What's $1^3+2^3+3^3+cdots+99^3$ modulo $3$?
$begingroup$
What's the remainder when the sum
$1^3+2^3+3^3+cdots+99^3$ is divided by $3$?
Background:
I saw this question on MSE but it was closed and I wanted to learn how to approach it. The help given to the asker had failed to bring the question up to an acceptable standard.
I can solve it but I doubt my methods are efficient.
One way is to start by removing every third term. Then the first two terms of the remaining sequence can be removed, and so on... a pattern may emerge.
Another way is to look at the expansion of $(x+1)^3$ and see what it does to the residues for each $x$, then sum over that by induction.
But I'm sure my inventions aren't very efficient.
algebra-precalculus elementary-number-theory modular-arithmetic
asked Dec 27 '18 at 5:40
user334732user334732
4,28411240
$endgroup$
add a comment |
$begingroup$
What's the remainder when the sum
$1^3+2^3+3^3+cdots+99^3$ is divided by $3$?
Background:
I saw this question on MSE but it was closed and I wanted to learn how to approach it. The help given to the asker had failed to bring the question up to an acceptable standard.
I can solve it but I doubt my methods are efficient.
One way is to start by removing every third term. Then the first two terms of the remaining sequence can be removed, and so on... a pattern may emerge.
Another way is to look at the expansion of $(x+1)^3$ and see what it does to the residues for each $x$, then sum over that by induction.
But I'm sure my inventions aren't very efficient.
algebra-precalculus elementary-number-theory modular-arithmetic
asked Dec 27 '18 at 5:40
user334732user334732
4,28411240
$endgroup$
5
$begingroup$
Use $n^3equiv npmod3$.
$endgroup$
– Lord Shark the Unknown
Dec 27 '18 at 5:42
$begingroup$
math.stackexchange.com/questions/1328798
$endgroup$
– Andrei
Dec 27 '18 at 5:43
1
$begingroup$
In general, note that you can handle the terms in groups of $3$ as $n + 3 equiv n pmod 3$, and also use that $n^3 equiv n pmod 3$ as Lord Shark the Unknown stated above.
$endgroup$
– John Omielan
Dec 27 '18 at 5:46
$begingroup$
@LordSharktheUnknown ah thanks. I guess the $(x+1)^3$ method would've revealed that.
$endgroup$
– user334732
Dec 27 '18 at 5:46
add a comment |
$begingroup$
What's the remainder when the sum
$1^3+2^3+3^3+cdots+99^3$ is divided by $3$?
Background:
I saw this question on MSE but it was closed and I wanted to learn how to approach it. The help given to the asker had failed to bring the question up to an acceptable standard.
I can solve it but I doubt my methods are efficient.
One way is to start by removing every third term. Then the first two terms of the remaining sequence can be removed, and so on... a pattern may emerge.
Another way is to look at the expansion of $(x+1)^3$ and see what it does to the residues for each $x$, then sum over that by induction.
But I'm sure my inventions aren't very efficient.
algebra-precalculus elementary-number-theory modular-arithmetic
asked Dec 27 '18 at 5:40
user334732user334732
4,28411240
$endgroup$
What's the remainder when the sum
$1^3+2^3+3^3+cdots+99^3$ is divided by $3$?
Background:
I saw this question on MSE but it was closed and I wanted to learn how to approach it. The help given to the asker had failed to bring the question up to an acceptable standard.
I can solve it but I doubt my methods are efficient.
One way is to start by removing every third term. Then the first two terms of the remaining sequence can be removed, and so on... a pattern may emerge.
Another way is to look at the expansion of $(x+1)^3$ and see what it does to the residues for each $x$, then sum over that by induction.
But I'm sure my inventions aren't very efficient.
algebra-precalculus elementary-number-theory modular-arithmetic
algebra-precalculus elementary-number-theory modular-arithmetic
asked Dec 27 '18 at 5:40
user334732user334732
4,28411240
asked Dec 27 '18 at 5:40
user334732user334732
4,28411240
asked Dec 27 '18 at 5:40
user334732user334732
4,28411240
asked Dec 27 '18 at 5:40
user334732user334732
4,28411240
asked Dec 27 '18 at 5:40
user334732user334732
4,28411240
4,28411240
5
$begingroup$
Use $n^3equiv npmod3$.
