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$10x+7\equiv 2$ mod 25

Solve for x, Correct answer is 2,7,12,17,22

What method can we use to solve this problem?

I tried and got this far:

$10x+7\equiv 2$ mod 25

$10 x + 7 = 2 + 25n$

$10x + 5 = 25n$

$2x = -1$ mod 5

Daniel Andersson
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  • You're on the right track. Now find the multiplicative inverse of $2$ mod $5$, i.e. the number $a$ such that $2a \equiv 1 , (\text{mod }5)$, and multiply both sides by it. – MSDG May 04 '19 at 19:32
  • $!\bmod 5!:,\ 2x\equiv -1,\equiv 4\iff x\equiv, \ldots\ \ \ $ – Bill Dubuque May 04 '19 at 19:34

2 Answers2

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Note that for each integer $x$ the equation $10x \equiv_{25} 10(x+5)$ holds, and that $10x \not \equiv_{25} 10(x+i)$ for each $i=1,2,3,4$.

Thus if $x_0$ is an integer satisfying $10x_0 + 7 \equiv_{25} 2$ then the set of integers $x$ satisfying $10x + 7 \equiv_{25} 2$ is precisely the set $\{x: x = x_0 \pm 5n$ for some nonnegative integer $n\}$. Note that $x_0=2$ satifies the equation $10x_0 + 7 \equiv_{25} 2$. So the set of integers $x$ satisfying $10x + 7 \equiv_{25} 2$ is precisely the set $\{x: x = 2 \pm 5n$ for some nonnegative integer $n\}$.

This is precisely the set of integers $x$ satisfying $2x \equiv_5 -1 \equiv_5 4$ or equivalently $x \equiv_5 2$, relating to your solution.

Mike
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  • i.e. by linearity the general solution of $,10x\equiv -5\equiv 20,$ is the sum of a particular solution $,x\equiv 2,$ plus the general solution of the associated homogeneous equation, which is $,10x\equiv 0\pmod {!25}\iff x\equiv 0\pmod{!5},\ $ by $\ 25\mid 10x\iff 5\mid 2x\iff 5\mid x\iff x = 5n$ $\ \ \ $ – Bill Dubuque May 04 '19 at 19:57
  • Hmm okay sounds great but im sorry to say that i don't understand what you mean right now. Maybe if you showed your solution mathematicly? – Daniel Andersson May 04 '19 at 20:24
  • @DanielAndersson Follow the link I gave. See also here. – Bill Dubuque May 04 '19 at 20:51
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$10x+7\equiv2\bmod 25\\10x\equiv 20\bmod 25\\2x\equiv 4\bmod 5\\x\equiv 2\bmod 5$