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Solve $x \equiv 32 \pmod{81}$ and $x \equiv 59 \pmod{64}$.

$32 + 81k = 59 + 64n \implies 81k - 64n = 27$

$17k \equiv 27 \pmod{64}$.

$64 = 3(17) + 13$ $17 = 1(13) + 4$ $13 = 3(4) + 1$

So $1 = 13 - 3(4) = 13 - 3[17 - 13] = 4(13) - 3(17) = 4(64 - 3*17) - 3*17 = 4(64) - 15(17)$. Thus $k \equiv 1 \pmod{64} \implies k = 1 + 64y$ so we have $n = \frac{54 + 5184y}{64}$

But this is not possible . Help?

EDIT $k\equiv 43 \pmod{64}$ thus,

$k = 43 + 64y$ thus $x = 32 + 81(43 + 64y) = 3515 + 5824y$ so then $x \equiv 3515 \pmod{5824}$.

Amad27
  • 10,465
  • Your first mistake is when you said $k\equiv 1\pmod{64}$. The equations you use in your Euclidean algorithm tell you that $17^{-1}\equiv -15\pmod{64}$, so $ k\equiv 27\cdot (-15)\equiv 43\not\equiv 1\pmod{64}$. – user236182 Aug 05 '16 at 12:45

5 Answers5

1

You've correctly used the extended Euclidean algorithm to deduce the Bezout identity

$$ 4(64) -15(17)\, =\, 1$$

Reduced modulo $\,64\,$ the above implies that $\, -15(17) \equiv 1,\ $ i.e. $\ 17^{-1}\!\equiv -15\pmod{64}$

So $\, 17k\equiv 27\,$ times $\,17^{-1}\!\equiv -15\,$ $\Rightarrow$ $\,k\equiv 27(-15)\equiv (7(4)\!-\!1)(-15)\equiv 7(4)\!+\!15\equiv 43 $

But you deduced that $\ k \equiv 1\, $ vs. the correct $\ k\equiv 43.\,$ Fixing that, the rest goes smoothly.

Simpler: $\ 81k - 64n = 27\,\Rightarrow\, 27\mid n,\, $ so with $\,\bar n = n/27\,$ cancelling $\,27\,$ $\Rightarrow$ $\, 3k-64\bar n = 1\,$ so $\,{\rm mod}\ 3\!:\ {-}\bar n\equiv 1\ $ so $\ \bar n = -1+ 3j\ $ so $\ k = (1+64(-1\!+\!3j))/3 = -21+64j$

Remark $\ $ You can avoid the tedious error-prone back-substitution in the extended Euclidean algorithm by instead using this more convenient version of the algorithm.

Community
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Bill Dubuque
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0

So $$x = 81 \cdot a + 64 \cdot b$$ is what you need; where $81a \equiv 59\pmod{64}$ which you can do by finding $k$ such that $81k \equiv 1 \pmod{64}$ and then multiplying both sides by $59$.

Then you also need $64b \equiv 32\pmod{81}$ which you can do by finding $m$ such that $64m \equiv 1 \pmod{81}$ and multiplying both sides by $32$. Effectively you've reduced the problem to computing multiplicative inverses modulo $64$ and $81$, which is easy to do via trial and error or via the Euclidean algorithm.

Zain Patel
  • 16,802
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$x=32+81k\equiv 32+17k \equiv 59 \pmod {64}$

$k\equiv \frac{27}{17}\equiv\frac{27}{81}\equiv \frac{1}{3}\equiv \frac{129}{3}\equiv43 \pmod {64}$

$x=32+81(43+64k')=3515+5184k'$

Evariste
  • 2,511
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$$x = 59 + 64k$$

$$x = 32 + 81j$$

$\text{lcm}(81, 64) = 5184$

so

$$81x = 4779 + 5184k$$

$$64x = 2048 + 5184j$$

Subtract equations

$$17x = 2731 + 5184(k-j)$$

or

$$17x \equiv 2731 \bmod 5184$$

$\gcd(17, 5184)=1$, so:

$$x \equiv 2731 \cdot 17^{-1} \bmod 5184$$

$$x \equiv 2731 \cdot 305 \equiv 832955 \equiv 3515 \bmod 5184$$

Answer is

$$x \equiv 3515 \bmod 5184$$

Sean Hill
  • 683
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$x\equiv 32 (\mod 81) \ and \ x\equiv 59 (\mod 64) \Rightarrow x=81a+32 \ and \ x=64b+59 \\ $

then $81a+32=64b+59 \\ 81a-64b=27 \ ...I \\$

with the euclidean algorithm we get a solution $ (a,b)=(-405,-513) \\ 81(-405)-64(-513)=27 \ ...II \\ $

subtracting II from I we get $ 81(a+405)-64(b+513)=0 \Leftrightarrow 81(a+405)=64(b+513) \ ...III \\ $

$64\mid a+405\Rightarrow \boxed{a=64k-405} \\ $

plugging $a$ into III gives $\boxed{b=81k-513} \\$

plugging $a$ into $x=81a+32$ or $b$ into $x=64b+59\\$

$\boxed{x=5184k-32773} \ \ k\in \mathbb{Z}$

Student
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