Mixing
Sum of last digits of a sum
Clash Royale CLAN TAG#URR8PPP
up vote
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Let $d_n$ be the last digit of $S_n$ where $S_n = left(1 + 2 + 3 + .... + nright)$. Find the remainder when $sum_i=1^2017d_i$ is divided by 1000.
My attempt : I found the following
$$S_1 =1 , d_1 =1
\S_2 =3 , d_2 =3
\S_3 =6 , d_3 =6
\S_4 =10 , d_4 =0
\S_5 =15 , d_5 =5
\S_6 =21 , d_6 =1
\S_7 =28 , d_7 =8
\S_8 =36 , d_8 =6
\S_9 =45 , d_9 =5
\S_10 =55 , d_10 =5
$$
I was expecting to get a sequence, but did not find any. Please help me to solve this question.
sequences-and-series summation divisibility
asked Jul 31 at 6:39
MathsLearner
657213
add a comment |Â
up vote
3
down vote
favorite
Let $d_n$ be the last digit of $S_n$ where $S_n = left(1 + 2 + 3 + .... + nright)$. Find the remainder when $sum_i=1^2017d_i$ is divided by 1000.
My attempt : I found the following
$$S_1 =1 , d_1 =1
\S_2 =3 , d_2 =3
\S_3 =6 , d_3 =6
\S_4 =10 , d_4 =0
\S_5 =15 , d_5 =5
\S_6 =21 , d_6 =1
\S_7 =28 , d_7 =8
\S_8 =36 , d_8 =6
\S_9 =45 , d_9 =5
\S_10 =55 , d_10 =5
$$
I was expecting to get a sequence, but did not find any. Please help me to solve this question.
sequences-and-series summation divisibility
asked Jul 31 at 6:39
MathsLearner
657213
Finding the last digit of $S_n$ is the same as calculating $S_n mod 10$, and $d_1 mod 10 + d_2 mod 10 = (d_1+d_2) mod 10$. Does that help?
– postmortes
Jul 31 at 6:52
1
I believe $sum^nd_i$ is tabulated at oeis.org/A111072 and for $n=2017$ the table in the link at that page gives 7069.
– Gerry Myerson
Jul 31 at 7:19
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $d_n$ be the last digit of $S_n$ where $S_n = left(1 + 2 + 3 + .... + nright)$. Find the remainder when $sum_i=1^2017d_i$ is divided by 1000.
My attempt : I found the following
$$S_1 =1 , d_1 =1
\S_2 =3 , d_2 =3
\S_3 =6 , d_3 =6
\S_4 =10 , d_4 =0
\S_5 =15 , d_5 =5
\S_6 =21 , d_6 =1
\S_7 =28 , d_7 =8
\S_8 =36 , d_8 =6
\S_9 =45 , d_9 =5
\S_10 =55 , d_10 =5
$$
I was expecting to get a sequence, but did not find any. Please help me to solve this question.
sequences-and-series summation divisibility
asked Jul 31 at 6:39
MathsLearner
657213
Let $d_n$ be the last digit of $S_n$ where $S_n = left(1 + 2 + 3 + .... + nright)$. Find the remainder when $sum_i=1^2017d_i$ is divided by 1000.
My attempt : I found the following
$$S_1 =1 , d_1 =1
\S_2 =3 , d_2 =3
\S_3 =6 , d_3 =6
\S_4 =10 , d_4 =0
\S_5 =15 , d_5 =5
\S_6 =21 , d_6 =1
\S_7 =28 , d_7 =8
\S_8 =36 , d_8 =6
\S_9 =45 , d_9 =5
\S_10 =55 , d_10 =5
$$
I was expecting to get a sequence, but did not find any. Please help me to solve this question.
sequences-and-series summation divisibility
asked Jul 31 at 6:39
MathsLearner
657213
asked Jul 31 at 6:39
MathsLearner
657213
asked Jul 31 at 6:39
MathsLearner
657213
asked Jul 31 at 6:39
MathsLearner
657213
657213
Finding the last digit of $S_n$ is the same as calculating $S_n mod 10$, and $d_1 mod 10 + d_2 mod 10 = (d_1+d_2) mod 10$. Does that help?
