Question

Prove that the nth prime number $p_n$ satisfies $p_n\leq 2^{2n-1}$

score 6.7k · Answer 1 · 2018-02-15 04:13:23

6.7k

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Prove that the nth prime number $p_n$ satisfies $p_n\leq 2^{2n-1}$

Published on 15 Feb 2018 - 4:13

score 56 · Answer 2 · 2018-02-15 18:42:40

56

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Prove that if $a|b_1b_2$, $a|c_1c_2$ and $a|b_1c_1+b_2c_2$, then $a|b_1c_1$ and $a|b_2c_2$

Published on 15 Feb 2018 - 18:42

score 142 · Answer 3 · 2026-03-24 23:44:49

142

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The Chinese hypothesis revisited

Published on 24 Mar 2026 - 23:44

score 2.8k · Answer 4 · 2026-03-17 16:00:03

2.8k

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Greatest common divisor of $X^n-1$ and $X^m-1$

Published on 17 Mar 2026 - 16:00

score 321 · Answer 5 · 2018-02-17 16:10:41

321

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Prove $15\mid 4^{2n}-1$ without induction

Published on 17 Feb 2018 - 16:10

score 71 · Answer 6 · 2018-02-17 22:03:08

71

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Prove: If $ (a^2 \mid (2b^2))$, then $ (a \mid b). $

Published on 17 Feb 2018 - 22:03

score 74 · Answer 7 · 2018-02-18 05:12:45

74

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A sequence+ divisibility proof contest math

Published on 18 Feb 2018 - 5:12

score 70 · Answer 8 · 2026-03-25 01:28:00

70

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Find divisibility for $n \in \mathbb {Z+}, 2\nmid n, 2^n-1, 2^n +1$ w.r.t. $5$ using congruence arithmetic.

Published on 25 Mar 2026 - 1:28

score 78 · Answer 9 · 2018-02-18 17:16:05

78

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Find $m\in \mathbb{R}$ so that $P(x)$ is divisible by $X^2+X+1$

Published on 18 Feb 2018 - 17:16

score 4.3k · Answer 10 · 2018-02-20 00:39:36

4.3k

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Is $0$ divisible by $0$?

Published on 20 Feb 2018 - 0:39

score 1.8k · Answer 11 · 2026-03-23 01:12:09

1.8k

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Test for divisibility by 2 for a number written in base 9

Published on 23 Mar 2026 - 1:12

score 883 · Answer 12 · 2018-02-20 17:05:56

883

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Prove that $76$ raised to any integer power and then divided by $100$ (ignoring the remainder) is always divisible by $57$

Published on 20 Feb 2018 - 17:05

score 393 · Answer 13 · 2018-02-21 05:55:19

393

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The $2n$-digit number divisible by the product of two numbers within itself

Published on 21 Feb 2018 - 5:55

score 745 · Answer 14 · 2018-02-21 17:46:33

745

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Prove by Induction if x is an odd integer, then for every integer n ≥ 0, $x^{2^n}$ ≡ 1 (mod $2^{n + 1}$)

Published on 21 Feb 2018 - 17:46

score 68 · Answer 15 · 2018-02-21 19:25:32

68

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Sequences and divisibility

Published on 21 Feb 2018 - 19:25

List Question

Prove that the nth prime number $p_n$ satisfies $p_n\leq 2^{2n-1}$

Prove that if $a|b_1b_2$, $a|c_1c_2$ and $a|b_1c_1+b_2c_2$, then $a|b_1c_1$ and $a|b_2c_2$

The Chinese hypothesis revisited

Greatest common divisor of $X^n-1$ and $X^m-1$

Prove $15\mid 4^{2n}-1$ without induction

Prove: If $ (a^2 \mid (2b^2))$, then $ (a \mid b). $

A sequence+ divisibility proof contest math

Find divisibility for $n \in \mathbb {Z+}, 2\nmid n, 2^n-1, 2^n +1$ w.r.t. $5$ using congruence arithmetic.

Find $m\in \mathbb{R}$ so that $P(x)$ is divisible by $X^2+X+1$

Is $0$ divisible by $0$?

Test for divisibility by 2 for a number written in base 9

Prove that $76$ raised to any integer power and then divided by $100$ (ignoring the remainder) is always divisible by $57$

The $2n$-digit number divisible by the product of two numbers within itself

Prove by Induction if x is an odd integer, then for every integer n ≥ 0, $x^{2^n}$ ≡ 1 (mod $2^{n + 1}$)

Sequences and divisibility

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