List questions from chinese-remainder-theorem

Published on 17 Feb 2015 - 8:13

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Converse to Chinese Remainder Theorem

Published on 17 Feb 2015 - 14:00

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Non-linear congruence equation

Published on 29 Mar 2026 - 3:19

95

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Does $x^2=83\pmod{101}$ have solutions? without calculating them

Published on 21 Feb 2015 - 22:57

119

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Using CRT to prove a single congruence relation

Published on 25 Feb 2015 - 3:11

431

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Another Chinese Remainder Theorem Problem

Published on 29 Mar 2026 - 4:34

616

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Solving a system of congruences using the Chinese Remainder Theorem

Published on 29 Mar 2026 - 4:34

424

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If every pair of congruence equations admits solutions, then the entire system admits solutions

Published on 08 Mar 2015 - 14:06

385

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Show that $15\mid 21n^5 + 10n^3 + 14n$ for all integers $n$.

Published on 29 Mar 2026 - 7:40

535

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Solution congruence system $ x \equiv 11\pmod{36},\,x \equiv 7\pmod{40}, \,x \equiv 32\pmod{75}$

Published on 20 Mar 2015 - 22:38

992

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Chinese Remainder Theorem for $x\equiv 0 \pmod{y}$

Published on 24 Mar 2015 - 2:39

61

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Chinese remainder - Error in my solution

Published on 29 Mar 2026 - 11:00

564

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Simultaneusly solving $2x \equiv 11 \pmod{15}$ and $3x \equiv 6 \pmod 8$

Published on 28 Mar 2015 - 1:03

138

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To Find $a$ such that $2^{1990} \equiv a\pmod {1990}$.

Published on 29 Mar 2015 - 1:58

586

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There exist arbitrarily long sequences of consecutive integers that are not square-free

Published on 05 Apr 2015 - 21:19

List Question

Find $x$ given remainders mod $2, 3, 4, 5, 6, 7$

Converse to Chinese Remainder Theorem

Non-linear congruence equation

Does $x^2=83\pmod{101}$ have solutions? without calculating them

Using CRT to prove a single congruence relation

Another Chinese Remainder Theorem Problem

Solving a system of congruences using the Chinese Remainder Theorem

If every pair of congruence equations admits solutions, then the entire system admits solutions

Show that $15\mid 21n^5 + 10n^3 + 14n$ for all integers $n$.

Solution congruence system $ x \equiv 11\pmod{36},\,x \equiv 7\pmod{40}, \,x \equiv 32\pmod{75}$

Chinese Remainder Theorem for $x\equiv 0 \pmod{y}$

Chinese remainder - Error in my solution

Simultaneusly solving $2x \equiv 11 \pmod{15}$ and $3x \equiv 6 \pmod 8$

To Find $a$ such that $2^{1990} \equiv a\pmod {1990}$.

There exist arbitrarily long sequences of consecutive integers that are not square-free

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