Sum of Squares is Divisible by Sum

Number Theory Level 3

Find the sum of all positive integers $1\leq n\leq 100$ such that $\dfrac{1^2+2^2+3^2+\cdots+n^2}{1+2+3+\cdots +n}$ is an integer.

The answer is 1717.

7 solutions

Ahaan Rungta
Apr 18, 2014

Note that $\dfrac {\displaystyle\sum_{k=1}^{n} k^2}{\displaystyle\sum_{k=1}^{n} k} = \dfrac {\dfrac {n(n+1)(2n+1)}{6}}{\dfrac {n(n+1)}{2}} = \dfrac {2n+1}{3} \in \mathbb{Z}$ Thus, $3 \mid (2n+1) \implies 2n + 1 \equiv 0 \pmod 3 \implies 2n \equiv 2 \pmod 3 \implies n \equiv 1 \pmod 3$ .

Thus, the set of working $n$ is $\{ 1, 4, \cdots, 100 \}$ , so the sum is $1 + 4 + \cdots + 100 = \dfrac {101 \cdot 34}{2}.$

$= \boxed {1717}$ . I wonder why I didn't write that.

Ahaan Rungta - 7 years, 1 month ago

You can directly edit your solution, by clicking on the pencil that's just to the left of the up vote count.

Calvin Lin Staff - 7 years, 1 month ago

Sorry guys, but is there a method to get the result of a sum of non-consecutive numbers? Because this time i had to count all the numerbs and then multiply the half by the sum of two of them..

Antonio Giordano - 7 years, 1 month ago

@Antonio Giordano – That's the way I evaluated the sum while working out the problem myself. Of course, if you really don't want to do that, you can always write that the sum is equal to $\begin{aligned}&(3\cdot 1-2)+(3\cdot 2-2)+\cdots+(3\cdot 33-2)+(3\cdot 34-2)\\&=3(1+2+\cdots+34)-2\cdot 34\\&=3\left(\dfrac{34\cdot 35}2\right)-68=1717.\end{aligned}$ In my mind, though, using the formula $S=\tfrac{n(a+z)}2$ where $n$ is the number of terms and $a,z$ are the first and last terms respectively is the easiest way to go.

David Altizio - 7 years, 1 month ago

@David Altizio – I write C code to find the result :))

ナルトウズマキ - 7 years, 1 month ago

@ナルトウズマキ – Where is the Code Pal..

Thiliban Varadharajan - 7 years, 1 month ago

@Antonio Giordano – Learn the formulas of "SEQUENCES AND SERIES"......ie AP , GP and HP...

Vighnesh Raut - 7 years, 1 month ago

@Antonio Giordano – For Antonio, I believe this would answer your inquiry. You can use the formula for the nth term. $a_n = a_1 + (n-1)d,$ where $a_n$ is the nth term, $a_1$ is the first term, n is the number of terms (which you wanted to get) and d is the common difference. So, you substitute: 100 = 1 + (n-1) * 3 and you'll get n = 34 :)

Sas Estoperes - 7 years, 1 month ago

1156 is correct ans

Akhlaque Ahmad - 7 years, 1 month ago

Lol I got the correct Answer :)

Brent Nico Tan - 7 years, 1 month ago

I did it this way too but instead of beginning with the known formula of the sum of squares I had to derive it first ;p

Alaa Qarooni - 7 years, 1 month ago

N = Sum{k = 1; n (k^2)} = n(n + 1)(2n + 1)/6;

D = Sum{k = 1; n (k)} = n(n + 1)/2;

N/D = n(n + 1)(2n + 1)/6 x 2/(n(n + 1)) = (2n + 1)/3;

N/D is an integer if:

2n + 1 = 3m; 1 ≤ n ≤ 100; m ∈ ℕ

2n + 1 is odd, so

2n + 1 = 3, 9, 15, 21, ..., 3 + 6(k - 1), 201; k ∈ ℕ;

2n + 1 = 3 + 6(k - 1); n - 1 = 3(k - 1); n = 3(k - 1) + 1 = 3k - 2

n = 1, 4, 7, 10,...,3k - 2,..., 100;

3kmax - 2 = 100; kmax = 102/3 = 34;

we must find:

S = Sum{k = 1; 34(3k - 2)};

sum of first 34 terms of an arithmetIc progression an, with difference:

d = 3, and:

a(1) = 1, a(34) = 100, that is:

