Question

If $A_{K} = \frac{K\left(K -1\right)}{2}$ cos $\frac{K\left(K -1\right)}{2} \pi$ and $\sum\limits^{105}_{K =1} A_{K} =S$ then

• A. number of divisors of $S$ are $12$
• B. Sum of digits of $S$ is one of the divisor of $S$
• C. The number of divisors of $S$ are $16$
• D. Sum of divisors of $S = 11340$

Answer 1

Answer: (B), (C), (D)

$A_{K} = \frac{K\left(K -1\right)}{2} cos \frac{K\left(K -1\right)}{2} \pi, K =1, 2,....,105$
Now $A_{1} = 0, A_{2} =$ cos $\pi = -A_{3} =3$ cos $3\pi = -3$
$\therefore A_{2} +A_{3} = -2^{2}$
Similarly, $A_{4} +A_{5} =6$ cos $6\pi + 10$ cos $10\pi = 16 = 4^{2}$
and $A_{6}+A_{7} = 15$ cos $15\pi + 21$ cos $21\pi = - 36 = - 6^{2}$
$\therefore A_{104} + A_{105} = 5356$ cos $\left(5356\pi\right) + 5460$ cos $\left(5460\pi\right) = 10816 =\left(104\right)^{2}$
$\therefore S =\sum\limits^{105}_{K = 1}A_{K} = A_{1} +\left(A_{2} + A_{3}\right) +\left(A_{4} +A_{5}\right) +....+A_{104} +A_{105}$
$S = -2^{2} +4^{2} -6^{2} + .....-\left(102\right)^{2} +\left(104\right)^{2}$
$= -4\left(1^{2} -2^{2} + 3^{2}-4^{2} +....+ 51^{2} -52^{2}\right)$
$= 4\left(1 + 2 + 3 +....+ 52\right)$
$= \frac{4 \times 52 \times53}{2} = 5512$
$\therefore $ Prime factors of $S = 2^{3} \times 13^{1} \times 53^{1}$So, number of divisors of $S$
$= 4 \times 2 \times 2 = 16$
Sum of the digits of $S = 5 + 5 + 1 + 2 = 13$ which is one of the divisor of 5 and sum of all divisors
$= \frac{2^{4} -1}{2 -1} \times\frac{13^{2} -1}{13 -1} \times \frac{53^{2} -1}{53 -1} = 15 \times14 \times54=11340$