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# given that 1^2+2^2+3^2+.....+k^2=k*(k+1)*(2k+1)/6,what is the smallest integer k such that?

given that 1^2+2^2+3^2+.....+k^2=k*(k+1)*(2k+1)/6,what is the smallest integer k such that 1^2+2^2+3^2+.....+k^2 is divisible by 100.

### 4 Answers

- MikeLv 77 years agoFavorite Answer
k(k+1)(2k+1)/6 must be divisible by 100 means k(k+1)(2k+1) divisible by 600 which is 5x5x2x2x2x3

2k+1 is odd, and only one of k or k+1 can be even

so you need k or k+1 divisible by 8

one of k, k+1, and 2k+1 will always be divisible by 3

to be divisible by 5, k must be 0,2 or 4 mod 5

looking at mod 25, we see that k must be 12 or 24 mod 25

(checking 0,2,4,5,7,9,10,12,14,15,17,19,20,22,24)

so 12 or 24 mod 25, and 7 or 8 mod 8

12 24 37 49 62 74 87 99 112 124 137 149 162 174 187 and 199 mod 200 are possibilities

checking for mod 8, yields answers

24, 87, 112, and 199 mod 200

- pc-5Lv 67 years ago
We want the smallest integer k such that k(k+1)(2k+1) is divisible by 600.

note that 600 = 24*25.

so k = 24 is a possible solution with a sum of 24*25*49/6 = 4900

In solving this problem, I have noticed that only one of the three factors is divisible by 5.

But k(k+1)(2k+1) is divisible by 25.

So only one of the three factors is divisible by 25.

Let's start the search by setting each factor equal to 25.

k = 25, sum = 25*26*51/6 = 25*13*17, No

k+1 = 25, sum = 24*25*49/6 = 4900, Yes

2k+1 = 25, sum = 12*13*25/6 = 25*26, No

It is now clear that k = 24 is the solution.

- grunfeldLv 77 years ago
100c = k ( k + 1 ) ( 2k + 1 ) / 6

600c = k ( k + 1 ) ( 2k + 1 )

600c = ( k^2 + k ) ( 2k + 1 )

600c = 2k^3 + 3k^2 + k

0 = 2k^3 + 3k^2 + k - 600c

solve this cubic.

Source(s): my brain