Last $3$ digits of addition, Olympiad problem.
Find the last three digits of the following:
$$1!+3!+5!+7!+ldots+2013!+2015!+2016!$$
I have tried to divide by $1000$ up to $15!$, but the number is too high and confused. So I wanna know if there is other way to solve this problem.
Thanks in advance!
real-analysis modular-arithmetic contest-math factorial
edited Apr 20 '17 at 4:39
Ángel Mario Gallegos
18.5k11230
asked Apr 20 '17 at 4:26
November ft Blue
1036
|
show 2 more comments
Find the last three digits of the following:
$$1!+3!+5!+7!+ldots+2013!+2015!+2016!$$
I have tried to divide by $1000$ up to $15!$, but the number is too high and confused. So I wanna know if there is other way to solve this problem.
Thanks in advance!
real-analysis modular-arithmetic contest-math factorial
edited Apr 20 '17 at 4:39
Ángel Mario Gallegos
18.5k11230
asked Apr 20 '17 at 4:26
November ft Blue
1036
The pattern is 2n-1 ! So 2014 isn't included ! And 2016 just added to confused the candidates becuz it is in grade 10 exercise
– November ft Blue
Apr 20 '17 at 4:28
2
You seem to have already figured out that you don't need to care about $n!$ for $n ge 15$. So just compute the last three digits of $n!$ for $n=1,3,ldots,13$ and add them up.
– angryavian
Apr 20 '17 at 4:31
So, there is only one way to do it !? It doesn't related to the fumula 2n-1 !?
– November ft Blue
Apr 20 '17 at 4:33
Becuz I don't really want to multiply a lot of digits XD ! If there is an easy way to solve XD
– November ft Blue
Apr 20 '17 at 4:33
1
1000 divides n! For all n >= 15 so you only have to find the last three digits of 1!+3!+5!+7!+...+13!
– fleablood
Apr 20 '17 at 7:55
|
show 2 more comments
Find the last three digits of the following:
$$1!+3!+5!+7!+ldots+2013!+2015!+2016!$$
I have tried to divide by $1000$ up to $15!$, but the number is too high and confused. So I wanna know if there is other way to solve this problem.
Thanks in advance!
real-analysis modular-arithmetic contest-math factorial
edited Apr 20 '17 at 4:39
Ángel Mario Gallegos
18.5k11230
asked Apr 20 '17 at 4:26
November ft Blue
1036
Find the last three digits of the following:
$$1!+3!+5!+7!+ldots+2013!+2015!+2016!$$
I have tried to divide by $1000$ up to $15!$, but the number is too high and confused. So I wanna know if there is other way to solve this problem.
Thanks in advance!
real-analysis modular-arithmetic contest-math factorial
real-analysis modular-arithmetic contest-math factorial
edited Apr 20 '17 at 4:39
Ángel Mario Gallegos
18.5k11230
asked Apr 20 '17 at 4:26
November ft Blue
1036
edited Apr 20 '17 at 4:39
Ángel Mario Gallegos
18.5k11230
asked Apr 20 '17 at 4:26
November ft Blue
1036
edited Apr 20 '17 at 4:39
Ángel Mario Gallegos
18.5k11230
edited Apr 20 '17 at 4:39
Ángel Mario Gallegos
18.5k11230
edited Apr 20 '17 at 4:39
Ángel Mario Gallegos
18.5k11230
18.5k11230
asked Apr 20 '17 at 4:26
November ft Blue
1036
asked Apr 20 '17 at 4:26
November ft Blue
1036
asked Apr 20 '17 at 4:26
November ft Blue
1036
1036
The pattern is 2n-1 ! So 2014 isn't included ! And 2016 just added to confused the candidates becuz it is in grade 10 exercise
– November ft Blue
Apr 20 '17 at 4:28
2
You seem to have already figured out that you don't need to care about $n!$ for $n ge 15$. So just compute the last three digits of $n!$ for $n=1,3,ldots,13$ and add them up.
– angryavian
Apr 20 '17 at 4:31
So, there is only one way to do it !? It doesn't related to the fumula 2n-1 !?