$endgroup$
– Lord Shark the Unknown
Dec 27 '18 at 5:42
$begingroup$
math.stackexchange.com/questions/1328798
$endgroup$
– Andrei
Dec 27 '18 at 5:43
1
$begingroup$
In general, note that you can handle the terms in groups of $3$ as $n + 3 equiv n pmod 3$, and also use that $n^3 equiv n pmod 3$ as Lord Shark the Unknown stated above.
$endgroup$
– John Omielan
Dec 27 '18 at 5:46
$begingroup$
@LordSharktheUnknown ah thanks. I guess the $(x+1)^3$ method would've revealed that.
$endgroup$
– user334732
Dec 27 '18 at 5:46
add a comment |
5
$begingroup$
Use $n^3equiv npmod3$.
$endgroup$
– Lord Shark the Unknown
Dec 27 '18 at 5:42
$begingroup$
math.stackexchange.com/questions/1328798
$endgroup$
– Andrei
Dec 27 '18 at 5:43
1
$begingroup$
In general, note that you can handle the terms in groups of $3$ as $n + 3 equiv n pmod 3$, and also use that $n^3 equiv n pmod 3$ as Lord Shark the Unknown stated above.
$endgroup$
– John Omielan
Dec 27 '18 at 5:46
$begingroup$
@LordSharktheUnknown ah thanks. I guess the $(x+1)^3$ method would've revealed that.
$endgroup$
– user334732
Dec 27 '18 at 5:46
5
5
$begingroup$
Use $n^3equiv npmod3$.
$endgroup$
– Lord Shark the Unknown
Dec 27 '18 at 5:42
$begingroup$
Use $n^3equiv npmod3$.
$endgroup$
– Lord Shark the Unknown
Dec 27 '18 at 5:42
$begingroup$
math.stackexchange.com/questions/1328798
$endgroup$
– Andrei
Dec 27 '18 at 5:43
$begingroup$
math.stackexchange.com/questions/1328798
$endgroup$
– Andrei
Dec 27 '18 at 5:43
1
1
$begingroup$
In general, note that you can handle the terms in groups of $3$ as $n + 3 equiv n pmod 3$, and also use that $n^3 equiv n pmod 3$ as Lord Shark the Unknown stated above.
$endgroup$
– John Omielan
Dec 27 '18 at 5:46
$begingroup$
In general, note that you can handle the terms in groups of $3$ as $n + 3 equiv n pmod 3$, and also use that $n^3 equiv n pmod 3$ as Lord Shark the Unknown stated above.
$endgroup$
– John Omielan
Dec 27 '18 at 5:46
$begingroup$
@LordSharktheUnknown ah thanks. I guess the $(x+1)^3$ method would've revealed that.
$endgroup$
– user334732
Dec 27 '18 at 5:46
$begingroup$
@LordSharktheUnknown ah thanks. I guess the $(x+1)^3$ method would've revealed that.
$endgroup$
– user334732
Dec 27 '18 at 5:46
add a comment |
9 Answers
9
active
oldest
votes
$begingroup$
By Fermat's little theorem $a^3equiv apmod3$. Then you have $sum_{k=1}^{99} k=frac{(99)(100)}2=50cdot 99equiv 0pmod3$.
