 5s is 10!.
So y can be at most 15 to give (25 - 15)! = 10!
This will be divisible by 25.

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Intern  B
Joined: 16 Oct 2016
Posts: 5
Re: Let f(x,y) be defined as the remainder when (x–y)! is divided by x. If  [#permalink]

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f(x,y)=(x-y)!/x

since x= 25
we need atleast two multiples of 5 in (x-y)! i.e. 5,10 which is possible in 10!
so x-y=10
y=x-10=25-10=15
Max value of y is 15 Re: Let f(x,y) be defined as the remainder when (x–y)! is divided by x. If   [#permalink] 11 Feb 2019, 22:49
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