– November ft Blue
Apr 20 '17 at 4:33
Becuz I don't really want to multiply a lot of digits XD ! If there is an easy way to solve XD
– November ft Blue
Apr 20 '17 at 4:33
1
1000 divides n! For all n >= 15 so you only have to find the last three digits of 1!+3!+5!+7!+...+13!
– fleablood
Apr 20 '17 at 7:55
|
show 2 more comments
The pattern is 2n-1 ! So 2014 isn't included ! And 2016 just added to confused the candidates becuz it is in grade 10 exercise
– November ft Blue
Apr 20 '17 at 4:28
2
You seem to have already figured out that you don't need to care about $n!$ for $n ge 15$. So just compute the last three digits of $n!$ for $n=1,3,ldots,13$ and add them up.
– angryavian
Apr 20 '17 at 4:31
So, there is only one way to do it !? It doesn't related to the fumula 2n-1 !?
– November ft Blue
Apr 20 '17 at 4:33
Becuz I don't really want to multiply a lot of digits XD ! If there is an easy way to solve XD
– November ft Blue
Apr 20 '17 at 4:33
1
1000 divides n! For all n >= 15 so you only have to find the last three digits of 1!+3!+5!+7!+...+13!
– fleablood
Apr 20 '17 at 7:55
The pattern is 2n-1 ! So 2014 isn't included ! And 2016 just added to confused the candidates becuz it is in grade 10 exercise
– November ft Blue
Apr 20 '17 at 4:28
The pattern is 2n-1 ! So 2014 isn't included ! And 2016 just added to confused the candidates becuz it is in grade 10 exercise
– November ft Blue
Apr 20 '17 at 4:28
2
2
You seem to have already figured out that you don't need to care about $n!$ for $n ge 15$. So just compute the last three digits of $n!$ for $n=1,3,ldots,13$ and add them up.
– angryavian
Apr 20 '17 at 4:31
You seem to have already figured out that you don't need to care about $n!$ for $n ge 15$. So just compute the last three digits of $n!$ for $n=1,3,ldots,13$ and add them up.
– angryavian
Apr 20 '17 at 4:31
So, there is only one way to do it !? It doesn't related to the fumula 2n-1 !?
– November ft Blue
Apr 20 '17 at 4:33
So, there is only one way to do it !? It doesn't related to the fumula 2n-1 !?
– November ft Blue
Apr 20 '17 at 4:33
Becuz I don't really want to multiply a lot of digits XD ! If there is an easy way to solve XD
– November ft Blue
Apr 20 '17 at 4:33
Becuz I don't really want to multiply a lot of digits XD ! If there is an easy way to solve XD
– November ft Blue
Apr 20 '17 at 4:33
1
1
1000 divides n! For all n >= 15 so you only have to find the last three digits of 1!+3!+5!+7!+...+13!
– fleablood
Apr 20 '17 at 7:55
1000 divides n! For all n >= 15 so you only have to find the last three digits of 1!+3!+5!+7!+...+13!
– fleablood
Apr 20 '17 at 7:55
|
show 2 more comments
1 Answer
1
active
oldest
votes
You can work via modular arithmetic, simplifying mod $1000$ after each step.
$1! = 1, 3! = 6, 5! =120$.
$7! = 7 times 6 times 5! = 5040 equiv 40 mod 100$ (by computation)
$9! = 9 times 8 times 40 equiv 880$ (here we do not have to compute $9!$).
$11! = 11 times 10 times 880 = 11 times 800 equiv 800 $
$13! = 13 times 12 times 800 = 13*600 equiv 800$
$15!$ is a multiple of $1000$, and so are the others after this, so the answer would be $800 + 800 + 880 + 40 + 120+6+1 = 647 $ modulo $1000$. Important to note that no factorial above $7!$ was computed explicitly.
Note : You cannot avoid simple computations. All the above computations were atmost of four-digit complexity.
Wolfram verifies the the sum up to $13!$ as $6267305647$, so the above answer is correct.
EDIT : About the question asked in the comments, we have to essentially compute $1! + ... + 14!$ modulo $1000$. Again, we can work by hand :
You will get, by hand, the sum up to $6!$, as $720 + 120 + 24 + 6 + 2 +1 = 873$.