answered Dec 27 '18 at 5:46
Chris CusterChris Custer
13.8k3827
$endgroup$
add a comment |
$begingroup$
Hint:
$n^3equiv npmod3$ as $n^3-n=n(n-1)(n+1)$ is the product of three consecutive integers
We have $sum_{r=1}^mr^3equivsum_{r=1}^mrpmod3$
$equiv dfrac{m(m+1)}2$
Alternatively $$sum_{r=1}^m r^3=dfrac{m^2(m+1)^2}4$$
answered Dec 27 '18 at 5:44
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
add a comment |
$begingroup$
begin{align}
sum_{i=1}^{99}i^3 &= sum_{i=0}^{32} (3i+1)^3 + sum_{i=0}^{32} (3i+2)^3 + sum_{i=0}^{32} (3i+3)^3 \
&equiv sum_{i=0}^{32} 1^3 + sum_{i=0}^{32} 2^3 + sum_{i=0}^{32} 0^3 pmod{3}\
&equiv sum_{i=0}^{32} 1^3 + sum_{i=0}^{32} (-1)^3 + sum_{i=0}^{32} 0^3 pmod{3}\
&equiv 0 pmod{3}
end{align}
answered Dec 27 '18 at 5:48
Siong Thye GohSiong Thye Goh
102k1466118
$endgroup$
add a comment |
$begingroup$
Split the numbers from $1,ldots, 99$ in
$33$ with remainder $1$
$33$ with remainder $2$
$33$ with remainder $0$
$$1^3 + 2^3 + cdots + 99^3 equiv 33cdot (1^3+2^3+0^3) equiv 0 mod 3$$
answered Dec 27 '18 at 5:51
trancelocationtrancelocation
12.3k1826
$endgroup$
add a comment |
$begingroup$
Another method is to use the identity
$$
sum_{k=1}^nk^3=left(sum_{k=1}^nkright)^2
$$
So,
$$
sum_{k=1}^{99}k^3equivleft(sum_{k=1}^{99}kright)^2equivfrac{(99)^2(100)^2}{4}equiv0 (mod3)
$$
answered Dec 27 '18 at 6:18
GödelGödel
1,440319
$endgroup$
add a comment |
$begingroup$
Yet another method is to use that $(x-1)^3=x^3-3x^2+3x-1$ and $(x+1)^3=x^3+3x^2+3x+1$, so $(x-1)^3+x^3+(x+1)^3 = 3x^3+6x$, which is divisible by three. This shows that any three consecutive numbers, when cubed, add up to zero modulus three.
edited Dec 27 '18 at 6:19
abiessu
6,68721541
answered Dec 27 '18 at 6:08
AcccumulationAcccumulation
7,0192619
$endgroup$
1
$begingroup$
@Accumulation A small issue is that in the second equation, $left(x - 1right)^3$ should be $left(x + 1right)^3$ instead.
$endgroup$
– John Omielan
Dec 27 '18 at 6:18
add a comment |
$begingroup$
From $a^{3} + b^{3} = (a+b)(a^{2}-ab+b^{2})$, you have $a+b|a^{3} +b^{3}$ and
begin{align}
&1^{3} + 2^{3} + 3^{3} + cdots + 97^{3} + 98^{3} + 99^{3} \
&= (1^{3} + 2^{3}) + (4^{3} + 5^{3}) + cdots + (97^{3}+98^{3}) + (3^{3}+6^{3} + cdots + 99^{3})
end{align}
is a multiple of 3.
answered Dec 27 '18 at 5:49
Seewoo LeeSeewoo Lee
6,954927
$endgroup$
add a comment |
$begingroup$
You can also calculate that sum using
$sum_{i=0}^{i=n}i^3=dfrac{n^2(n+1)^2}{4}$
Here: $sum_{i=0}^{i=99}i^3=dfrac{99^2(99+1)^2}{4}$ which is clearly divisible by 3.
answered Dec 27 '18 at 7:46
harshit54harshit54
348113
$endgroup$
add a comment |
$begingroup$
consider these observations:
$$1 = 1 pmod 3$$
$$2 = -1 pmod 3$$
$$3 = 0 pmod 3$$
This becomes:
$$33(1^3+(-1)^3+0) equiv 33(1+(-1)+0) equiv 33(0) equiv 0 pmod 3$$
answered Dec 27 '18 at 14:57
Maged SaeedMaged Saeed
8671417
$endgroup$
add a comment |
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9 Answers
9
active
oldest
votes
9 Answers
9
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
By Fermat's little theorem $a^3equiv apmod3$. Then you have $sum_{k=1}^{99} k=frac{(99)(100)}2=50cdot 99equiv 0pmod3$.
answered Dec 27 '18 at 5:46
Chris CusterChris Custer
13.8k3827
$endgroup$
add a comment |
$begingroup$
By Fermat's little theorem $a^3equiv apmod3$. Then you have $sum_{k=1}^{99} k=frac{(99)(100)}2=50cdot 99equiv 0pmod3$.
answered Dec 27 '18 at 5:46
Chris CusterChris Custer
13.8k3827
$endgroup$
add a comment |
$begingroup$
By Fermat's little theorem $a^3equiv apmod3$. Then you have $sum_{k=1}^{99} k=frac{(99)(100)}2=50cdot 99equiv 0pmod3$.
