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.

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  • Mike
    Lv 7
    7 years ago
    Favorite 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-5
    Lv 6
    7 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.

  • 7 years ago

    k=24 the sum is 4900. k*(k+1)*(2k+1)/6 = 24*25*49/6 = 4*25*49 = 100*49.

  • 7 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
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