How many positive integers x ≤ 10000 are there such that the difference 2 x − x 2 is not divisible by 7? Options: A . &nbsp 6662 B . &nbsp 7142 C . &nbsp 3687 D . &nbsp 7562 E . &nbsp 123

Answer: Option B : B 2 x − x 2 is divisible by 7 if and only if 2 x and x 2 both leave the same remainder when divided by 7. x 2 x R e m a i n d e r w i t h 7 1 2 2 2 4 4 3 8 1 4 16 2 5 32 4 6 64 1 7 128 2 8 256 4 x x 2 R e m a i n d e r w i t h 7 1 1 1 2 4 4 3 9 2 4 16 2 5 25 4 6 36 1 7 49 0 8 64 1 9 81 4 10 100 2 Remainder of 2 x repeats after every 3 and remainder of x 2 repeats after 7. The remainders of both 2 x and x 2 will repeat after a period of length 3 × 7 = 21 x 1 2 3 4 5 6 7 8 9 1 0 1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 2 0 2 1 2 x 2 4 1 2 4 1 2 4 1 2 4 1 2 4 1 2 4 1 2 4 1 x 2 1 4 2 2 4 1 0 1 4 2 2 4 1 0 1 4 2 2 4 1 0 In this set of 21 values, we see that there are 6 values of x for which the remainders are same. As per the periodicity, within every consecutive 21 values, 6 will repeat. 10,000 = 21 × 476 + 4 The 476 groups contribute 476 × 6 = 2856 values and the remaining 4 will give 2 values. So, total 2858 integers ≤ 10000 are divisible. Thus, 10000- 2858 = 7142 integers are not divisible.

How many positive integers x≤10000 are there such that the difference 2

Question

How many positive integers $x \leq 10000$ are there such that the difference $2^{x} - x^{2}$ is not divisible by 7?

Answer: Option B
:
B

$2^{x} - x^{2}$ is divisible by 7 if and only if $2^{x}$ and $x^{2}$ both leave the same remainder when divided by 7.
$\begin{matrix} x & 2^{x} & R e m a i n d e r w i t h 7 1 & 2 & 2 2 & 4 & 4 3 & 8 & 1 4 & 16 & 2 5 & 32 & 4 6 & 64 & 1 7 & 128 & 2 8 & 256 & 4 \end{matrix}$
$\begin{matrix} x & x^{2} & R e m a i n d e r w i t h 7 1 & 1 & 1 2 & 4 & 4 3 & 9 & 2 4 & 16 & 2 5 & 25 & 4 6 & 36 & 1 7 & 49 & 0 8 & 64 & 1 9 & 81 & 4 10 & 100 & 2 \end{matrix}$

Remainder of $2^{x}$ repeats after every 3 and remainder of $x^{2}$ repeats after 7. The remainders of both $2^{x}$ and $x^{2}$ will repeat after a period of length 3 $\times$ 7 = 21
$\begin{matrix} x & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & \frac{1}{0} & \frac{1}{1} & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & \frac{1}{6} & \frac{1}{7} & \frac{1}{8} & \frac{1}{9} & \frac{2}{0} & \frac{2}{1} 2 x & 2 & 4 & 1 & 2 & 4 & 1 & 2 & 4 & 1 & 2 & 4 & 1 & 2 & 4 & 1 & 2 & 4 & 1 & 2 & 4 & 1 x^{2} & 1 & 4 & 2 & 2 & 4 & 1 & 0 & 1 & 4 & 2 & 2 & 4 & 1 & 0 & 1 & 4 & 2 & 2 & 4 & 1 & 0 \end{matrix}$

In this set of 21 values, we see that there are 6 values of x for which the remainders are same.

As per the periodicity, within every consecutive 21 values, 6 will repeat.

10,000 = 21 $\times$ 476 + 4

The 476 groups contribute 476 $\times$ 6 = 2856 values and the remaining 4 will give 2 values.
So, total 2858 integers $\leq 10000$ are divisible. Thus, 10000- 2858 = 7142 integers are not divisible.

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