answered Dec 27 '18 at 5:46
Chris CusterChris Custer
13.8k3827
$endgroup$
By Fermat's little theorem $a^3equiv apmod3$. Then you have $sum_{k=1}^{99} k=frac{(99)(100)}2=50cdot 99equiv 0pmod3$.
answered Dec 27 '18 at 5:46
Chris CusterChris Custer
13.8k3827
answered Dec 27 '18 at 5:46
Chris CusterChris Custer
13.8k3827
answered Dec 27 '18 at 5:46
Chris CusterChris Custer
13.8k3827
answered Dec 27 '18 at 5:46
Chris CusterChris Custer
13.8k3827
13.8k3827
add a comment |
add a comment |
$begingroup$
Hint:
$n^3equiv npmod3$ as $n^3-n=n(n-1)(n+1)$ is the product of three consecutive integers
We have $sum_{r=1}^mr^3equivsum_{r=1}^mrpmod3$
$equiv dfrac{m(m+1)}2$
Alternatively $$sum_{r=1}^m r^3=dfrac{m^2(m+1)^2}4$$
answered Dec 27 '18 at 5:44
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
add a comment |
$begingroup$
Hint:
$n^3equiv npmod3$ as $n^3-n=n(n-1)(n+1)$ is the product of three consecutive integers
We have $sum_{r=1}^mr^3equivsum_{r=1}^mrpmod3$
$equiv dfrac{m(m+1)}2$
Alternatively $$sum_{r=1}^m r^3=dfrac{m^2(m+1)^2}4$$
answered Dec 27 '18 at 5:44
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
add a comment |
$begingroup$
Hint:
$n^3equiv npmod3$ as $n^3-n=n(n-1)(n+1)$ is the product of three consecutive integers
We have $sum_{r=1}^mr^3equivsum_{r=1}^mrpmod3$
$equiv dfrac{m(m+1)}2$
Alternatively $$sum_{r=1}^m r^3=dfrac{m^2(m+1)^2}4$$
answered Dec 27 '18 at 5:44
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
Hint:
$n^3equiv npmod3$ as $n^3-n=n(n-1)(n+1)$ is the product of three consecutive integers
We have $sum_{r=1}^mr^3equivsum_{r=1}^mrpmod3$
$equiv dfrac{m(m+1)}2$
Alternatively $$sum_{r=1}^m r^3=dfrac{m^2(m+1)^2}4$$
answered Dec 27 '18 at 5:44
lab bhattacharjeelab bhattacharjee
226k15157275
answered Dec 27 '18 at 5:44
lab bhattacharjeelab bhattacharjee
226k15157275
answered Dec 27 '18 at 5:44
lab bhattacharjeelab bhattacharjee
226k15157275
answered Dec 27 '18 at 5:44
lab bhattacharjeelab bhattacharjee
226k15157275
226k15157275
add a comment |
add a comment |
$begingroup$
begin{align}
sum_{i=1}^{99}i^3 &= sum_{i=0}^{32} (3i+1)^3 + sum_{i=0}^{32} (3i+2)^3 + sum_{i=0}^{32} (3i+3)^3 \
&equiv sum_{i=0}^{32} 1^3 + sum_{i=0}^{32} 2^3 + sum_{i=0}^{32} 0^3 pmod{3}\
&equiv sum_{i=0}^{32} 1^3 + sum_{i=0}^{32} (-1)^3 + sum_{i=0}^{32} 0^3 pmod{3}\
&equiv 0 pmod{3}
end{align}
answered Dec 27 '18 at 5:48
Siong Thye GohSiong Thye Goh
102k1466118
$endgroup$
add a comment |
$begingroup$
begin{align}
sum_{i=1}^{99}i^3 &= sum_{i=0}^{32} (3i+1)^3 + sum_{i=0}^{32} (3i+2)^3 + sum_{i=0}^{32} (3i+3)^3 \
&equiv sum_{i=0}^{32} 1^3 + sum_{i=0}^{32} 2^3 + sum_{i=0}^{32} 0^3 pmod{3}\
&equiv sum_{i=0}^{32} 1^3 + sum_{i=0}^{32} (-1)^3 + sum_{i=0}^{32} 0^3 pmod{3}\
&equiv 0 pmod{3}
end{align}
answered Dec 27 '18 at 5:48
Siong Thye GohSiong Thye Goh
102k1466118
$endgroup$
add a comment |
$begingroup$
begin{align}
sum_{i=1}^{99}i^3 &= sum_{i=0}^{32} (3i+1)^3 + sum_{i=0}^{32} (3i+2)^3 + sum_{i=0}^{32} (3i+3)^3 \
