Question

If $A_K=\frac{K\left(K-1\right)}{2}$ cos $\frac{K\left(K-1\right)}{2}\pi$ and $\sum _{K=1}^{105}{A}_K=S$ 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 $\left(5460\pi \right)=10816={\left(104\right)}^2$
$\therefore S=\sum _{K=1}^{105}{A}_K=A_1+\left(A_2+A_3\right)+\left(A_4+A_5\right)+....+A_{104}+A_{105}$
$S=-2^2+4^2-6^2+.....-{\left(102\right)}^2+{\left(104\right)}^2$
$=-4\left(1^2-2^2+3^2-4^2+....+51^2-52^2\right)$
$=4\left(1+2+3+....+52\right)$
$=\frac{4\times 52\times 53}{2}=5512$
$\therefore$ 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
$=\frac{2^4-1}{2-1}\times \frac{13^2-1}{13-1}\times \frac{53^2-1}{53-1}=15\times 14\times 54=11340$