how-many-positive-x-10-000-integers-are-such-that-2-x-x-2-is-divisible-by-7-

Question Number 154672 by talminator2856791 last updated on 20/Sep/21

how many positive x ⩽ 10 000 integers are

such that 2^{x} - x^{2} is divisible by 7 ?

Answered by MJS_new last updated on 20/Sep/21

2^{3 n} = 7 k + 1

2^{3 n + 1} = 7 k + 2

2^{3 n + 2} = 7 k + 4

{(7 n)}^{2} = 7 k + 0

{(7 n + 1)}^{2} = 7 k + 1

{(7 n + 2)}^{2} = 7 k + 4

{(7 n + 3)}^{2} = 7 k + 2

{(7 n + 4)}^{2} = 7 k + 2

{(7 n + 5)}^{2} = 7 k + 4

{(7 n + 6)}^{2} = 7 k + 1

\Rightarrow

n = 0, 1, 2, 3, \dots : (2^{n} - n^{2}) / 7 = 7 k +

1 1 0 6 0 0 0 2 3 4 0 2 4 1 4 0 5 2 6 5 3 1 1 0 6 \dots

\Rightarrow

remainders are 0 for

n = 21 k + {2, 4, 5, 6, 10, 15}

21 \times 476 + 5 = 10001 \Rightarrow

\Rightarrow answer is 475 \times 6 + 2 = 2852

Commented by mathdanisur last updated on 20/Sep/21

very nice