Determine the number of all positive integers less than 4183444 if each is divisible by 29, 7, 17.

What is its sum?

Correct result:

n =  1212
S =  2536754178


a1=29717=3451  n=4183444a1=1212a_1 = 29\cdot 7\cdot 17= 3451 \ \\ \ \\ n = \lfloor \dfrac{ 4183444}{ a_1 } \rfloor = 1212
S=n2(a1+an)=n2(a1+na1)=n(n+1)2a1=2536754178S = \dfrac{n}{2} (a_1 + a_n) = \dfrac{n}{2} (a_1 + n a_1 ) = \dfrac{n( n + 1)}{2} \cdot a_1 = 2536754178

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