&equiv sum_{i=0}^{32} 1^3 + sum_{i=0}^{32} 2^3 + sum_{i=0}^{32} 0^3 pmod{3}\
&equiv sum_{i=0}^{32} 1^3 + sum_{i=0}^{32} (-1)^3 + sum_{i=0}^{32} 0^3 pmod{3}\
&equiv 0 pmod{3}
end{align}
answered Dec 27 '18 at 5:48
Siong Thye GohSiong Thye Goh
102k1466118
$endgroup$
begin{align}
sum_{i=1}^{99}i^3 &= sum_{i=0}^{32} (3i+1)^3 + sum_{i=0}^{32} (3i+2)^3 + sum_{i=0}^{32} (3i+3)^3 \
&equiv sum_{i=0}^{32} 1^3 + sum_{i=0}^{32} 2^3 + sum_{i=0}^{32} 0^3 pmod{3}\
&equiv sum_{i=0}^{32} 1^3 + sum_{i=0}^{32} (-1)^3 + sum_{i=0}^{32} 0^3 pmod{3}\
&equiv 0 pmod{3}
end{align}
answered Dec 27 '18 at 5:48
Siong Thye GohSiong Thye Goh
102k1466118
answered Dec 27 '18 at 5:48
Siong Thye GohSiong Thye Goh
102k1466118
answered Dec 27 '18 at 5:48
Siong Thye GohSiong Thye Goh
102k1466118
answered Dec 27 '18 at 5:48
Siong Thye GohSiong Thye Goh
102k1466118
102k1466118
add a comment |
add a comment |
$begingroup$
Split the numbers from $1,ldots, 99$ in
$33$ with remainder $1$
$33$ with remainder $2$
$33$ with remainder $0$
$$1^3 + 2^3 + cdots + 99^3 equiv 33cdot (1^3+2^3+0^3) equiv 0 mod 3$$
answered Dec 27 '18 at 5:51
trancelocationtrancelocation
12.3k1826
$endgroup$
add a comment |
$begingroup$
Split the numbers from $1,ldots, 99$ in
$33$ with remainder $1$
$33$ with remainder $2$
$33$ with remainder $0$
$$1^3 + 2^3 + cdots + 99^3 equiv 33cdot (1^3+2^3+0^3) equiv 0 mod 3$$
answered Dec 27 '18 at 5:51
trancelocationtrancelocation
12.3k1826
$endgroup$
add a comment |
$begingroup$
Split the numbers from $1,ldots, 99$ in
$33$ with remainder $1$
$33$ with remainder $2$
$33$ with remainder $0$
$$1^3 + 2^3 + cdots + 99^3 equiv 33cdot (1^3+2^3+0^3) equiv 0 mod 3$$
answered Dec 27 '18 at 5:51
trancelocationtrancelocation
12.3k1826
$endgroup$
Split the numbers from $1,ldots, 99$ in
$33$ with remainder $1$
$33$ with remainder $2$
$33$ with remainder $0$
$$1^3 + 2^3 + cdots + 99^3 equiv 33cdot (1^3+2^3+0^3) equiv 0 mod 3$$
answered Dec 27 '18 at 5:51
trancelocationtrancelocation
12.3k1826
answered Dec 27 '18 at 5:51
trancelocationtrancelocation
12.3k1826
answered Dec 27 '18 at 5:51
trancelocationtrancelocation
12.3k1826
answered Dec 27 '18 at 5:51
trancelocationtrancelocation
12.3k1826
12.3k1826
add a comment |
add a comment |
$begingroup$
Another method is to use the identity
$$
sum_{k=1}^nk^3=left(sum_{k=1}^nkright)^2
$$
So,
$$
sum_{k=1}^{99}k^3equivleft(sum_{k=1}^{99}kright)^2equivfrac{(99)^2(100)^2}{4}equiv0 (mod3)
$$
answered Dec 27 '18 at 6:18
GödelGödel
1,440319
$endgroup$
add a comment |
$begingroup$
Another method is to use the identity
$$
sum_{k=1}^nk^3=left(sum_{k=1}^nkright)^2
$$
So,
$$
sum_{k=1}^{99}k^3equivleft(sum_{k=1}^{99}kright)^2equivfrac{(99)^2(100)^2}{4}equiv0 (mod3)
$$
answered Dec 27 '18 at 6:18
GödelGödel
1,440319
$endgroup$
add a comment |
$begingroup$
Another method is to use the identity
$$
sum_{k=1}^nk^3=left(sum_{k=1}^nkright)^2
$$
So,
$$
