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x is a three-digit natural number,

$$2x \equiv 3 \pmod 5$$

$$3x \equiv 1 \pmod 4 $$

what is the smallest value that x can take?

What is the method to solve this kind of questions?

Clyde Kertzer
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DrFu
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    Maybe first rewrite our congruences as $x\equiv 4\pmod{5}$, $x\equiv 3\pmod{4}$. – André Nicolas Dec 05 '15 at 18:11
  • @AndréNicolas And then what? I guess I am just stupid... – DrFu Dec 05 '15 at 18:42
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    https://en.wikipedia.org/wiki/Chinese_remainder_theorem – Sammy Black Dec 05 '15 at 20:06
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    I was just about to post that. In particular, you want this section: https://en.wikipedia.org/wiki/Chinese_remainder_theorem#Finding_the_solution_with_basic_algebra_and_modular_arithmetic – Paul Sinclair Dec 05 '15 at 20:07
  • Multiplying by the inverses of the lead coefs yields the equivalent system $,x\equiv -1,$ both $\bmod 5\ &\ 4,$ which is equivalent to $x\equiv -1\pmod{20},$ by linked CCRT (or generally, by Easy CRT. See also here for a fractional generalization of CRT. – Bill Dubuque Feb 23 '23 at 23:02

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More generally, you can use the Chinese Remainder Theorem, but in a case like this, you can probably just guess and check because the congruences are small.

Let's find the smallest solution to $2x \equiv 3 \pmod 5$. Looks like $4,9,14,19,24$ works so $x \equiv 4 \pmod 5$ is a solution.

Let's do the same thing for $3x \equiv 1 \pmod 4$. Looks like $3, 7, 11, 15, 19$ works so $x \equiv 3 \pmod 4$ is a solution.

Thus the smallest such solution looks to be $x = 19$. If you want a higher number (not sure why you would want a specific 3-digit number, but that's not important), you can use the general congruences.

Clyde Kertzer
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