2

I have this problem, I need to find the smallest possible solution

$$x \equiv 3 \pmod{10}, \\ x \equiv 11 \pmod{13}, \\ x \equiv 15 \pmod{17}.$$

I used Chinese remainder theorem and found that: $x_1=3, x_2=11$ and $x_3= -45$??

The solutions would be congruent to $v= (13\times17)\times3 + (10\times17)\times11 + (10\times13)\times (-45) = -3317 \pmod{10\times13\times17}$. Is that the smaller solution? Is $x_3= -45$ correct?

NYN
  • 107
  • It is hard to know whether your values of $x_1,x_2,x_3$ are correct when you haven't told us what those symbols stand for. – Gerry Myerson Nov 12 '15 at 06:10
  • Also, when you write $\bmod{10\times13\times15}$, is 15 a typo for 17? – Gerry Myerson Nov 12 '15 at 06:11
  • And you can add or subtract any multiple of $10\times13\times17$ to a solution to get another solution, so you ought to be able to figure out whether there's a smaller solution (I take it you mean smaller in absolute value?). – Gerry Myerson Nov 12 '15 at 06:13
  • If I know what you are doing the $-45$ is fine. It could equally well have been $(14)(15)$. – André Nicolas Nov 12 '15 at 06:21
  • Gerry it was a type is 17. I am not sure what "symbols" you refer to. Is just a simple congruence system. – NYN Nov 12 '15 at 14:46
  • Andre, why (14)(15) ? My question is how to reduce -3317 to a positive number to have as a final answer. – NYN Nov 12 '15 at 14:49
  • Just add 2210 until you get a positive number. As another way to check you have the right answer, fire up Mathematica or go to Wolfram Alpha and do ChineseRemainder[{3, 11, 15}, {10, 13, 17}] – Robert Soupe Nov 13 '15 at 02:42
  • You say you found $x_1=3$, $x_2=11$, and $x_3=-45$, but you do not tell us what those symbols, $x_1,x_2,x_3$ stand for. What is $x_1$ supposed to mean? How does it relate to the three congruences, none of which mentions an $x_1$? Similarly for $x_2$ and $x_3$. – Gerry Myerson Nov 13 '15 at 09:57

2 Answers2

2

Let's see: $10 \times 13 \times 17 = 2210$. Then we find $$1 \times \frac{2210}{10} = 221 \equiv 1 \pmod{10},$$ $$1 \times \frac{2210}{13} = 170 \equiv 1 \pmod{13},$$ $$14 \times \frac{2210}{10} = 1820 \equiv 1 \pmod{17}.$$ Then we add up $$3 \times 1 \times 221 + 11 \times 1 \times 170 + 15 \times 14 \times 130 = 663 + 1870 + 27300 = 29833.$$ Now, $29833$ is a solution of the congruences above, but it's not the smallest positive solution. It's the remainder of this: $$29833 = \frac{28730}{2210} + 1103.$$ Now check that $$1103 \equiv 3 \pmod{10},$$ $$1103 \equiv 11 \pmod{13},$$ $$1103 \equiv 15 \pmod{17}.$$

So you did your calculations kind of different from what I would have done, but you can still get at the right answer just the same. Note that $1103 - (-3317) = 4420$, which is twice $2210$. This means that $-3317 + 4420 = 1103$ still gives you the smallest positive answer which I found with a different $x_3$.

Mr. Brooks
  • 1,156
0

First, we deal with the last two, which state $x\equiv-2\pmod{13}$ $x\equiv-2\pmod{17}$ So by the Chinese Remainder Theorem, $x\equiv-2\pmod{221}.$ We have to solve this for $x\equiv3\pmod{10}.$ First, we set it to $x\equiv219\pmod{221}$ and we need to add enough $221$'s (note that $219$ is the least integer that satisfies the last two equations) so that we finally have the first modular equation satisfied. This leads to $x\equiv219+221+221+221+221=119+884=\boxed{1103}\pmod{2210}.$ The smallest positive integer that satisfies this is $\boxed{1103},$ evidently.

Jason Kim
  • 932