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Retired Moderator
Joined: 19 Nov 2020
Posts: 326
Given Kudos: 64
GRE 1: Q160 V152
Function f(x) is defined as f(x)=1! + 2! + 3! + ...
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28 Jul 2021, 15:29
28 Jul 2021, 15:29
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64% (01:53) correct
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Function f(x) is defined as f(x)=1! + 2! + 3! + ... x!.
When x=218, the function f(x) is defined as f(218) = 1! + 2! + 3! ... 218! What may be a remainder, if x=218 and f(x)/5?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
When x=218, the function f(x) is defined as f(218) = 1! + 2! + 3! ... 218! What may be a remainder, if x=218 and f(x)/5?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
Re: Function f(x) is defined as f(x)=1! + 2! + 3! + ...
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28 Jul 2021, 21:48
28 Jul 2021, 21:48
5
Well, the answer doesn't really require us to care about the factorials after 4! since all of them are multiples of 5 anyway and would have remainder 0 when divided by 5. For example,
9! = 9*8*7*6*5*(4*3*2*1) has '5' in it, and hence a multiple of 5.
So the remainder of f(218)/5 is simply going to be the remainder of [ f(4)/5 ]. Since f(4) equals 33 ( 1! + 2! + 3! + 4!) , we have our remainder as 3. Hence, the answer is D
9! = 9*8*7*6*5*(4*3*2*1) has '5' in it, and hence a multiple of 5.
So the remainder of f(218)/5 is simply going to be the remainder of [ f(4)/5 ]. Since f(4) equals 33 ( 1! + 2! + 3! + 4!) , we have our remainder as 3. Hence, the answer is D
General Discussion
Function f(x) is defined as f(x)=1! + 2! + 3! + ...
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Updated on: 29 Jul 2021, 18:17
Updated on: 29 Jul 2021, 18:17
2
My train of thought here was to look for some sort of pattern as I add factorials and use that to deduce f(218)
1! is 1
2! is 2... 2! + 1! is 3
3! is 6... 3! + 2! + 1! is 9
4! is 24... 4! + 3! + 2! + 1! is 33
5! is 120... 5! + 4! + 3! + 2! + 1! is 153
6! is 720... We can stop right here since we can see a pattern where the ones digit is now 0.
Multiplying a number with a 0 at the ones digit is not going to change the ones digit; it will remain as a 0.
So even though we could add 8! and 9!, they will simply add some large (and annoying to calculate) number that has a 0 at the ones digit.
Ok, since we now know that 5!, 6!, 7!, ..., 218! are going to have a 0 at the ones digit, we know that adding those to 153, while yielding some large number, is simply adding 0 to the ones digit every time.
Thus, the ones will stay as 3 with no change. What is the remainder of a number with a ones digit of 3? It can only be 3 of course!!! (13%5 = 3, 14553%5 = 3, and 12314344983442342423423412348790213%3 = 3)
So the answer here is C
I hope this helps. This is the first question I answered on here.
1! is 1
2! is 2... 2! + 1! is 3
3! is 6... 3! + 2! + 1! is 9
4! is 24... 4! + 3! + 2! + 1! is 33
5! is 120... 5! + 4! + 3! + 2! + 1! is 153
6! is 720... We can stop right here since we can see a pattern where the ones digit is now 0.
Multiplying a number with a 0 at the ones digit is not going to change the ones digit; it will remain as a 0.
So even though we could add 8! and 9!, they will simply add some large (and annoying to calculate) number that has a 0 at the ones digit.
Ok, since we now know that 5!, 6!, 7!, ..., 218! are going to have a 0 at the ones digit, we know that adding those to 153, while yielding some large number, is simply adding 0 to the ones digit every time.
Thus, the ones will stay as 3 with no change. What is the remainder of a number with a ones digit of 3? It can only be 3 of course!!! (13%5 = 3, 14553%5 = 3, and 12314344983442342423423412348790213%3 = 3)
So the answer here is C
I hope this helps. This is the first question I answered on here.
Originally posted by jagjitsatti on 28 Jul 2021, 16:59.
Last edited by jagjitsatti on 29 Jul 2021, 18:17, edited 1 time in total.
Last edited by jagjitsatti on 29 Jul 2021, 18:17, edited 1 time in total.