sum_{k=1}^{99}k^3equivleft(sum_{k=1}^{99}kright)^2equivfrac{(99)^2(100)^2}{4}equiv0 (mod3)
$$
answered Dec 27 '18 at 6:18
GödelGödel
1,440319
$endgroup$
Another method is to use the identity
$$
sum_{k=1}^nk^3=left(sum_{k=1}^nkright)^2
$$
So,
$$
sum_{k=1}^{99}k^3equivleft(sum_{k=1}^{99}kright)^2equivfrac{(99)^2(100)^2}{4}equiv0 (mod3)
$$
answered Dec 27 '18 at 6:18
GödelGödel
1,440319
answered Dec 27 '18 at 6:18
GödelGödel
1,440319
answered Dec 27 '18 at 6:18
GödelGödel
1,440319
answered Dec 27 '18 at 6:18
GödelGödel
1,440319
1,440319
add a comment |
add a comment |
$begingroup$
Yet another method is to use that $(x-1)^3=x^3-3x^2+3x-1$ and $(x+1)^3=x^3+3x^2+3x+1$, so $(x-1)^3+x^3+(x+1)^3 = 3x^3+6x$, which is divisible by three. This shows that any three consecutive numbers, when cubed, add up to zero modulus three.
edited Dec 27 '18 at 6:19
abiessu
6,68721541
answered Dec 27 '18 at 6:08
AcccumulationAcccumulation
7,0192619
$endgroup$
1
$begingroup$
@Accumulation A small issue is that in the second equation, $left(x - 1right)^3$ should be $left(x + 1right)^3$ instead.
$endgroup$
– John Omielan
Dec 27 '18 at 6:18
add a comment |
$begingroup$
Yet another method is to use that $(x-1)^3=x^3-3x^2+3x-1$ and $(x+1)^3=x^3+3x^2+3x+1$, so $(x-1)^3+x^3+(x+1)^3 = 3x^3+6x$, which is divisible by three. This shows that any three consecutive numbers, when cubed, add up to zero modulus three.
edited Dec 27 '18 at 6:19
abiessu
6,68721541
answered Dec 27 '18 at 6:08
AcccumulationAcccumulation
7,0192619
$endgroup$
1
$begingroup$
@Accumulation A small issue is that in the second equation, $left(x - 1right)^3$ should be $left(x + 1right)^3$ instead.
$endgroup$
– John Omielan
Dec 27 '18 at 6:18
add a comment |
$begingroup$
Yet another method is to use that $(x-1)^3=x^3-3x^2+3x-1$ and $(x+1)^3=x^3+3x^2+3x+1$, so $(x-1)^3+x^3+(x+1)^3 = 3x^3+6x$, which is divisible by three. This shows that any three consecutive numbers, when cubed, add up to zero modulus three.
edited Dec 27 '18 at 6:19
abiessu
6,68721541
answered Dec 27 '18 at 6:08
AcccumulationAcccumulation
7,0192619
$endgroup$
Yet another method is to use that $(x-1)^3=x^3-3x^2+3x-1$ and $(x+1)^3=x^3+3x^2+3x+1$, so $(x-1)^3+x^3+(x+1)^3 = 3x^3+6x$, which is divisible by three. This shows that any three consecutive numbers, when cubed, add up to zero modulus three.
edited Dec 27 '18 at 6:19
abiessu
6,68721541
answered Dec 27 '18 at 6:08
AcccumulationAcccumulation
7,0192619
edited Dec 27 '18 at 6:19
abiessu
6,68721541
edited Dec 27 '18 at 6:19
abiessu
6,68721541
edited Dec 27 '18 at 6:19
abiessu
6,68721541
6,68721541
answered Dec 27 '18 at 6:08
AcccumulationAcccumulation
7,0192619
answered Dec 27 '18 at 6:08
AcccumulationAcccumulation
7,0192619
answered Dec 27 '18 at 6:08
AcccumulationAcccumulation
7,0192619
7,0192619
1
$begingroup$
@Accumulation A small issue is that in the second equation, $left(x - 1right)^3$ should be $left(x + 1right)^3$ instead.
