We want to find an integer $k$ such that the expression is divisible by $100$ .

We can say the following: $\frac{k(k+1)(2k+1)}{6} = 100n$ , where $n$ is a positive integer.

Simplifying the above equation, we get: $k(k+1)(2k+1) = 600n$ . This means that the product needs to be divisible by 600 times an integer, $n$ .

It might be helpful to prime factorize $600$ , which can be written as $2^{3} * 3 * 5^{2}$ . We can see that $k$ must contain at least 3 powers of 2, since both $k+1$ and $2k+1$ will be odd if $k$ is even. Trying out the value $k = 24$ works, since $k$ contains 3 powers of 2, and 1 power of three. $k+1$ contains 2 powers of 5. This meets our constraints, so the least possible value for $k$ is $\boxed{24}$ .

The question is equivalent to finding the smallest integer $k$ such that $k(k+1)(2k+1)$ is divisible by $600=2^{3} \times 3^{1} \times 5^{2}$ .

We note that the integers $k, (k+1), (2k+1)$ are coprime. Therefore, exactly one of these integers must be divisible by $25$ since they do not share any common factors.

The smallest possible value of $k$ such that one of the three integers is divisible by $25$ is $k=12$ , where $2k+1=25$ . However, we find that the product $k(k+1)(2k+1)=12 \times 13 \times 25=3900$ is not divisible by $600$ .

The next smallest possible value of $k$ is $k=24$ , where $k+1=25$ . In this case, we find that $k(k+1)(2k+1)=24 \times 25 \times 49=29400$ is in fact divisible by $600$ .

Thus, the smallest integer $k=24$ .

k(k+1)(2k+1)= 600N = 8 25 3*an integer

k(k+1)(2k+1) is always divisible by 3. This can be proved by method of exhaustion. If k = 0 mod(3), it is obviously true. If k = 1 mod(3), then 2k +1 = 0 mod(3). If k = 2 mod(3), then k+1 = 0 mod(3).

2k +1 is odd and 8 = 2^3. Hence k(k+1)(2k+1) is divisible by 8 iff k(k+1) is. Hence, k = 0 mod(8) or k = -1 mod(8) = 7 mod(8).

It is not possible for k(k+1)(2k+1) to be divisible by 25 by having any two the three factors divisible by 5. This can be shown again by method of exhaustion. If k = 0 mod(5) , then k+1= 1 mod(5) and 2k+1 = 1 mod (5). If k +1 = 0 mod(5), then k = -1 mod(5) and 2k +1= -1 mod(5). If 2k +1 = 0 mod(5), then 2k = -1 mod(5) = 4 mod(5). Hence, k = 2 mod(5) and k+1 = 3 mod(5).

Hence, k(k+1)(2k+1) is divisible by 25 if k = 0 mod(25) or 24 mod(25) or 12 mod(25).

Solving by using the Chinese remainder theorem, the simultaneous solutions are 0, 24, 87, 112, 175, or 199 modulo 200. The smallest positive integer is 24.

Solution 1 Let $T_n=1^2+2^2+3^2+...+n^2$ . When $1\leq{n}\leq23$ , $T_n$ are not divisible by $100$ . When $n=24$ , $T_n=4900$ , which is divisible by 100. Therefore, $24$ is the smallest integer required.

Solution 2 Since $25|100$ and $gcd(25,6)=1$ , divide into 2 cases:
Case 1 : At least two of $k,k+1,2k+1$ are divisible by $5$ . If $k\equiv0\pmod{5}, k+1\equiv1\pmod{5}, 2k+1\equiv1\pmod{5}$ . If $k\equiv1\pmod{5}, k+1\equiv2\pmod{5}, 2k+1\equiv3\pmod{5}$ . If $k\equiv2\pmod{5}, k+1\equiv3\pmod{5}, 2k+1\equiv0\pmod{5}$ . If $k\equiv3\pmod{5}, k+1\equiv4\pmod{5}, 2k+1\equiv2\pmod{5}$ . If $k\equiv4\pmod{5}, k+1\equiv0\pmod{5}, 2k+1\equiv4\pmod{5}$ . Therefore, there are no solution for Case 1.
Case 2 : At least one of $k,k+1,2k+1$ are divisible by $25$ . If $25|k+1$ , $25a=k+1$ for some positive integer $a$ , $k=25a-1$ , when $a=1, k=24$ and this is indeed a solution, when $a>1, k>24$ and contradicts the minimal of $k$ . If $25|k$ , $25b=k$ for some positive integer $b$ and for all positive integer $b$ , $k>24$ , again, contradicts the minimal of $k$ . If $25|2k+1$ , $25c=2k+1$ for some positive integer $c$ , $k=\frac{25c-1}{2}$ , when $c=1, k=12$ (does not satisfy the conditions of the problem), when $c\geq2$ , $k>24$ , again, contradicts the minimal of $k$ . Therefore, $24$ is the required minimum.

