All About Factorials!
f ( x ) = m = 1 ∑ x m !
Let f be the function of the sum of all factorials under x as shown above. If n is a positive integer such that it satisfies the constraint n = f ( 2 a ) − f ( a ) = ( a ! ) 2 ( b ! ) for some positive integers a and b greater than 1, what is the value of n ?
The answer is 864.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
2 solutions
I'm not certain what you are trying to show here. I disagree with your proof of " E 5 ( f ( 2 a ) − f ( a ) ) = E 5 ( ( a + 1 ) ! ) ". What you have shown is that E 5 ( f ( 2 a ) − f ( a ) ) ≥ E 5 ( ( a + 1 ) ! ) . You have not explained why we cannot get another factor of 5 in there.
As an explicit counter-example, E 5 ( f ( 1 4 ) − f ( 7 ) ) = 2 , but E 5 ( 8 ! ) = 1 . This is because 8 ! + 9 ! is a multiple of 25, even though we only have 5 ∣ 8 ! and 5 ∣ 9 ! . Clearly 2 5 ∣ 1 0 ! , 1 1 ! , 1 2 ! , 1 4 ! , which establishes the first equation. More generally, because ( 5 n + 3 ) + ( 5 n + 3 ) ( 5 n + 4 ) = 5 ( n + 1 ) ( 5 n + 3 ) introduces another factor of 5, the statement is not true for a ≡ 2 ( m o d 5 ) (with a > 2 ).
I am unable to understand the rest of your proof, or what you are trying to do. Using paragraphs to split out the main ideas will allow me to figure out what you are trying to express, and see if various parts could be fixed.
If I'm not mistaken, your function works like this: E(50) = E(2(5^2)) = 2. However, I'm not quite sure how E(f(2a) - f(a)) = E((a+1)!(1+(a+2)+(a+2)(a+3)+...+(2a)!/(a+1)!) = E((a+1)!) because what if 1+(a+2)+(a+2)(a+3)+...+(2a)!/(a+1)! is divisible by 5, shouldn't the maximum degree in E(f(2a) - f(a)) be increased? Please correct me if I've got any part wrong.
Log in to reply
Please read my solution once again. That is what i have proved for the case a is congruent to 1 mod 5 we need to prove that E 5 ( [ ( a + 1 ) ! + . . . . . ( a + 3 ) ! ] = [ E 5 ( ( a + 1 ) ) ! ] because if this happens so when we take 5 j common from our whole sum of factorials from (a+1) to 2a where j is the maximum power of 5 dividing (a+1)! . Then we get 5 j ( 2 a ! / 5 j + . . . . . . . ( a + 4 ) ! / 5 j + ( a + 3 ) 2 ∗ ( a + 1 ) ! / 5 j ) and it is clear that 5 j + 1 ∣ ( a + 4 ) ! As a is congruent to 1 mod 5. So now its proved for a is congruent to 1 mod 5 similarly it can be proved for other cases. swe have [ ( a + 1 ) ! + ( a + 2 ) ! + ( a + 3 ) ! ] = ( a + 3 ) 2 . ( a + 1 ) ! And because a is congruent to 1 mod 5 , 5 does not divide ( a + 3 ) 2 Is it correct?
Log in to reply
@Nihar Mahajan , @Dev Sharma , @Calvin Lin please check if my solution is correct and post a solution for this
f (6) - f (3) = (3!)^2 (4!) = 864
Can you stop posting final answers? Nobody, not even you can benefit from this solution.
Log in to reply
Can you stop posting final answers? Nobody, not even you can benefit from this solution.
Well, people can benefit a little from it.
Incidentally, I notice that today the problem is classified as Level 5 rather than Level 4.
Giving hint is a good solution to reserve a trial for readers. You can open yours to write and this doesn't matter that you don't have to bother about others.
Correct! Can you explain more on how you got the answer?
Log in to reply
Using Excel, observe a list of figures for f (x) and think. Good guess made a fast finding.
Alright, but can you show that that is the only solution?
It seems to me that you should use the fact that, since a > 1 , at least one of a + 2 , a + 4 , a + 6 must be composite. (Unless there's an easier way than I'm thinking of.)
Log in to reply
.
It seems to me that you should use the fact that, since a > 1 , at least one of a + 2 , a + 4 , a + 6 must be composite.
Does anyone have a different method?
Let others to check for other possibilities.
The easy way to think is: If this is not the answer, then it must be a responsibility of the question for making question with ambiguous answers.
Log in to reply
The easy way to think is: If this is not the answer, then it must be a responsibility of the question for making question with ambiguous answers.
Alright, but I see it as a two-part challenge: figure out what the answer is, and show that there aren't any others.
Log in to reply
@Peter Byers – "what is the value of n?" was the challenge. To avoid something, I only answer to an extend which can satisfy requirement. Now, you make me another challenge. Basically we start by finding possibility which comes first. So, I got a small one. Rough idea was likeliness is reducing when difference is becoming more when x is getting greater, just like many other examples I have ever met. Usually, once an initial chance of interception has been passed by, the remained shall leave with no more to occur. To be honest, I must check further. You may have found the fact before I answer you with my finding later.
Let E 5 ( n ) denote the maximum power of 5 in n. Let (a,b) > 4, now if E 5 ( a ! ) = c , E 5 ( b ! ) = e , E 5 ( a + 1 ) ! = d E 5 ( f ( 2 a ) − f ( a ) ) = E 5 ( ( a + 1 ) ! ) because when we expand f(2a) - f(a) then we get 2a! + (2a - 1)! +...... + (a+1)! So whatever be the maximum power of 5 dividing (a+1)! That will surely divide the rest of the terms now we will prove the claim stated above ; if a is congruent to 1 mod 5. E 5 ( a + 1 ) ! = E 5 ( a + 2 ) ! = E 5 ( a + 3 ) ! now (a+1)! + (a+2)! + (a+3)! = (a+3)^{2} . (a+1)! This equation clearly shows the result that E 5{n} = E {5}(a+1)! Similarly it can be proved for the other cases. So d = 2c + e,, so d > 2c this is only possible if (a+1) contains a large power of 5 or itself is a complete power of 5 then only it is possibe that d becomes so large that it is greater than 2c. So let a + 1 = 5 n we can now easily find E 5 ( 5 n ! ) and E 5 ( 5 n − 1 ) ! and we observe that the inequality d > 2c is not obeyed. Also we observe that at a = 9 the relation d = 2c holds and at further greater values of a d< 2c is found true. Now we conclude that a < 5 so by trying out a = 4,3,2. We get a = 3, b = 4 and n = 864. I am not sure whether what i wrote is completely correct. Please feel free to point out any mistakes.