S = 34(a(1) + a(34))/2 = 17(1 + 100) = 1717

Antonio Fanari - 6 years, 9 months ago

Adarsh Kumar
Apr 25, 2014

Sum of consecutive number's squares=1^2+2^2+....n^2=n(n+1)(2n+1)\6.If we put this value in the question we get the following result:(2n+1)/3.Now,this has to be an integer.By hit and trial we get that 1,4,7,11 satisfy this equation.Thus,forming an A.P.sum=1717.

x=(n(n+1)((2n+1) /6)/(n(n+1)/2) =(2n+1)/3 .. n can take values 1 to 100 at n=100 x=201/3=67

what sum are we finding... if we are finding some of numbers for which this holds.. then it yields sum of first 100 natural numbers which is 5050...

Kushal Walia - 7 years, 1 month ago

1156 is correct ans

Akhlaque Ahmad - 7 years, 1 month ago

Vishnudatt Gupta
May 16, 2014

1^2+2^2+...+n^2=n(n+1)(2n+1)/6 and 1+2+...+n=n(n+1)/2

Therefore, the fraction is [n(n+1)(2n+1)/6]/[n(n+1)/2] = (2n+1)/3

Possible values of n are 1, 4, 7,..., 100.

So the sum is (1+100)*[(100-1)/3+1]/2=1717.

Sujay Sheth
May 4, 2014

Sum of $1^{st}$ n square numbers = $\frac {n(n+1)(2n+1)}{6}$

Sum of $1^{st}$ n numbers = $\frac {n(n+1)}{2}$

Thus, $\frac {\ 1^2 + 2^2 + 3^2 + .... n^2 }{1 + 2 + 3 + .... + n}$ = $\frac{n(n+1)(2n+1)/6}{n(n+1)/2}$ = $\frac {2n+1}{3}$ where n ( Arithmetic series) = 1, 4, 7, 10, 13, ...., 100 (Substituting n in the formula gives perfect integers) (We have to find the sum of all this numbers)

Now for Arithmetic Progression (A.P), a= 1, d= 3 and n= ? (Here n is for numbers in series)

x $n$ = a + d(n-1)

So, 100 = 1 + 3 (n-1)

n = 34

Thus n = $34^{th}$ which is 100 in the series.

Sum of A.P.= $\frac {n( 2a+ (n -1)d )}{2}$ = $\frac {34( 2(1)+ (34 -1)3 )}{2}$ = $\boxed{1717}$

Ramasubramaniyan Gunasridharan
Apr 29, 2014

[n(n+1)(2n+1)/6]*[1/[n(n+1)/2]]= (2n+1)/3. The possible values of n where, (2n+1)/3 will be a real number are, 1,4,7,........,100. In this series, a =1; d =3; l=100; n=(l-a)/d +1 = 34. Thus, Sn = n/2[a+l] = 1717

Anika .
Apr 28, 2014

1^2+2^2+...+n^2=n(n+1)(2n+1)/6 and 1+2+...+n=n(n+1)/2

Therefore, the fraction is [n(n+1)(2n+1)/6]/[n(n+1)/2] = (2n+1)/3

Possible values of n are 1, 4, 7,..., 100.

So the sum is (1+100)*[(100-1)/3+1]/2=1717.

Abhishek Bisht
Apr 28, 2014

Ans: 1717

Sol: 1^2+2^2+...+n^2=n(n+1)(2n+1)/6 and 1+2+...+n=n(n+1)/2 Therefore, the fraction is [n(n+1)(2n+1)/6]/[n(n+1)/2] = (2n+1)/3 Possible values of n are 1, 4, 7,..., 100. So the sum is (1+100)*[(100-1)/3+1]/2=1717.

here is a small matlab code that u can use ;) clear all; close all; sum=0; num=zeros(101,1); den=zeros(101,1); j=1; k=1; number=zeros(100,1); while(j<101) for i=1:j num(j)=num(j)+i^2; den(j)=den(j)+i; end frac=num(j)/den(j) if((rem(frac,1)==0))

number(k)=j;
k=k+1;
sum=sum+j;

end j=j+1 end

Aaditya Rcs - 7 years, 1 month ago

Sum of Squares is Divisible by Sum

The answer is 1717.

7 solutions

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