$endgroup$
– John Omielan
Dec 27 '18 at 6:18
add a comment |
1
$begingroup$
@Accumulation A small issue is that in the second equation, $left(x - 1right)^3$ should be $left(x + 1right)^3$ instead.
$endgroup$
– John Omielan
Dec 27 '18 at 6:18
1
1
$begingroup$
@Accumulation A small issue is that in the second equation, $left(x - 1right)^3$ should be $left(x + 1right)^3$ instead.
$endgroup$
– John Omielan
Dec 27 '18 at 6:18
$begingroup$
@Accumulation A small issue is that in the second equation, $left(x - 1right)^3$ should be $left(x + 1right)^3$ instead.
$endgroup$
– John Omielan
Dec 27 '18 at 6:18
add a comment |
$begingroup$
From $a^{3} + b^{3} = (a+b)(a^{2}-ab+b^{2})$, you have $a+b|a^{3} +b^{3}$ and
begin{align}
&1^{3} + 2^{3} + 3^{3} + cdots + 97^{3} + 98^{3} + 99^{3} \
&= (1^{3} + 2^{3}) + (4^{3} + 5^{3}) + cdots + (97^{3}+98^{3}) + (3^{3}+6^{3} + cdots + 99^{3})
end{align}
is a multiple of 3.
answered Dec 27 '18 at 5:49
Seewoo LeeSeewoo Lee
6,954927
$endgroup$
add a comment |
$begingroup$
From $a^{3} + b^{3} = (a+b)(a^{2}-ab+b^{2})$, you have $a+b|a^{3} +b^{3}$ and
begin{align}
&1^{3} + 2^{3} + 3^{3} + cdots + 97^{3} + 98^{3} + 99^{3} \
&= (1^{3} + 2^{3}) + (4^{3} + 5^{3}) + cdots + (97^{3}+98^{3}) + (3^{3}+6^{3} + cdots + 99^{3})
end{align}
is a multiple of 3.
answered Dec 27 '18 at 5:49
Seewoo LeeSeewoo Lee
6,954927
$endgroup$
add a comment |
$begingroup$
From $a^{3} + b^{3} = (a+b)(a^{2}-ab+b^{2})$, you have $a+b|a^{3} +b^{3}$ and
begin{align}
&1^{3} + 2^{3} + 3^{3} + cdots + 97^{3} + 98^{3} + 99^{3} \
&= (1^{3} + 2^{3}) + (4^{3} + 5^{3}) + cdots + (97^{3}+98^{3}) + (3^{3}+6^{3} + cdots + 99^{3})
end{align}
is a multiple of 3.
answered Dec 27 '18 at 5:49
Seewoo LeeSeewoo Lee
6,954927
$endgroup$
From $a^{3} + b^{3} = (a+b)(a^{2}-ab+b^{2})$, you have $a+b|a^{3} +b^{3}$ and
begin{align}
&1^{3} + 2^{3} + 3^{3} + cdots + 97^{3} + 98^{3} + 99^{3} \
&= (1^{3} + 2^{3}) + (4^{3} + 5^{3}) + cdots + (97^{3}+98^{3}) + (3^{3}+6^{3} + cdots + 99^{3})
end{align}
is a multiple of 3.
answered Dec 27 '18 at 5:49
Seewoo LeeSeewoo Lee
6,954927
answered Dec 27 '18 at 5:49
Seewoo LeeSeewoo Lee
6,954927
answered Dec 27 '18 at 5:49
Seewoo LeeSeewoo Lee
6,954927
answered Dec 27 '18 at 5:49
Seewoo LeeSeewoo Lee
6,954927
6,954927
add a comment |
add a comment |
$begingroup$
You can also calculate that sum using
$sum_{i=0}^{i=n}i^3=dfrac{n^2(n+1)^2}{4}$
Here: $sum_{i=0}^{i=99}i^3=dfrac{99^2(99+1)^2}{4}$ which is clearly divisible by 3.