– postmortes
Jul 31 at 6:52
1
I believe $sum^nd_i$ is tabulated at oeis.org/A111072 and for $n=2017$ the table in the link at that page gives 7069.
– Gerry Myerson
Jul 31 at 7:19
add a comment |Â
Finding the last digit of $S_n$ is the same as calculating $S_n mod 10$, and $d_1 mod 10 + d_2 mod 10 = (d_1+d_2) mod 10$. Does that help?
– postmortes
Jul 31 at 6:52
1
I believe $sum^nd_i$ is tabulated at oeis.org/A111072 and for $n=2017$ the table in the link at that page gives 7069.
– Gerry Myerson
Jul 31 at 7:19
Finding the last digit of $S_n$ is the same as calculating $S_n mod 10$, and $d_1 mod 10 + d_2 mod 10 = (d_1+d_2) mod 10$. Does that help?
– postmortes
Jul 31 at 6:52
Finding the last digit of $S_n$ is the same as calculating $S_n mod 10$, and $d_1 mod 10 + d_2 mod 10 = (d_1+d_2) mod 10$. Does that help?
– postmortes
Jul 31 at 6:52
1
1
I believe $sum^nd_i$ is tabulated at oeis.org/A111072 and for $n=2017$ the table in the link at that page gives 7069.
– Gerry Myerson
Jul 31 at 7:19
I believe $sum^nd_i$ is tabulated at oeis.org/A111072 and for $n=2017$ the table in the link at that page gives 7069.
– Gerry Myerson
Jul 31 at 7:19
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
8
down vote
accepted
The $S_n$ are the triangular numbers and their last digit is given by
$$fracn(n+1)2bmod10,$$ which is
$$left(frac n2bmod10right)((n+1)bmod10)bmod10$$ for even $n$ and $$left(frac n+12bmod10right)(nbmod10)bmod10$$ for odd $n$. Hence the sequence has period $20$, with a sum of $70$.
Then the sum
$$sum_i=1^2017fracn(n+1)2bmod10$$
has $100$ complete periods and $17$ remaining terms, $7069$ in total.
answered Jul 31 at 8:04
Yves Daoust
110k665203
This is excellent. I found the period by inspection and then constructed the solution around it. +1
– Piyush Divyanakar
Jul 31 at 9:41
add a comment |Â
up vote
3
down vote
Ok So note the following things for $n=20m$
$$d_n+1+d_n+2...+d_n+4=10 \
d_n+5+d_n+6...+d_n+15=50 \
d_n+16+d_n+16...+d_n+20=10$$
Can you use this to calculate the result?
- Note that $n=20mimplies mod(S_n, 10)=0$. Now $$S_n+1=S_n+20m+1 \ S_n+1=1mod(10) \ S_n+2=3mod(10) \ S_n+3=6mod(10) \ S_n+4=0mod(10)$$
- Note that $n=20m+15impliesmod(S_n,10)=0$, so similar to above argument we can find the sequence form $S_n+5$ to $S_n+15$.
- Same argument work for $S_n+16$ to $S_n+20$.
So to answer the question $sum_1^2000d_i=70timesfrac200020=7000$, $sum_2001^2015d_i=60$, $d_2016=6, d_2017=3$
Hence $T=sum_2001^2017d_i=7069$, $T%1000=69$
answered Jul 31 at 7:17
Piyush Divyanakar
3,258122
add a comment |Â
up vote
1
down vote
continuing
$$0,1,3,6,0,5,1,8,6,5,5,6,8,1,5,0,6,3,1,0,0,1・・・$$
You can find something.