Now, it is a question of computation and reduction:
$7! = 7times 6! = 5040 = 40$
$8! = 8 times 40 = 320$
$9! = 9 times 320 = 2880 = 880$
$10! = 10 times 880 = 8800 = 800$
$11! = 800 times 11 = 8800 = 800$
$12! = 800 times 12 = 9600 = 600$
$13! = 600 times 13 = 7800 = 800$
$14! = 800 times 14 = 11200 = 200$
Now, add these up : $100(8+8+8+2+6) + 880 + 320+40+873 = 273$ (after reduction).
However, I would feel sympathetic if you were allotted four minutes for this question. Probably it is worth more marks than the others, but that is scant consolation for a problem nearly involving outright computation.
Furthermore, there isn't any sum of factorials formula, by my reckoning, that is capable of producing the remainder mod $1000$ (and if there is some complicated formula involving exponentials and gamma functions, then that is a roof too high for tenth grade, certainly). Hence (unfortunately) this problem is done only by computation.
answered Apr 20 '17 at 4:36
астон вілла олоф мэллбэрг
37.4k33376
Thank you sir! What if the problem is 1!+2!+3!+...+2016!. We have to add up to 14! And it takes really long duration! So I wanna know if we can use the fumula(sigma of n!) to find the last 3 digits !
– November ft Blue
Apr 20 '17 at 4:40
@NovemberftBlue: How much time do you get per problem? They try to size the computation to the time limit. Do you know the small factorials? Even if you only know up to $5!=120$ you are down to two digit already. Doing this up to $14$ wouldn't take long. You are down to a single digit once you pass $9$ so you only need your times tables.
– Ross Millikan
Apr 20 '17 at 4:43
1
@NovemberftBlue You are welcome! Also, I am not aware of the formula which you are considering (summation of $n!$). If it is easily computable, certainly we are in a good position to take the remainder when we divide by $1000$. If you like, I will attempt the second variation also.
– астон вілла олоф мэллбэрг
Apr 20 '17 at 4:48
1
Please see the edit. Also, while I am extremely pleased, calling me Sir will be disrespecting the many people better than me on this website. You should keep it for them. Also, I am not in tenth grade yet, so I am younger than you. Hence I should call you Sir, if at all.
– астон вілла олоф мэллбэрг
Apr 20 '17 at 5:03
1
The only formula I know for sum of factorials is this one: math.stackexchange.com/questions/301615/… , but I don't think it applies here.
– N74
Apr 20 '17 at 5:23
|
show 5 more comments
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1 Answer
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You can work via modular arithmetic, simplifying mod $1000$ after each step.
$1! = 1, 3! = 6, 5! =120$.
$7! = 7 times 6 times 5! = 5040 equiv 40 mod 100$ (by computation)
$9! = 9 times 8 times 40 equiv 880$ (here we do not have to compute $9!$).
$11! = 11 times 10 times 880 = 11 times 800 equiv 800 $
$13! = 13 times 12 times 800 = 13*600 equiv 800$
$15!$ is a multiple of $1000$, and so are the others after this, so the answer would be $800 + 800 + 880 + 40 + 120+6+1 = 647 $ modulo $1000$. Important to note that no factorial above $7!$ was computed explicitly.
Note : You cannot avoid simple computations. All the above computations were atmost of four-digit complexity.
Wolfram verifies the the sum up to $13!$ as $6267305647$, so the above answer is correct.
EDIT : About the question asked in the comments, we have to essentially compute $1! + ... + 14!$ modulo $1000$. Again, we can work by hand :
You will get, by hand, the sum up to $6!$, as $720 + 120 + 24 + 6 + 2 +1 = 873$.
Now, it is a question of computation and reduction:
$7! = 7times 6! = 5040 = 40$
$8! = 8 times 40 = 320$
$9! = 9 times 320 = 2880 = 880$
$10! = 10 times 880 = 8800 = 800$
$11! = 800 times 11 = 8800 = 800$
$12! = 800 times 12 = 9600 = 600$
$13! = 600 times 13 = 7800 = 800$
$14! = 800 times 14 = 11200 = 200$
Now, add these up : $100(8+8+8+2+6) + 880 + 320+40+873 = 273$ (after reduction).