$100 \mid \frac{k(k+1)(2k+1)}{6} \Rightarrow 25 \mid k(k+1)(2k+1)$ which is a product of 3 factors.

Assume all 3 factors are not divisible by 25. At least 2 of the factors must then be divisible by 5.

Case 1: $5 \mid k$ . Clearly $5 \nmid k+1$ and $5 \nmid 2k+1$ which fails.

Case 2: $5 \nmid k$ so $5 \mid k+1$ and $5 \mid 2k+1$ but this means that $5 \mid (2k+1)-(k+1)=k$ which is a contradiction.

Therefore, our initial assumption is wrong and at least 1 factor is divisible by 25.

Case 1: $25 \mid k+1$ Therefore, $k+1 \geq 25 \Rightarrow k \geq 24$ . For $k=24$ , $\frac{k(k+1)(2k+1)}{6} = 4900$ which is divisible by 100.

Case 2: $25 \mid k$ Therefore, $k \geq 25$ which will not give a smaller integer than in Case 1.

Case 3: $25 \mid 2k+1$ Therefore, $2k+1 \geq 25 \Rightarrow k \geq 12$ For $k=12$ , $\frac{k(k+1)(2k+1)}{6} = 650$ which is not divisible by 100. Any other number smaller than 24 will not satisfy $25 \mid 2k+1$ .

Therefore, the required answer is 24.

Note that $k$ , $k+1$ and $2k+1$ are pairwise coprime.

For $100|1^2+2^2+3^3+...+k^2=\frac{k(k+1)(2k+1)}{6}$ , one of $k$ , $k+1$ and $2k+1$ have to be a multiple of 25.

The three smallest numbers $k$ that have that property are 12, 24, and 25, which are easily found by letting $k, k+1, 2k+1=25$ .

Testing these numbers, we see that $k=24$ is the smallest possible, letting $1^2+2^2+3^2+...+24^2=\frac{24\times 25\times 49}{6}=4900$

First, concentrate on the value of k to be put so k(k+1)(2k+1) can be divided by 5. This can be achieved when either k, k+1 , or 2k+1 can be divided by 5. Checking the modulo 5 of k, k+1, and 2k+1 when k = 5m, (k+1)mod5 = 1 ,(2k+1)mod5 = 1 and k = 5m+4 , (k+1)mod5 = 0 ,(2k+1)mod5 = 4 => proof that we need either k or k+1 or 2k+1 to be divisible by 25

from that, the smallest k will be 12, making 2k+1 = 25. But this way, 1^2+2^2+...+12^2 = 650 which is not divisible by 100.

so, we move to the next smallest k, which is 24, giving k+1 a value of 25. 1^2+2^2+...+24^2 = 4900 which is divisible by 100.

In conclusion, the smallest k is 24.

For the sum of squares to be divisible by 100, the term $k(k+1)(2k+1)$ must be divisible by 8 and 25.

Only one of k, k+1, or 2k+1 can be even, and only one can be divisible by 5.

The smallest k that would work is k = 12 such that 2k + 1 is divisible by 25. However, no term is divisible by 8.

The next choice, k = 24, works because k is divisible by 8 and k + 1 is divisible by 25, so that is our answer.

Lets say that $\dfrac{k(k+1)(2k+1)}{6}=100m$ for some positive integer $m$ .

So:

$k(k+1)(2k+1)=600m$ $k(k+1)(2k+1)=2^3\cdot 3\cdot 5^2m$

If $k$ is odd, $k+1$ is even and $2k+1$ is odd. If $k$ is even, $k+1$ and $2k+1$ are odd

So, only $k$ or $k+1$ will be even. Then, one and only one of the three numbers willl be a multiple of $2^3=8$ .

1. If $k$ is multiple of $8$ , lets say that $k=8p$ for some positive integer $p$ . Then:

$k(k+1)(2k+1)=8p(8p+1)(16p+1)$ $8p(8p+1)(16p+1)=2^3\cdot3\cdot5^2\cdot m$ $p(8p+1)(16p+1)=3\cdot5^2\cdot m$

• For $p=1$ :

$p(8p+1)(16p+1)=3^2\cdot17$ $3^2\cdot 17=3\cdot5^2\cdot m$

And $m$ will not be integer.