answered Jul 31 at 7:44
Takahiro Waki
1,964520
add a comment |Â
up vote
0
down vote
In general, if $f(x)inmathbbQ[x]$ is an integer-valued polynomial of degree $d$ and $m$ is a positive integer, then
$$fbig(n+d!,mbig)equiv f(n)pmodm$$
for every $ninmathbbZ$. Hence, $d!,m$ is a (not necessarily minimal) period of $big(f(n)big)_ninmathbbZ$ modulo $m$. To prove this, it is well known that
$$f(x)=sum_k=0^d,t_k,binomxk$$
for some $t_0,t_1,t_2,ldots,t_k inmathbbZ$. Thus, $F(x):=d!,f(x)$ is a polynomial with integral coefficients. It is also commonly known that, for $g(x)inmathbbZ[x]$, $g(n+N)equiv g(n)pmodN$ for any $ninmathbbZ$ and $NinmathbbZ_>0$. That is,
$$F(n+d!,m)equiv F(n)pmodd!,m,.$$
Dividing both sides by $d!$, noting that $d!mid F(n)$ and $d!mid F(n+d!,m)$, we obtain the desired result.
Now, since $S_n=dfracn(n+1)2$ is an integer-valued polynomial of degree $2$, we see that $S_n$ in modulo $10$ is periodic with period $2!cdot 10=20$. The rest is just as other users have concluded.
answered Jul 31 at 9:15
Batominovski
22.8k22776
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
The $S_n$ are the triangular numbers and their last digit is given by
$$fracn(n+1)2bmod10,$$ which is
$$left(frac n2bmod10right)((n+1)bmod10)bmod10$$ for even $n$ and $$left(frac n+12bmod10right)(nbmod10)bmod10$$ for odd $n$. Hence the sequence has period $20$, with a sum of $70$.
Then the sum
$$sum_i=1^2017fracn(n+1)2bmod10$$
has $100$ complete periods and $17$ remaining terms, $7069$ in total.
answered Jul 31 at 8:04
Yves Daoust
110k665203
This is excellent. I found the period by inspection and then constructed the solution around it. +1
– Piyush Divyanakar
Jul 31 at 9:41
add a comment |Â
up vote
8
down vote
accepted
The $S_n$ are the triangular numbers and their last digit is given by
$$fracn(n+1)2bmod10,$$ which is
$$left(frac n2bmod10right)((n+1)bmod10)bmod10$$ for even $n$ and $$left(frac n+12bmod10right)(nbmod10)bmod10$$ for odd $n$. Hence the sequence has period $20$, with a sum of $70$.
Then the sum
$$sum_i=1^2017fracn(n+1)2bmod10$$
has $100$ complete periods and $17$ remaining terms, $7069$ in total.
answered Jul 31 at 8:04
Yves Daoust
110k665203
This is excellent. I found the period by inspection and then constructed the solution around it. +1
– Piyush Divyanakar
Jul 31 at 9:41
add a comment |Â
up vote
8
down vote
accepted
up vote
8
down vote
accepted
The $S_n$ are the triangular numbers and their last digit is given by
$$fracn(n+1)2bmod10,$$ which is
$$left(frac n2bmod10right)((n+1)bmod10)bmod10$$ for even $n$ and $$left(frac n+12bmod10right)(nbmod10)bmod10$$ for odd $n$. Hence the sequence has period $20$, with a sum of $70$.
Then the sum
$$sum_i=1^2017fracn(n+1)2bmod10$$
has $100$ complete periods and $17$ remaining terms, $7069$ in total.
answered Jul 31 at 8:04
Yves Daoust
110k665203
The $S_n$ are the triangular numbers and their last digit is given by
$$fracn(n+1)2bmod10,$$ which is
$$left(frac n2bmod10right)((n+1)bmod10)bmod10$$ for even $n$ and $$left(frac n+12bmod10right)(nbmod10)bmod10$$ for odd $n$. Hence the sequence has period $20$, with a sum of $70$.
Then the sum
$$sum_i=1^2017fracn(n+1)2bmod10$$
has $100$ complete periods and $17$ remaining terms, $7069$ in total.