However, I would feel sympathetic if you were allotted four minutes for this question. Probably it is worth more marks than the others, but that is scant consolation for a problem nearly involving outright computation.
Furthermore, there isn't any sum of factorials formula, by my reckoning, that is capable of producing the remainder mod $1000$ (and if there is some complicated formula involving exponentials and gamma functions, then that is a roof too high for tenth grade, certainly). Hence (unfortunately) this problem is done only by computation.
answered Apr 20 '17 at 4:36
астон вілла олоф мэллбэрг
37.4k33376
Thank you sir! What if the problem is 1!+2!+3!+...+2016!. We have to add up to 14! And it takes really long duration! So I wanna know if we can use the fumula(sigma of n!) to find the last 3 digits !
– November ft Blue
Apr 20 '17 at 4:40
@NovemberftBlue: How much time do you get per problem? They try to size the computation to the time limit. Do you know the small factorials? Even if you only know up to $5!=120$ you are down to two digit already. Doing this up to $14$ wouldn't take long. You are down to a single digit once you pass $9$ so you only need your times tables.
– Ross Millikan
Apr 20 '17 at 4:43
1
@NovemberftBlue You are welcome! Also, I am not aware of the formula which you are considering (summation of $n!$). If it is easily computable, certainly we are in a good position to take the remainder when we divide by $1000$. If you like, I will attempt the second variation also.
– астон вілла олоф мэллбэрг
Apr 20 '17 at 4:48
1
Please see the edit. Also, while I am extremely pleased, calling me Sir will be disrespecting the many people better than me on this website. You should keep it for them. Also, I am not in tenth grade yet, so I am younger than you. Hence I should call you Sir, if at all.
– астон вілла олоф мэллбэрг
Apr 20 '17 at 5:03
1
The only formula I know for sum of factorials is this one: math.stackexchange.com/questions/301615/… , but I don't think it applies here.
– N74
Apr 20 '17 at 5:23
|
show 5 more comments
You can work via modular arithmetic, simplifying mod $1000$ after each step.
$1! = 1, 3! = 6, 5! =120$.
$7! = 7 times 6 times 5! = 5040 equiv 40 mod 100$ (by computation)
$9! = 9 times 8 times 40 equiv 880$ (here we do not have to compute $9!$).
$11! = 11 times 10 times 880 = 11 times 800 equiv 800 $
$13! = 13 times 12 times 800 = 13*600 equiv 800$
$15!$ is a multiple of $1000$, and so are the others after this, so the answer would be $800 + 800 + 880 + 40 + 120+6+1 = 647 $ modulo $1000$. Important to note that no factorial above $7!$ was computed explicitly.
Note : You cannot avoid simple computations. All the above computations were atmost of four-digit complexity.
Wolfram verifies the the sum up to $13!$ as $6267305647$, so the above answer is correct.
EDIT : About the question asked in the comments, we have to essentially compute $1! + ... + 14!$ modulo $1000$. Again, we can work by hand :
You will get, by hand, the sum up to $6!$, as $720 + 120 + 24 + 6 + 2 +1 = 873$.
Now, it is a question of computation and reduction:
$7! = 7times 6! = 5040 = 40$
$8! = 8 times 40 = 320$
$9! = 9 times 320 = 2880 = 880$
$10! = 10 times 880 = 8800 = 800$
$11! = 800 times 11 = 8800 = 800$
$12! = 800 times 12 = 9600 = 600$
$13! = 600 times 13 = 7800 = 800$
$14! = 800 times 14 = 11200 = 200$
Now, add these up : $100(8+8+8+2+6) + 880 + 320+40+873 = 273$ (after reduction).
However, I would feel sympathetic if you were allotted four minutes for this question. Probably it is worth more marks than the others, but that is scant consolation for a problem nearly involving outright computation.
Furthermore, there isn't any sum of factorials formula, by my reckoning, that is capable of producing the remainder mod $1000$ (and if there is some complicated formula involving exponentials and gamma functions, then that is a roof too high for tenth grade, certainly). Hence (unfortunately) this problem is done only by computation.
answered Apr 20 '17 at 4:36
астон вілла олоф мэллбэрг
37.4k33376
Thank you sir! What if the problem is 1!+2!+3!+...+2016!. We have to add up to 14! And it takes really long duration! So I wanna know if we can use the fumula(sigma of n!) to find the last 3 digits !
