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Use the method of back substitution to find all integers $x$ such that $x ≡ 1 \mod 5$, $x ≡ 2 \mod 6 $, and $x ≡ 3 \mod 7$. Solution: The first congruence can be rewritten as an equality, $x = 5t + 1$, where $t$ is an integer. Substituting this expression for $x$ into the second congruence tells us that $5t + 1 ≡ 2 \mod 6$, which can be solved to show that $t ≡ 5 $ (mod $6$) (as the reader should verify).

Can someone verify that and show me how to do it?

J. W. Tanner
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Scarlet
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  • If $5t+1\equiv2$, then (subtract $1$ from both sides) $5t\equiv1$, so (multiply both sides by $5$ or $-1$, which is its own inverse) $t\equiv5\pmod6$ – J. W. Tanner Aug 26 '20 at 14:03
  • Thank you, Tanner. I am supposed to do this without the knowledge of inverse since the textbook is arriving at that. So I took a long way, found a multiple of 5, greater than multiple of 6 by 1, which is 25 = 5*5 and that gave me t = 5. and later I realised that is basically finding inverse but like brute force. – Scarlet Aug 26 '20 at 14:31
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    Easier to use CCRT = Constant case CRT: $\ x\equiv -4\pmod{!5,6,7}\iff x\equiv -4\pmod{!5\cdot 6\cdot 7}\ \ $ – Bill Dubuque Aug 26 '20 at 15:05

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To solve $$5t+1\equiv2\pmod6,$$

subtract $1$ from both sides: $$5t\equiv1\pmod6.$$

To get $t\equiv5\pmod6$ from that, multiply both sides by $5$.

In general, if $a\equiv b\pmod n$, then $a+c\equiv b+c\pmod n$ and $a\times c\equiv b\times c\pmod n$.

J. W. Tanner
  • 60,406