answered Jul 31 at 8:04
Yves Daoust
110k665203
edited Jul 31 at 14:44
answered Jul 31 at 8:04
Yves Daoust
110k665203
answered Jul 31 at 8:04
Yves Daoust
110k665203
answered Jul 31 at 8:04
Yves Daoust
110k665203
110k665203
This is excellent. I found the period by inspection and then constructed the solution around it. +1
– Piyush Divyanakar
Jul 31 at 9:41
add a comment |Â
This is excellent. I found the period by inspection and then constructed the solution around it. +1
– Piyush Divyanakar
Jul 31 at 9:41
This is excellent. I found the period by inspection and then constructed the solution around it. +1
– Piyush Divyanakar
Jul 31 at 9:41
This is excellent. I found the period by inspection and then constructed the solution around it. +1
– Piyush Divyanakar
Jul 31 at 9:41
add a comment |Â
up vote
3
down vote
Ok So note the following things for $n=20m$
$$d_n+1+d_n+2...+d_n+4=10 \
d_n+5+d_n+6...+d_n+15=50 \
d_n+16+d_n+16...+d_n+20=10$$
Can you use this to calculate the result?
- Note that $n=20mimplies mod(S_n, 10)=0$. Now $$S_n+1=S_n+20m+1 \ S_n+1=1mod(10) \ S_n+2=3mod(10) \ S_n+3=6mod(10) \ S_n+4=0mod(10)$$
- Note that $n=20m+15impliesmod(S_n,10)=0$, so similar to above argument we can find the sequence form $S_n+5$ to $S_n+15$.
- Same argument work for $S_n+16$ to $S_n+20$.
So to answer the question $sum_1^2000d_i=70timesfrac200020=7000$, $sum_2001^2015d_i=60$, $d_2016=6, d_2017=3$
Hence $T=sum_2001^2017d_i=7069$, $T%1000=69$
answered Jul 31 at 7:17
Piyush Divyanakar
3,258122
add a comment |Â
up vote
3
down vote
Ok So note the following things for $n=20m$
$$d_n+1+d_n+2...+d_n+4=10 \
d_n+5+d_n+6...+d_n+15=50 \
d_n+16+d_n+16...+d_n+20=10$$
Can you use this to calculate the result?
- Note that $n=20mimplies mod(S_n, 10)=0$. Now $$S_n+1=S_n+20m+1 \ S_n+1=1mod(10) \ S_n+2=3mod(10) \ S_n+3=6mod(10) \ S_n+4=0mod(10)$$
- Note that $n=20m+15impliesmod(S_n,10)=0$, so similar to above argument we can find the sequence form $S_n+5$ to $S_n+15$.
- Same argument work for $S_n+16$ to $S_n+20$.
So to answer the question $sum_1^2000d_i=70timesfrac200020=7000$, $sum_2001^2015d_i=60$, $d_2016=6, d_2017=3$
Hence $T=sum_2001^2017d_i=7069$, $T%1000=69$
answered Jul 31 at 7:17
Piyush Divyanakar
3,258122
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Ok So note the following things for $n=20m$
$$d_n+1+d_n+2...+d_n+4=10 \
d_n+5+d_n+6...+d_n+15=50 \
d_n+16+d_n+16...+d_n+20=10$$
Can you use this to calculate the result?
- Note that $n=20mimplies mod(S_n, 10)=0$. Now $$S_n+1=S_n+20m+1 \ S_n+1=1mod(10) \ S_n+2=3mod(10) \ S_n+3=6mod(10) \ S_n+4=0mod(10)$$
- Note that $n=20m+15impliesmod(S_n,10)=0$, so similar to above argument we can find the sequence form $S_n+5$ to $S_n+15$.
- Same argument work for $S_n+16$ to $S_n+20$.
So to answer the question $sum_1^2000d_i=70timesfrac200020=7000$, $sum_2001^2015d_i=60$, $d_2016=6, d_2017=3$
Hence $T=sum_2001^2017d_i=7069$, $T%1000=69$
answered Jul 31 at 7:17
Piyush Divyanakar
3,258122
Ok So note the following things for $n=20m$
$$d_n+1+d_n+2...+d_n+4=10 \
d_n+5+d_n+6...+d_n+15=50 \
d_n+16+d_n+16...+d_n+20=10$$
Can you use this to calculate the result?
- Note that $n=20mimplies mod(S_n, 10)=0$. Now $$S_n+1=S_n+20m+1 \ S_n+1=1mod(10) \ S_n+2=3mod(10) \ S_n+3=6mod(10) \ S_n+4=0mod(10)$$
- Note that $n=20m+15impliesmod(S_n,10)=0$, so similar to above argument we can find the sequence form $S_n+5$ to $S_n+15$.