– November ft Blue
Apr 20 '17 at 4:40
@NovemberftBlue: How much time do you get per problem? They try to size the computation to the time limit. Do you know the small factorials? Even if you only know up to $5!=120$ you are down to two digit already. Doing this up to $14$ wouldn't take long. You are down to a single digit once you pass $9$ so you only need your times tables.
– Ross Millikan
Apr 20 '17 at 4:43
1
@NovemberftBlue You are welcome! Also, I am not aware of the formula which you are considering (summation of $n!$). If it is easily computable, certainly we are in a good position to take the remainder when we divide by $1000$. If you like, I will attempt the second variation also.
– астон вілла олоф мэллбэрг
Apr 20 '17 at 4:48
1
Please see the edit. Also, while I am extremely pleased, calling me Sir will be disrespecting the many people better than me on this website. You should keep it for them. Also, I am not in tenth grade yet, so I am younger than you. Hence I should call you Sir, if at all.
– астон вілла олоф мэллбэрг
Apr 20 '17 at 5:03
1
The only formula I know for sum of factorials is this one: math.stackexchange.com/questions/301615/… , but I don't think it applies here.
– N74
Apr 20 '17 at 5:23
|
show 5 more comments
You can work via modular arithmetic, simplifying mod $1000$ after each step.
$1! = 1, 3! = 6, 5! =120$.
$7! = 7 times 6 times 5! = 5040 equiv 40 mod 100$ (by computation)
$9! = 9 times 8 times 40 equiv 880$ (here we do not have to compute $9!$).
$11! = 11 times 10 times 880 = 11 times 800 equiv 800 $
$13! = 13 times 12 times 800 = 13*600 equiv 800$
$15!$ is a multiple of $1000$, and so are the others after this, so the answer would be $800 + 800 + 880 + 40 + 120+6+1 = 647 $ modulo $1000$. Important to note that no factorial above $7!$ was computed explicitly.
Note : You cannot avoid simple computations. All the above computations were atmost of four-digit complexity.
Wolfram verifies the the sum up to $13!$ as $6267305647$, so the above answer is correct.
EDIT : About the question asked in the comments, we have to essentially compute $1! + ... + 14!$ modulo $1000$. Again, we can work by hand :
You will get, by hand, the sum up to $6!$, as $720 + 120 + 24 + 6 + 2 +1 = 873$.
Now, it is a question of computation and reduction:
$7! = 7times 6! = 5040 = 40$
$8! = 8 times 40 = 320$
$9! = 9 times 320 = 2880 = 880$
$10! = 10 times 880 = 8800 = 800$
$11! = 800 times 11 = 8800 = 800$
$12! = 800 times 12 = 9600 = 600$
$13! = 600 times 13 = 7800 = 800$
$14! = 800 times 14 = 11200 = 200$
Now, add these up : $100(8+8+8+2+6) + 880 + 320+40+873 = 273$ (after reduction).
However, I would feel sympathetic if you were allotted four minutes for this question. Probably it is worth more marks than the others, but that is scant consolation for a problem nearly involving outright computation.
Furthermore, there isn't any sum of factorials formula, by my reckoning, that is capable of producing the remainder mod $1000$ (and if there is some complicated formula involving exponentials and gamma functions, then that is a roof too high for tenth grade, certainly). Hence (unfortunately) this problem is done only by computation.
answered Apr 20 '17 at 4:36
астон вілла олоф мэллбэрг
37.4k33376
You can work via modular arithmetic, simplifying mod $1000$ after each step.
$1! = 1, 3! = 6, 5! =120$.
$7! = 7 times 6 times 5! = 5040 equiv 40 mod 100$ (by computation)
$9! = 9 times 8 times 40 equiv 880$ (here we do not have to compute $9!$).
$11! = 11 times 10 times 880 = 11 times 800 equiv 800 $
$13! = 13 times 12 times 800 = 13*600 equiv 800$
$15!$ is a multiple of $1000$, and so are the others after this, so the answer would be $800 + 800 + 880 + 40 + 120+6+1 = 647 $ modulo $1000$. Important to note that no factorial above $7!$ was computed explicitly.