• For $p=2$ :

$p(8p+1)(16p+1)=2\cdot17\cdot33=2\cdot3\cdot11\cdot17$ $2\cdot3\cdot11\cdot17=3\cdot5^2\cdot m$

And $m$ will not be integer.

• For $p=3$ :

$p(8p+1)(16p+1)=3\cdot25\cdot49=3\cdot5^2\cdot7^2$ $3\cdot5^2\cdot7^2=3\cdot5^2\cdot m\implies m=7^2$

So, for $k$ even, the smallest value of $k$ such $1^2+2^2+\cdots+k^2$ is divisible by 100 is $8\cdot3=24$ .

1. If $k+1$ is multiple of $8$ , lets say that $k+1=8q$ for some positive integer $q$ . Then:

$k(k+1)(2k+1)=(8q-1)8q(2(8q-1)+1)$ $k(k+1)(2k+1)=(8q-1)8q(16q-1)$ $(8q-1)8q(16q-1)=2^3\cdot3\cdot5^2\cdot m$ $q(8q-1)(16q-1)=3\cdot5^2\cdot m$

• For $q=1$ :

$q(8q-1)(16q-1)=7\cdot15=3\cdot5\cdot7$ $3\cdot5\cdot7=3\cdot5^2\cdot m$

And $m$ will not be integer.

• For $q=2$ :

$q(8q-1)(16q-1)=2\cdot15\cdot31=2\cdot3\cdot5\cdot31$ $2\cdot3\cdot5\cdot31=3\cdot5^2\cdot m$

And $m$ will not be integer.

• For $q=3$ :

$q(8q-1)(16q-1)=3\cdot23\cdot47$ $3\cdot23\cdot47=3\cdot5^2\cdot m$

And $m$ will not be integer.

• For $q\ge 4$ , we will have $k=8q-1\ge31$

So, the smallest value of $k$ such $1^2+2^2+\cdots+k^2$ is when $k=24$ .

k(k+1)(2k+1)/6 must be divisible by 100. Therefore, k(k+1)(2k+1) must be divisible by 600 = (2^3)(3)(5^2) = (8)(3)(25). Consider the factor 25, and the possibility that there are two factors each divisible by 5: either (k, k+1), (k, 2k+1), (k+1, 2k+1). If k is congruent to 0 mod 5, k+1 and 2k+1 are congruent to 1 mod 5, so the first two pairs cannot consist of two numbers divisible by 5. For the third pair, if k+1 is congruent to 0 mod 5, then k is congruent to 4 mod 5, and 2k+1 is congruent to 4 mod 5 as well. Therefore, if no two factors can both be divisible by 5, there must be one factor divisible by 25. To achieve the smallest value of k, let the greatest factor 2k+1=25. This yields k=12. However, (12)(12+1)(2 12+1) = 3900, which is not divisible by 600. The next possibility is to let the second greatest factor k+1=25. This leads to k=24, and since (24)(24+1)(2 24+1) = (24)(25)(49) = (600)(49), the expression is divisible by 600. Therefore, k=24.

Let $S = k(k+1)(2k+1)$ . Since $(1^2 + 2^2 + 3^2 + \ldots + k^2)$ is a natural number, therefore $\frac{S}{6}$ is a natural number and $6|S$ . $\Rightarrow (2*3)|S$

3 can divide any of $k, k+1$ and $2k+1$ . 2 cannot divide $2k+1$ . So 2 can divide either of $k$ or $k+1$ . We want k such that 100 divides S. So $(5^2*2^2)|S$ . $\Rightarrow 8|S.$ We have two cases:

Case 1: $8|k$

We put in multiples of 8 in S. By trial we see 24 is the smallest value that satisfies 100|S.

Case 2: $8|k+1$

The possible values of k are $7, 15, 23, \ldots$ . None of these values satisfy 100|S.

Therefore, the smallest possible value of k is $\boxed{24}$ .

Obviously the condition is equivalent to k(k+1)(2k+1)|600. Only one of the three things we are multiplying can be 0 mod 5, so whichever one it is must be 0 mod 25. Checking k where one of these is divisible by 25, we find that k=24 gives 24(25)(49)=600*49.

k(k+1)(2k+1) = 6 x 100 x n = 2^3 x 3 x 5^2 x n 2k + 1 is an odd number. As seen, there is 2^3 x 3 = 24 and 5^2 = 25 which fits k and k+1, therefore, k = 24. Also, to double confirm the answer, 15 and 25 is tried on 2k+1 which is odd and it would not work. Therefore, 24 is the smallest integer for k.

