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Philippine Mathematical Olympiad

Number theory Difficulty 4.9 AIME Find the answer Philippines

Problem:

Consider addition ⊕\oplus and multiplication ⊗\otimes modulo 7 of the numbers in S={0,1,2,3,4,5,6}S=\{0,1,2,3,4,5,6\}. This means that
m⊕n= remainder when m+n is divided by 7m⊗n= remainder when m×n is divided by 7.  \begin{aligned} & m \oplus n=\text{ remainder when } m+n \text{ is divided by } 7 \\ & m \otimes n=\text{ remainder when } m \times n \text{ is divided by } 7 \text{. } \end{aligned}
Then 1 is the multiplicative identity and each element a∈Sa \in S has a multiplicative inverse 1a\frac{1}{a}. Find the value of 14⊕(2⊗13)\frac{1}{4} \oplus\left(2 \otimes \frac{1}{3}\right) in this number system.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Solution:

First, we need to find 14\frac{1}{4} and 13\frac{1}{3} in SS under multiplication modulo 7.

14\frac{1}{4} is the number xx such that 4⊗x≡1(mod7)4 \otimes x \equiv 1 \pmod{7}.
Try x=2x = 2: 4×2=8≡1(mod7)4 \times 2 = 8 \equiv 1 \pmod{7}, since 8−7=18 - 7 = 1.
So 14=2\frac{1}{4} = 2.

13\frac{1}{3} is the number yy such that 3⊗y≡1(mod7)3 \otimes y \equiv 1 \pmod{7}.
Try y=5y = 5: 3×5=15≡1(mod7)3 \times 5 = 15 \equiv 1 \pmod{7}, since 15−2×7=115 - 2 \times 7 = 1.
So 13=5\frac{1}{3} = 5.

Now compute 2⊗13=2⊗5=10≡3(mod7)2 \otimes \frac{1}{3} = 2 \otimes 5 = 10 \equiv 3 \pmod{7}.

Now compute 14⊕(2⊗13)=2⊕3=5\frac{1}{4} \oplus (2 \otimes \frac{1}{3}) = 2 \oplus 3 = 5 (since 2+3=5<72 + 3 = 5 < 7).

Therefore, the value is 5\boxed{5}.

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