- Same argument work for $S_n+16$ to $S_n+20$.
So to answer the question $sum_1^2000d_i=70timesfrac200020=7000$, $sum_2001^2015d_i=60$, $d_2016=6, d_2017=3$
Hence $T=sum_2001^2017d_i=7069$, $T%1000=69$
answered Jul 31 at 7:17
Piyush Divyanakar
3,258122
edited Jul 31 at 7:35
answered Jul 31 at 7:17
Piyush Divyanakar
3,258122
answered Jul 31 at 7:17
Piyush Divyanakar
3,258122
answered Jul 31 at 7:17
Piyush Divyanakar
3,258122
3,258122
add a comment |Â
add a comment |Â
up vote
1
down vote
continuing
$$0,1,3,6,0,5,1,8,6,5,5,6,8,1,5,0,6,3,1,0,0,1・・・$$
You can find something.
answered Jul 31 at 7:44
Takahiro Waki
1,964520
add a comment |Â
up vote
1
down vote
continuing
$$0,1,3,6,0,5,1,8,6,5,5,6,8,1,5,0,6,3,1,0,0,1・・・$$
You can find something.
answered Jul 31 at 7:44
Takahiro Waki
1,964520
add a comment |Â
up vote
1
down vote
up vote
1
down vote
continuing
$$0,1,3,6,0,5,1,8,6,5,5,6,8,1,5,0,6,3,1,0,0,1・・・$$
You can find something.
answered Jul 31 at 7:44
Takahiro Waki
1,964520
continuing
$$0,1,3,6,0,5,1,8,6,5,5,6,8,1,5,0,6,3,1,0,0,1・・・$$
You can find something.
answered Jul 31 at 7:44
Takahiro Waki
1,964520
answered Jul 31 at 7:44
Takahiro Waki
1,964520
answered Jul 31 at 7:44
Takahiro Waki
1,964520
answered Jul 31 at 7:44
Takahiro Waki
1,964520
1,964520
add a comment |Â
add a comment |Â
up vote
0
down vote
In general, if $f(x)inmathbbQ[x]$ is an integer-valued polynomial of degree $d$ and $m$ is a positive integer, then
$$fbig(n+d!,mbig)equiv f(n)pmodm$$
for every $ninmathbbZ$. Hence, $d!,m$ is a (not necessarily minimal) period of $big(f(n)big)_ninmathbbZ$ modulo $m$. To prove this, it is well known that
$$f(x)=sum_k=0^d,t_k,binomxk$$
for some $t_0,t_1,t_2,ldots,t_k inmathbbZ$. Thus, $F(x):=d!,f(x)$ is a polynomial with integral coefficients. It is also commonly known that, for $g(x)inmathbbZ[x]$, $g(n+N)equiv g(n)pmodN$ for any $ninmathbbZ$ and $NinmathbbZ_>0$. That is,
$$F(n+d!,m)equiv F(n)pmodd!,m,.$$
Dividing both sides by $d!$, noting that $d!mid F(n)$ and $d!mid F(n+d!,m)$, we obtain the desired result.
Now, since $S_n=dfracn(n+1)2$ is an integer-valued polynomial of degree $2$, we see that $S_n$ in modulo $10$ is periodic with period $2!cdot 10=20$. The rest is just as other users have concluded.
answered Jul 31 at 9:15
Batominovski
22.8k22776
add a comment |Â
up vote
0
down vote
In general, if $f(x)inmathbbQ[x]$ is an integer-valued polynomial of degree $d$ and $m$ is a positive integer, then
$$fbig(n+d!,mbig)equiv f(n)pmodm$$
for every $ninmathbbZ$. Hence, $d!,m$ is a (not necessarily minimal) period of $big(f(n)big)_ninmathbbZ$ modulo $m$. To prove this, it is well known that
$$f(x)=sum_k=0^d,t_k,binomxk$$
for some $t_0,t_1,t_2,ldots,t_k inmathbbZ$. Thus, $F(x):=d!,f(x)$ is a polynomial with integral coefficients. It is also commonly known that, for $g(x)inmathbbZ[x]$, $g(n+N)equiv g(n)pmodN$ for any $ninmathbbZ$ and $NinmathbbZ_>0$. That is,
$$F(n+d!,m)equiv F(n)pmodd!,m,.$$
Dividing both sides by $d!$, noting that $d!mid F(n)$ and $d!mid F(n+d!,m)$, we obtain the desired result.
