Modular equation I

Number Theory Level 3

$\large 4x+20 \equiv 27x-1 \pmod5$

Find all integers $x$ that satisfy the linear congruence above

Your answer can be written as $x\equiv a \pmod b$ , where $a$ and $b$ are positive primes . Submit it as $a+b$ .

2 solutions

Paola Ramírez
Apr 21, 2016

$4x+20 \equiv 27x-1 \text{ mod } 5$ is equal to $23x-21\equiv 0 \text{ mod } 5$ . The last means $23x-21=5n$ .

A number is divisible by $5$ if its last digit is $0$ or $5$ . So, the last digit of $23x$ minus $1$ must be $0$ or $5$ .

The last digit of $23x$ only could be $1$ or $6$ . So the last digit of $x$ must be $7$ or $2$ . Therefore $\boxed{x \equiv 2 \text{ mod }5}$

Otto Bretscher
Apr 21, 2016

Nice problem!

Modulo 5 we have $23x\equiv 21$ or $3x\equiv 1$ , or,doubling it, $x\equiv 2$ . The answer is $2+5=\boxed{7}$