Note : You cannot avoid simple computations. All the above computations were atmost of four-digit complexity.
Wolfram verifies the the sum up to $13!$ as $6267305647$, so the above answer is correct.
EDIT : About the question asked in the comments, we have to essentially compute $1! + ... + 14!$ modulo $1000$. Again, we can work by hand :
You will get, by hand, the sum up to $6!$, as $720 + 120 + 24 + 6 + 2 +1 = 873$.
Now, it is a question of computation and reduction:
$7! = 7times 6! = 5040 = 40$
$8! = 8 times 40 = 320$
$9! = 9 times 320 = 2880 = 880$
$10! = 10 times 880 = 8800 = 800$
$11! = 800 times 11 = 8800 = 800$
$12! = 800 times 12 = 9600 = 600$
$13! = 600 times 13 = 7800 = 800$
$14! = 800 times 14 = 11200 = 200$
Now, add these up : $100(8+8+8+2+6) + 880 + 320+40+873 = 273$ (after reduction).
However, I would feel sympathetic if you were allotted four minutes for this question. Probably it is worth more marks than the others, but that is scant consolation for a problem nearly involving outright computation.
Furthermore, there isn't any sum of factorials formula, by my reckoning, that is capable of producing the remainder mod $1000$ (and if there is some complicated formula involving exponentials and gamma functions, then that is a roof too high for tenth grade, certainly). Hence (unfortunately) this problem is done only by computation.
answered Apr 20 '17 at 4:36
астон вілла олоф мэллбэрг
37.4k33376
edited 2 days ago
answered Apr 20 '17 at 4:36
астон вілла олоф мэллбэрг
37.4k33376
answered Apr 20 '17 at 4:36
астон вілла олоф мэллбэрг
37.4k33376
answered Apr 20 '17 at 4:36
астон вілла олоф мэллбэрг
37.4k33376
37.4k33376
Thank you sir! What if the problem is 1!+2!+3!+...+2016!. We have to add up to 14! And it takes really long duration! So I wanna know if we can use the fumula(sigma of n!) to find the last 3 digits !
– November ft Blue
Apr 20 '17 at 4:40
@NovemberftBlue: How much time do you get per problem? They try to size the computation to the time limit. Do you know the small factorials? Even if you only know up to $5!=120$ you are down to two digit already. Doing this up to $14$ wouldn't take long. You are down to a single digit once you pass $9$ so you only need your times tables.
– Ross Millikan
Apr 20 '17 at 4:43
1
@NovemberftBlue You are welcome! Also, I am not aware of the formula which you are considering (summation of $n!$). If it is easily computable, certainly we are in a good position to take the remainder when we divide by $1000$. If you like, I will attempt the second variation also.
– астон вілла олоф мэллбэрг
Apr 20 '17 at 4:48
1
Please see the edit. Also, while I am extremely pleased, calling me Sir will be disrespecting the many people better than me on this website. You should keep it for them. Also, I am not in tenth grade yet, so I am younger than you. Hence I should call you Sir, if at all.
– астон вілла олоф мэллбэрг
Apr 20 '17 at 5:03
1
The only formula I know for sum of factorials is this one: math.stackexchange.com/questions/301615/… , but I don't think it applies here.
– N74
Apr 20 '17 at 5:23
|
show 5 more comments
Thank you sir! What if the problem is 1!+2!+3!+...+2016!. We have to add up to 14! And it takes really long duration! So I wanna know if we can use the fumula(sigma of n!) to find the last 3 digits !
– November ft Blue
Apr 20 '17 at 4:40
@NovemberftBlue: How much time do you get per problem? They try to size the computation to the time limit. Do you know the small factorials? Even if you only know up to $5!=120$ you are down to two digit already. Doing this up to $14$ wouldn't take long. You are down to a single digit once you pass $9$ so you only need your times tables.
– Ross Millikan
Apr 20 '17 at 4:43
1
@NovemberftBlue You are welcome! Also, I am not aware of the formula which you are considering (summation of $n!$). If it is easily computable, certainly we are in a good position to take the remainder when we divide by $1000$. If you like, I will attempt the second variation also.