answered Dec 27 '18 at 7:46
harshit54harshit54
348113
$endgroup$
add a comment |
$begingroup$
You can also calculate that sum using
$sum_{i=0}^{i=n}i^3=dfrac{n^2(n+1)^2}{4}$
Here: $sum_{i=0}^{i=99}i^3=dfrac{99^2(99+1)^2}{4}$ which is clearly divisible by 3.
answered Dec 27 '18 at 7:46
harshit54harshit54
348113
$endgroup$
add a comment |
$begingroup$
You can also calculate that sum using
$sum_{i=0}^{i=n}i^3=dfrac{n^2(n+1)^2}{4}$
Here: $sum_{i=0}^{i=99}i^3=dfrac{99^2(99+1)^2}{4}$ which is clearly divisible by 3.
answered Dec 27 '18 at 7:46
harshit54harshit54
348113
$endgroup$
You can also calculate that sum using
$sum_{i=0}^{i=n}i^3=dfrac{n^2(n+1)^2}{4}$
Here: $sum_{i=0}^{i=99}i^3=dfrac{99^2(99+1)^2}{4}$ which is clearly divisible by 3.
answered Dec 27 '18 at 7:46
harshit54harshit54
348113
answered Dec 27 '18 at 7:46
harshit54harshit54
348113
answered Dec 27 '18 at 7:46
harshit54harshit54
348113
answered Dec 27 '18 at 7:46
harshit54harshit54
348113
348113
add a comment |
add a comment |
$begingroup$
consider these observations:
$$1 = 1 pmod 3$$
$$2 = -1 pmod 3$$
$$3 = 0 pmod 3$$
This becomes:
$$33(1^3+(-1)^3+0) equiv 33(1+(-1)+0) equiv 33(0) equiv 0 pmod 3$$
answered Dec 27 '18 at 14:57
Maged SaeedMaged Saeed
8671417
$endgroup$
add a comment |
$begingroup$
consider these observations:
$$1 = 1 pmod 3$$
$$2 = -1 pmod 3$$
$$3 = 0 pmod 3$$
This becomes:
$$33(1^3+(-1)^3+0) equiv 33(1+(-1)+0) equiv 33(0) equiv 0 pmod 3$$
answered Dec 27 '18 at 14:57
Maged SaeedMaged Saeed
8671417
$endgroup$
add a comment |
$begingroup$
consider these observations:
$$1 = 1 pmod 3$$
$$2 = -1 pmod 3$$
$$3 = 0 pmod 3$$
This becomes:
$$33(1^3+(-1)^3+0) equiv 33(1+(-1)+0) equiv 33(0) equiv 0 pmod 3$$
answered Dec 27 '18 at 14:57
Maged SaeedMaged Saeed
8671417
$endgroup$
consider these observations:
$$1 = 1 pmod 3$$
$$2 = -1 pmod 3$$
$$3 = 0 pmod 3$$
This becomes:
$$33(1^3+(-1)^3+0) equiv 33(1+(-1)+0) equiv 33(0) equiv 0 pmod 3$$
answered Dec 27 '18 at 14:57
Maged SaeedMaged Saeed
8671417
answered Dec 27 '18 at 14:57
Maged SaeedMaged Saeed
8671417
answered Dec 27 '18 at 14:57
Maged SaeedMaged Saeed
8671417
answered Dec 27 '18 at 14:57
Maged SaeedMaged Saeed
8671417
8671417
add a comment |
add a comment |
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$begingroup$
Use $n^3equiv npmod3$.
$endgroup$
– Lord Shark the Unknown
Dec 27 '18 at 5:42
$begingroup$
math.stackexchange.com/questions/1328798
$endgroup$
– Andrei
Dec 27 '18 at 5:43
1
$begingroup$
In general, note that you can handle the terms in groups of $3$ as $n + 3 equiv n pmod 3$, and also use that $n^3 equiv n pmod 3$ as Lord Shark the Unknown stated above.
$endgroup$
– John Omielan
Dec 27 '18 at 5:46
$begingroup$
@LordSharktheUnknown ah thanks. I guess the $(x+1)^3$ method would've revealed that.
$endgroup$
– user334732
Dec 27 '18 at 5:46