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- Jan 26, 2012

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Thanks to those who participated in last week's POTW!! Here's this week's problem!

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From the description of the problem, one sets up the following system of congruences:

\[\begin{aligned}x &\equiv 32 \pmod{83}\\ x & \equiv 70\pmod{110}\\ x & \equiv 30 \pmod{135}\end{aligned}\qquad\qquad(1)\]

However, 83, 110, and 135 are not pairwise co-prime since $(110,135)\neq 1$, so one can't apply the CRT...yet. However, we can rewrite (1) as the following system of congruences:

\[\begin{aligned}x & \equiv 32 \pmod{83}\\ x & \equiv 70 \pmod{2} \\ x & \equiv 70 \pmod{5}\\ x & \equiv 70 \pmod{11}\\ x & \equiv 30 \pmod{27} \\ x & \equiv 30 \pmod{5}\end{aligned}\qquad\qquad (2)\]

The idea now is to remove the redundancies in the system (2). In the end, one should end up with a system of congruences (see spoiler if you want) in which one satisfies the pairwise co-prime condition in order to use the Chinese Remainder Theorem.

\[\begin{aligned}x & \equiv 32 \pmod{83}\\ x & \equiv 0 \pmod{10}\\ x & \equiv 4 \pmod{11}\\ x & \equiv 3 \pmod{27}\end{aligned}\qquad\qquad (3)\]

Remember to read the POTW submission guidelines to find out how to submit your answers!

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**Problem**: Three farmers divide equally the rice that they have grown. One goes to a market where an 83-pound weight is used, another to a market that uses a 110-pound weight, and the third to a market using a 135-pound weight. Each farmer sells as many full measures as possible, and when the three return home, the first had 32 pounds of rice left, the second 70 pounds, and the third 30 pounds. Find the total amount of rice they took to the market.-----

**Hint**:Use the Chinese Remainder Theorem.

**Remark**:\[\begin{aligned}x &\equiv 32 \pmod{83}\\ x & \equiv 70\pmod{110}\\ x & \equiv 30 \pmod{135}\end{aligned}\qquad\qquad(1)\]

However, 83, 110, and 135 are not pairwise co-prime since $(110,135)\neq 1$, so one can't apply the CRT...yet. However, we can rewrite (1) as the following system of congruences:

\[\begin{aligned}x & \equiv 32 \pmod{83}\\ x & \equiv 70 \pmod{2} \\ x & \equiv 70 \pmod{5}\\ x & \equiv 70 \pmod{11}\\ x & \equiv 30 \pmod{27} \\ x & \equiv 30 \pmod{5}\end{aligned}\qquad\qquad (2)\]

The idea now is to remove the redundancies in the system (2). In the end, one should end up with a system of congruences (see spoiler if you want) in which one satisfies the pairwise co-prime condition in order to use the Chinese Remainder Theorem.

Remember to read the POTW submission guidelines to find out how to submit your answers!

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