– астон вілла олоф мэллбэрг
Apr 20 '17 at 4:48
1
Please see the edit. Also, while I am extremely pleased, calling me Sir will be disrespecting the many people better than me on this website. You should keep it for them. Also, I am not in tenth grade yet, so I am younger than you. Hence I should call you Sir, if at all.
– астон вілла олоф мэллбэрг
Apr 20 '17 at 5:03
1
The only formula I know for sum of factorials is this one: math.stackexchange.com/questions/301615/… , but I don't think it applies here.
– N74
Apr 20 '17 at 5:23
Thank you sir! What if the problem is 1!+2!+3!+...+2016!. We have to add up to 14! And it takes really long duration! So I wanna know if we can use the fumula(sigma of n!) to find the last 3 digits !
– November ft Blue
Apr 20 '17 at 4:40
Thank you sir! What if the problem is 1!+2!+3!+...+2016!. We have to add up to 14! And it takes really long duration! So I wanna know if we can use the fumula(sigma of n!) to find the last 3 digits !
– November ft Blue
Apr 20 '17 at 4:40
@NovemberftBlue: How much time do you get per problem? They try to size the computation to the time limit. Do you know the small factorials? Even if you only know up to $5!=120$ you are down to two digit already. Doing this up to $14$ wouldn't take long. You are down to a single digit once you pass $9$ so you only need your times tables.
– Ross Millikan
Apr 20 '17 at 4:43
@NovemberftBlue: How much time do you get per problem? They try to size the computation to the time limit. Do you know the small factorials? Even if you only know up to $5!=120$ you are down to two digit already. Doing this up to $14$ wouldn't take long. You are down to a single digit once you pass $9$ so you only need your times tables.
– Ross Millikan
Apr 20 '17 at 4:43
1
1
@NovemberftBlue You are welcome! Also, I am not aware of the formula which you are considering (summation of $n!$). If it is easily computable, certainly we are in a good position to take the remainder when we divide by $1000$. If you like, I will attempt the second variation also.
– астон вілла олоф мэллбэрг
Apr 20 '17 at 4:48
@NovemberftBlue You are welcome! Also, I am not aware of the formula which you are considering (summation of $n!$). If it is easily computable, certainly we are in a good position to take the remainder when we divide by $1000$. If you like, I will attempt the second variation also.
– астон вілла олоф мэллбэрг
Apr 20 '17 at 4:48
1
1
Please see the edit. Also, while I am extremely pleased, calling me Sir will be disrespecting the many people better than me on this website. You should keep it for them. Also, I am not in tenth grade yet, so I am younger than you. Hence I should call you Sir, if at all.
– астон вілла олоф мэллбэрг
Apr 20 '17 at 5:03
Please see the edit. Also, while I am extremely pleased, calling me Sir will be disrespecting the many people better than me on this website. You should keep it for them. Also, I am not in tenth grade yet, so I am younger than you. Hence I should call you Sir, if at all.
– астон вілла олоф мэллбэрг
Apr 20 '17 at 5:03
1
1
The only formula I know for sum of factorials is this one: math.stackexchange.com/questions/301615/… , but I don't think it applies here.
– N74
Apr 20 '17 at 5:23
The only formula I know for sum of factorials is this one: math.stackexchange.com/questions/301615/… , but I don't think it applies here.
– N74
Apr 20 '17 at 5:23
|
show 5 more comments
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The pattern is 2n-1 ! So 2014 isn't included ! And 2016 just added to confused the candidates becuz it is in grade 10 exercise
– November ft Blue
Apr 20 '17 at 4:28
2
You seem to have already figured out that you don't need to care about $n!$ for $n ge 15$. So just compute the last three digits of $n!$ for $n=1,3,ldots,13$ and add them up.
– angryavian
Apr 20 '17 at 4:31
So, there is only one way to do it !? It doesn't related to the fumula 2n-1 !?
– November ft Blue
Apr 20 '17 at 4:33
Becuz I don't really want to multiply a lot of digits XD ! If there is an easy way to solve XD
– November ft Blue
Apr 20 '17 at 4:33
1
1000 divides n! For all n >= 15 so you only have to find the last three digits of 1!+3!+5!+7!+...+13!
– fleablood
Apr 20 '17 at 7:55