Expand the expression? No way!
Let f ( x ) denote a monic 2 0 1 5 th degree polynomial with integer coefficients and with roots r 1 , r 2 , … , r 2 0 1 5
And S n = ( − 1 ) n n 3 , where S n denote the n th symmetric sum of numbers r 1 , r 2 , … , r 2 0 1 5 .
What is the last three digits of
∣ ∣ ∣ ∣ ∣ k = 1 ∏ 2 0 1 5 ( r k + 1 ) ∣ ∣ ∣ ∣ ∣ ?
The answer is 855.
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1 solution
Isn't S n = ( − 1 ) n n 3 ? So shouldn't 1 + S 1 + S 2 + . . . + S 2 0 1 5 be 1 + ∑ ( − 1 ) j j 3 , not 1 + ∑ j 3
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Whoops! Fixed!
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One last question. What are you referring to by "symmetric sum"? Is S 1 = ∑ r i or − ∑ r i ?
It seems to me that you want to calculate − f ( − 1 ) instead right?
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No..... How so?
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f ( x ) = ∏ ( x − r i ) .
We want ∏ ( r i + 1 ) = ∏ − ( − 1 − r i ) = ( − 1 ) 2 0 1 5 ∏ ( − 1 − r i ) = − f ( − 1 ) .
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@Calvin Lin – ( Screams internally )
Well, this question is broken... Let me think how am I suppose to fix this. Or should I just delete this question? I blame my recent lack of practice in math.
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@Pi Han Goh – Well, I fixed the problem by adding absolute signs around the expression.
Technically, the last 3 digits of -1234 is 234, which is why I don't like asking for the last 3 digits of a negative number.
Can you update your solution and add an explanation for the equation? Thanks!
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@Calvin Lin – How does that fix the problem?
f ( − 1 ) = − 1 − S 1 − S 2 − S 3 − . . . − S 2 0 1 5
− f ( − 1 ) = 1 + S 1 + S 2 + S 3 + . . . + S 2 0 1 5
= 1 + ∑ S i
= 1 + ∑ ( − 1 ) n n 3 not 1 + ∑ n 3 .
There are only two fixes, either change S n = n 3 or have the question find ∏ ( 1 − r )
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@Siddhartha Srivastava – Ah thanks. I did not follow through the rest of the solution.
It looks like the value we want to calculate is
1 + ∑ ( − 1 ) n n 3 = 1 + 8 ( 1 − ( − 1 ) 2 0 1 5 ( 4 × 2 0 1 5 3 + 6 × 2 0 1 5 2 − 1 ) = − 4 0 9 3 7 2 1 8 5 6 .
@Pi Han Goh Can you confirm that I should update the answer so that 401 is incorrect, and either 856 or 144 is marked correct.
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@Calvin Lin – YESSSS, with the modulus sign there, the new answer is 8 5 6 .
Thanks you two!
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@Pi Han Goh – Thanks. Can you update the solution accordingly?
Those who answered 401 have been marked incorrect. Those who answered 855 or 145 have been marked correct. The correct answer is now 855.
@Pi Han Goh – NItpicking again, but its 8 5 5 . Wolfram Alpha
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@Siddhartha Srivastava – Curses, I wish you were wrong!
@Calvin Lin sorry to trouble you again, but....... I think I should just recreate a duplicate of this question.
@Siddhartha Srivastava – Oh, my fault, I forgot to add the 1. Fixed.
Thanks!!!
My solution is similar to yours.
Btw, its a nice problem, I also want to solve your algebra problems like this. Where can I find?
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Glad you enjoyed it. Unfortunately, I didn't make many of these problems. Why don't you create a similar question like this? =D
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well, i have created similar problem of sum, 2016 is coming, but i will try to create more problem like this
We have
f ( x ) = x 2 0 1 5 − S 1 x 2 0 1 4 + S 2 x 2 0 1 3 − … − S 2 0 1 3 x 2 + S 2 0 1 4 x − S 2 0 1 5
The desired expression is equals to
∏ ( r i + 1 ) = ∏ [ − ( − r i − 1 ) ] = ( − 1 ) 2 0 1 5 ∏ ( − 1 − r i ) = − f ( − 1 )
− f ( − 1 ) = = = = = = 1 + S 1 + S 2 + … + S 2 0 1 5 1 + ∑ S i 1 + ∑ ( − 1 ) n n 3 1 + ( − 1 3 + 2 3 − 3 3 + … − 2 0 1 5 3 ) 1 − ( 1 3 − 2 3 + 3 3 + … + 2 0 1 5 3 ) 1 − [ ( 1 3 + 2 3 + 3 3 + … + 2 0 1 5 3 ) − ( 2 3 + 4 3 + … + 2 0 1 4 3 ) ]
Use the fact that ∑ k = 1 n k 3 = ( 2 n ( n + 1 ) ) 2 , and use a little bit of modular arithmetic, plus some patience, we get 8 5 5