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Let f(x,y) be defined as the remainder when (x–y)! is divided by x. If
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Updated on: 06 Nov 2014, 04:26
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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
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Originally posted by prakash85 on 06 Nov 2014, 02:55.
Last edited by Bunuel on 06 Nov 2014, 04:26, edited 1 time in total.
Renamed the topic and edited the question.



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Re: Let f(x,y) be defined as the remainder when (x–y)! is divided by x. If
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06 Nov 2014, 03:43
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 (25y)!/ 25 yields zero remainder if y=20, then (2520)!=5!, which clearly is not divisible by 5^2 thus our answer will be less than 20 let's try 15 (2515)!=10!, which is clearly divisible by 5^2. hence C



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Re: Let f(x,y) be defined as the remainder when (x–y)! is divided by x. If
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06 Nov 2014, 03:50
Answer  D 1/25 no reminder
So simple
(Xy)!/x = no reminder If x = 25 Then (25x)!/25 = no reminder Let y = 25 Then 0!/25 , 0! = 1 So 1/25 which have 0 reminder So answer D



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Re: Let f(x,y) be defined as the remainder when (x–y)! is divided by x. If
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06 Nov 2014, 04:28



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Re: Let f(x,y) be defined as the remainder when (x–y)! is divided by x. If
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06 Nov 2014, 06:29
Totally agree. It was my mistake. Sorry
Posted from my mobile device



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Re: Let f(x,y) be defined as the remainder when (x–y)! is divided by x. If
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06 Nov 2014, 07:30
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 (25y)!/ 25 yields zero remainder if y=20, then (2520)!=5!, which clearly is not divisible by 5^2 thus our answer will be less than 20 let's try 15 (2515)!=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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Re: Let f(x,y) be defined as the remainder when (x–y)! is divided by x. If
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06 Nov 2014, 21:08
The question is in finding y to be (25y)! multiple of 25. That means we should have at least 5*5 in (25y)!
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



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Let f(x,y) be defined as the remainder when (x–y)! is divided by x. If
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10 Jul 2015, 05:32
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 (xy)! = x*n + f(x,y) f(x,y) = 0; x=25 > (25y)! = 25*n > (25y)! is divisible by 5*5. Among options, D and E make (25y)! not divisible by 5*5: eliminate Find Max y = find Min (25y) > among A, B, C, C offers the min (25y) and max y. > C



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Re: Let f(x,y) be defined as the remainder when (x–y)! is divided by x. If
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13 Oct 2017, 10:20
Bunuel why isn't it E? Please help!



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Re: Let f(x,y) be defined as the remainder when (x–y)! is divided by x. If
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15 Nov 2017, 21:54
rever08 wrote: Bunuel why isn't it E? Please help! I have the same question, 0/25.... reminder will be 0.. is it wrong?



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Re: Let f(x,y) be defined as the remainder when (x–y)! is divided by x. If
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15 Nov 2017, 22:08
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 Bunuel why isn't it E? Please help! 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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Re: Let f(x,y) be defined as the remainder when (x–y)! is divided by x. If
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11 Feb 2019, 21:43
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. Answer (C)
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Re: Let f(x,y) be defined as the remainder when (x–y)! is divided by x. If
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11 Feb 2019, 21:49
f(x,y)=(xy)!/x
since x= 25 we need atleast two multiples of 5 in (xy)! i.e. 5,10 which is possible in 10! so xy=10 y=x10=2510=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
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