For 1^2 +2^2 + 3^2 + ....... + K^2 to be divisible by 100, k(k+1)(2k+1)should be divisible by 600. therefore k(k+1)(2k+1) should be divisible by 2^2,5^2 and 3.

First, I showed that the given equation is divisible by 100.

\frac {k(k+1)(2k+1)}{6}=100n

I then simplified the numerator:

\frac {k(k+1)(2k+1)}{6} = \frac {(k^2+k)(2k+1)}{6} = \frac {2k^3+3k^2+k}{6} = 100n

Next, I multiplied both sides by six to get rid of the denominator:

2k^3+3k^2+k=600n

From this equation, I deduced that k must have at least two digits, and most likely would be a factor of 600. I tried 12 in the equation to see if it was divisible by 600. It did not, but the answer was 6.5:

2k^3+3k^2+k \rightarrow 2{12}^3+3{12}^2+{12} = 3900 \frac{3900}{600} =6.5

Because of this, I realized that 24 should be the answer:

2k^3+3k^2+k \rightarrow 2{24}^3+3{24}^2+{24} = 3900 \frac{29400}{600} =49

At most one of the factors in the top is a multiple of 5, so it must be a multiple of 25. Also, the numerator has 3 factors of 2. Thus we try: 12 => numerator has 2 factors of 2; 24 => numerator has 3 factors of 2, so answer is 24. (note this gives us 4900 as a check)

* k(k+1)(2k+1)=600x *

For smallest integer let,

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

Only k(k+1)=600 has a positive integer solution, where k=24.

When k=24,

k(k+1)(2k+1)=29400, divisible by 600.

Here is my simple approach :-

Let sum of squares of first 'n' natural numbers = $N$

We want smallest k for which N is divisible by 100. Now since square of any

two digit no. is $\geq 100$

$\Rightarrow$ only the digits in unit place of each number will govern the very last two digits N.

Now we know that,

$1^2 = 1$

$2^2 = 4$

$3^2 = 9$

$4^2 = 16$

$5^2 = 25$

$6^2 = 36$

$7^2 = 49$

$8^2 = 64$

$9^2 = 81$

$10^2 = 100$

Now if $k=10$ then $N=385$ . We will start from here.

Now to make N divisible by 100, We have to keep on adding these numbers in sequence untill we get a number divisible by 100. For Example- 400 is not possible since $385 + (1 + 4 +9 ) < 400 < 385 + (1 + 4 +9 + 16)$ .

Proceeding in this way we can clearly see it is not possible for 500, 600 and 700.

But if $k=20$ then $N=770$ to start with.

And we can clearly see that $770 + (1 +4 +9 +16) =800$

Thus, Required smallest value of $k=24.$

From the question, k(k+1)(2k+1)=600n(where n is any integer). Let us suppose k(k+1)=600 and n=2k+1.(Other two combinations are k(2k+1)=600 and k+1=n & (k+1)(2k+1)=600 &k=n which doesnot yield an integer value for k) Hence on solving the above equation k(k+1)=600, acceptable solution is k=24 and n=49

k(k+1)(2k+1)/6 so k/k+1 or 2k+1 should be divisble by 25 and rest of the two numbers should have 24 as a factor so that after division by 6 we get 100's factor Here first I tried to make 2k+1 as a factor of 25 wherein I got K =12 but then this came to be afctor of 50 but not 100 so then tried to make k+1 as a factor of 25 so k =24 and rest fit

Note that $\frac{k(k+1)(2k+1)}{6}=\frac{2k(2k+1)(2k+2)}{24}$ .

This means that the numerator, the product of 3 consecutive numbers, the first and last even and the middle odd, must be a multiple of 100 24=48 50.

This gives us one solution of 2k(2k+1)(2k+2) = 48 49 50, k = 24.

To prove that this solution is the smallest, note that in 3 consecutive numbers only 1 can be a multiple of 5, so exactly 1 term must be a multiple of 25. The solution found is the smallest one that uses 50 (compared to 49 50 51 and 50 51 52), and so a smaller solution would have to use 25 instead of 50. Since 25 is odd, the only such triple of consecutive numbers is 24 25 26, and this is clearly not a multiple of 2400.

Thus, the answer is $k=\boxed{24}$ .

//bruteforce approach

include<stdio.h>

int main() { int k; k=1; while(1) { if(((k (k+1) (2*k+1))/6)%100==0) { break; } k++; } printf("%d",k); return 0;
}

Trial and error may prove correct but its wrong. The elimination method would do. Firstly, any of those 3 above terms must be a multiple of 25. So, the answers will be 12, 24, etc... Now this luckily proved to be easier. Can someone tell me the perfect method??