KVPY 2015 4
What is the largest perfect square that divides
2 0 1 4 3 − 2 0 1 3 3 + 2 0 1 2 3 − 2 0 1 1 3 + … + 2 3 − 1 3 ?
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.
5 solutions
Great solution
Similar approach ... Nice solution btw !!
Awesome!!!
I did it by looking how the sequence bahaves when I try to divide by some number and what I get is: if the last posite bigger number of the sequence is 2 x , I can divide by the perfect square of haft of it. But I couldn't prove by algebra, and yet it's not prove, but by looking to this solution, it's really interesting. A lot of things to learn about it.
I'm still a little of bit lost on the fourth part though, in how to get it. If you could explaing for me, I'd be thank you.
Log in to reply
k = 1 ∑ n k 3 = ( 2 n ( n + 1 ) ) 2
I miss read what is required. My approach:-
L
e
t
n
=
1
0
0
7
.
T
a
k
i
n
g
p
a
i
r
s
o
f
(
2
x
)
3
−
(
2
x
−
1
)
3
=
1
2
x
2
−
6
x
+
1
,
f
(
x
)
=
1
∑
1
0
0
7
(
2
x
)
3
−
(
2
x
−
1
)
3
.
f
(
x
)
=
1
∑
n
1
2
x
2
−
6
x
+
1
.
f
(
x
)
=
1
2
6
n
(
n
+
1
)
(
2
n
+
1
)
−
6
2
n
(
n
+
1
+
n
=
2
n
(
n
+
1
)
(
2
n
+
1
)
−
3
n
(
n
+
1
)
+
n
=
n
{
(
n
+
1
)
(
4
n
+
2
−
3
)
+
1
}
=
n
{
(
n
+
1
)
(
4
n
−
1
)
+
1
}
=
n
∗
(
4
n
2
+
3
n
)
=
n
2
∗
(
4
n
+
3
)
.
1
0
0
7
2
∣
f
(
x
)
.
I did it in a bad way !! Observation... 2³-1³ is div by 1² which is (2/2)² 4³-3³+2³-1³ = 44 is div by 2² which is (4/2)² 6³-5³+4³-3³+2³-1³ = 135 is div by 3² which is (6/2)² So 2014³-2013³+2012³-2011³.......2³-1³ shud be div by (2014/2)²
nice observation man!!
use a^3-b^3=(a-b)(a^2+b^2+a.b)
its just a hint
So taking a look at the first part of the pattern e.g : 2 0 1 4 3 − 2 0 1 3 3 we can factor out the cube to make the following ( 2 0 1 4 − 2 0 1 3 ) 3 which is equal to 1 3 repeating this pattern equal to the number of pairs by dividing 2014 by 2, we get 1 0 0 7 3 which is divisible by 1 0 0 7 2 . This is my first time to write any solution in this website so feel free to give me your feedback.
i dont know if it´s right but i solve like this 2014^3-2013^3...-1^3=2014^3+2012^+2^3....-(2013^3+1^3...) x^3-y^3=x+y (x^2-xy+y^2) 2012^3+2^3+2010^3+4^3....=(2012+2) . (2012^2-4024+4)... so separing in even number pair we get 1017 pair that has the sum 2014 so the count can be rewrite as: 2014.(2012^2-4024+2010^2-8040+16...) 1007. 2014.((2012^2-4024+2010^2-8040+16...) we may do the same with the pairs of odd numbers 2013^3+1^3...=1017.2014.(2013^2-2013+1...) and then 1007.2014.(2012^2-4028+2^3-2013^2-2013+1...) + 2014^3 as 2014=1007.2 1007^2. 2(2012^2-4028+2^3-2013^2-2013+1...) + 1017.2^3 as (2013^2-2013+1...) results in an odd number and (2012^2-4024+2010^2-8040+16...) results in an even number we infer that (2012^2-4028+2^3-2013^2-2013+1...) is an even number so the answer can´t be 2014^2 because there´s no order factor even
Let the expression be S , then we have:
S = − 1 3 + 2 3 − 3 3 − . . . + 2 0 1 4 3 = − ( 1 3 + 2 3 + 3 3 + . . . + 2 0 1 4 3 ) + 2 ( 2 3 + 4 3 + 6 3 + . . . + 2 0 1 4 3 ) = − ( 1 3 + 2 3 + 3 3 + . . . + 2 0 1 4 3 ) + 2 4 ( 1 3 + 2 3 + 3 3 + . . . + 1 0 0 7 3 ) = − ( 2 2 0 1 4 ( 2 0 1 4 + 1 ) ) 2 + 2 4 ( 2 1 0 0 7 ( 1 0 0 7 + 1 ) ) 2 = − ( 1 0 0 7 ˙ 2 0 1 5 ) 2 + 2 2 ( 1 0 0 7 ˙ 1 0 0 8 ) 2 = − ( 1 0 0 7 ˙ 2 0 1 5 ) 2 + ( 1 0 0 7 ˙ 2 0 1 6 ) 2 = 1 0 0 7 2 ( 2 0 1 6 − 2 0 1 5 ) ( 2 0 1 6 + 2 0 1 5 ) = 1 0 0 7 2 ˙ 4 0 3 1 = 1 0 0 7 2 ˙ 2 9 ˙ 1 3 9
Therefore, 1 0 0 7 2 is the largest perfect square that divides S .