Now, since $S_n=dfracn(n+1)2$ is an integer-valued polynomial of degree $2$, we see that $S_n$ in modulo $10$ is periodic with period $2!cdot 10=20$. The rest is just as other users have concluded.
answered Jul 31 at 9:15
Batominovski
22.8k22776
add a comment |Â
up vote
0
down vote
up vote
0
down vote
In general, if $f(x)inmathbbQ[x]$ is an integer-valued polynomial of degree $d$ and $m$ is a positive integer, then
$$fbig(n+d!,mbig)equiv f(n)pmodm$$
for every $ninmathbbZ$. Hence, $d!,m$ is a (not necessarily minimal) period of $big(f(n)big)_ninmathbbZ$ modulo $m$. To prove this, it is well known that
$$f(x)=sum_k=0^d,t_k,binomxk$$
for some $t_0,t_1,t_2,ldots,t_k inmathbbZ$. Thus, $F(x):=d!,f(x)$ is a polynomial with integral coefficients. It is also commonly known that, for $g(x)inmathbbZ[x]$, $g(n+N)equiv g(n)pmodN$ for any $ninmathbbZ$ and $NinmathbbZ_>0$. That is,
$$F(n+d!,m)equiv F(n)pmodd!,m,.$$
Dividing both sides by $d!$, noting that $d!mid F(n)$ and $d!mid F(n+d!,m)$, we obtain the desired result.
Now, since $S_n=dfracn(n+1)2$ is an integer-valued polynomial of degree $2$, we see that $S_n$ in modulo $10$ is periodic with period $2!cdot 10=20$. The rest is just as other users have concluded.
answered Jul 31 at 9:15
Batominovski
22.8k22776
In general, if $f(x)inmathbbQ[x]$ is an integer-valued polynomial of degree $d$ and $m$ is a positive integer, then
$$fbig(n+d!,mbig)equiv f(n)pmodm$$
for every $ninmathbbZ$. Hence, $d!,m$ is a (not necessarily minimal) period of $big(f(n)big)_ninmathbbZ$ modulo $m$. To prove this, it is well known that
$$f(x)=sum_k=0^d,t_k,binomxk$$
for some $t_0,t_1,t_2,ldots,t_k inmathbbZ$. Thus, $F(x):=d!,f(x)$ is a polynomial with integral coefficients. It is also commonly known that, for $g(x)inmathbbZ[x]$, $g(n+N)equiv g(n)pmodN$ for any $ninmathbbZ$ and $NinmathbbZ_>0$. That is,
$$F(n+d!,m)equiv F(n)pmodd!,m,.$$
Dividing both sides by $d!$, noting that $d!mid F(n)$ and $d!mid F(n+d!,m)$, we obtain the desired result.
Now, since $S_n=dfracn(n+1)2$ is an integer-valued polynomial of degree $2$, we see that $S_n$ in modulo $10$ is periodic with period $2!cdot 10=20$. The rest is just as other users have concluded.
answered Jul 31 at 9:15
Batominovski
22.8k22776
answered Jul 31 at 9:15
Batominovski
22.8k22776
answered Jul 31 at 9:15
Batominovski
22.8k22776
answered Jul 31 at 9:15
Batominovski
22.8k22776
22.8k22776
add a comment |Â
add a comment |Â
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Finding the last digit of $S_n$ is the same as calculating $S_n mod 10$, and $d_1 mod 10 + d_2 mod 10 = (d_1+d_2) mod 10$. Does that help?
– postmortes
Jul 31 at 6:52
1
I believe $sum^nd_i$ is tabulated at oeis.org/A111072 and for $n=2017$ the table in the link at that page gives 7069.
– Gerry Myerson
Jul 31 at 7:19