Probability Divisible 35 Cbbc01

1. **State the problem:** We want to find the probability that $2^m + 3^n$ is divisible by $35$ when $m$ and $n$ are chosen from $\{1, 2, \dots, 600\}$ independently and uniformly. 2. **Key idea:** Since $35 = 5 \times 7$, $2^m + 3^n$ divisible by $35$ means divisible by both $5$ and $7$. 3. **Divisibility conditions:** - $2^m + 3^n \equiv 0 \pmod{5}$ - $2^m + 3^n \equiv 0 \pmod{7}$ 4. **Analyze modulo 5:** - The order of $2$ mod $5$ is $4$ because $2^4 = 16 \equiv 1 \pmod{5}$. - The order of $3$ mod $5$ is also $4$ because $3^4 = 81 \equiv 1 \pmod{5}$. So $2^m$ mod $5$ and $3^n$ mod $5$ cycle every $4$ steps. 5. **Possible residues mod 5:** - $2^m$ mod $5$ cycles as: $2, 4, 3, 1$ for $m \equiv 1,2,3,0 \pmod{4}$ respectively. - $3^n$ mod $5$ cycles as: $3, 4, 2, 1$ for $n \equiv 1,2,3,0 \pmod{4}$ respectively. 6. **Condition $2^m + 3^n \equiv 0 \pmod{5}$:** We want $2^m \equiv -3^n \equiv 5 - 3^n \pmod{5}$. Check all pairs $(m \bmod 4, n \bmod 4)$: - $(1,1): 2 + 3 = 5 \equiv 0$ - $(2,3): 4 + 2 = 6 \equiv 1$ - $(3,2): 3 + 4 = 7 \equiv 2$ - $(0,0): 1 + 1 = 2 \neq 0$ Check all systematically: | $m\bmod4$ | $2^m$ mod 5 | $n\bmod4$ | $3^n$ mod 5 | Sum mod 5 | |---|---|---|---|---| |1|2|1|3|0| |1|2|2|4|1| |1|2|3|2|4| |1|2|0|1|3| |2|4|1|3|2| |2|4|2|4|3| |2|4|3|2|1| |2|4|0|1|0| |3|3|1|3|1| |3|3|2|4|2| |3|3|3|2|0| |3|3|0|1|4| |0|1|1|3|4| |0|1|2|4|0| |0|1|3|2|3| |0|1|0|1|2| Pairs where sum mod 5 = 0 are: - $(1,1)$, $(2,0)$, $(3,3)$, $(0,2)$ 7. **Analyze modulo 7:** - The order of $2$ mod $7$ is $3$ because $2^3 = 8 \equiv 1 \pmod{7}$. - The order of $3$ mod $7$ is $6$ because $3^6 = 729 \equiv 1 \pmod{7}$. 8. **Possible residues mod 7:** - $2^m$ mod $7$ cycles every $3$: $2,4,1$ for $m \equiv 1,2,0 \pmod{3}$. - $3^n$ mod $7$ cycles every $6$: $3,2,6,4,5,1$ for $n \equiv 1,2,3,4,5,0 \pmod{6}$. 9. **Condition $2^m + 3^n \equiv 0 \pmod{7}$:** We want $2^m \equiv -3^n \equiv 7 - 3^n \pmod{7}$. Check all pairs $(m \bmod 3, n \bmod 6)$: - For $m=1$: $2^m=2$ - For $m=2$: $2^m=4$ - For $m=0$: $2^m=1$ Check $3^n$ values: | $n\bmod6$ | $3^n$ mod 7 | |---|---| |1|3| |2|2| |3|6| |4|4| |5|5| |0|1| Sum mod 7 = 0 means $2^m + 3^n \equiv 0$: Check each $m$: - $m=1$ ($2^m=2$): $2 + 3^n$ mod 7 = 0 means $3^n = 5$ (since $2 + 5 = 7$) - $m=2$ ($2^m=4$): $4 + 3^n = 0$ mod 7 means $3^n = 3$ - $m=0$ ($2^m=1$): $1 + 3^n = 0$ mod 7 means $3^n = 6$ From the table: - $3^n = 5$ when $n \equiv 5 \pmod{6}$ - $3^n = 3$ when $n \equiv 1 \pmod{6}$ - $3^n = 6$ when $n \equiv 3 \pmod{6}$ So valid pairs mod 7 are: - $(m,n) \equiv (1,5)$ - $(m,n) \equiv (2,1)$ - $(m,n) \equiv (0,3)$ 10. **Combine modulo 5 and modulo 7 conditions using CRT:** - Mod 5 valid pairs: $(1,1), (2,0), (3,3), (0,2)$ with mod 4 for $m,n$ - Mod 7 valid pairs: $(1,5), (2,1), (0,3)$ with mod 3 for $m$ and mod 6 for $n$ We want pairs $(m,n)$ such that: - $m \equiv a \pmod{4}$ and $m \equiv b \pmod{3}$ - $n \equiv c \pmod{4}$ and $n \equiv d \pmod{6}$ For each mod 5 pair $(a,c)$ and mod 7 pair $(b,d)$, solve: - $m \equiv a \pmod{4}$ and $m \equiv b \pmod{3}$ - $n \equiv c \pmod{4}$ and $n \equiv d \pmod{6}$ 11. **Solve for $m$:** - $m \equiv a \pmod{4}$ - $m \equiv b \pmod{3}$ Since $4$ and $3$ are coprime, $m$ mod 12 is unique. Check all $a \in \{0,1,2,3\}$ and $b \in \{0,1,2\}$: - For $(a,b)$ pairs from valid mod 5 and mod 7 $m$ residues: Mod 5 $m$ residues: $1,2,3,0$ Mod 7 $m$ residues: $1,2,0$ Calculate $m$ mod 12 for each pair: - $(1,1)$: Solve $m \equiv 1 \pmod{4}$ and $m \equiv 1 \pmod{3}$ - $m=1$ satisfies both - $(1,2)$: $m \equiv 1 \pmod{4}$ and $m \equiv 2 \pmod{3}$ - $m=10$ satisfies both - $(1,0)$: $m \equiv 1 \pmod{4}$ and $m \equiv 0 \pmod{3}$ - $m=4$ satisfies both Similarly for $a=2$: - $(2,1)$: $m=5$ - $(2,2)$: $m=2$ - $(2,0)$: $m=8$ For $a=3$: - $(3,1)$: $m=3$ - $(3,2)$: $m=9$ - $(3,0)$: $m=6$ For $a=0$: - $(0,1)$: $m=7$ - $(0,2)$: $m=11$ - $(0,0)$: $m=0$ Only $m$ values with $a$ from mod 5 and $b$ from mod 7 valid pairs: - Mod 5 $m$ residues: $1,2,3,0$ - Mod 7 $m$ residues: $1,2,0$ So valid $m$ mod 12 are: - $(1,1) \to 1$ - $(2,2) \to 2$ - $(3,0) \to 6$ - $(0,0) \to 0$ 12. **Solve for $n$:** - $n \equiv c \pmod{4}$ from mod 5 pairs: $1,0,3,2$ - $n \equiv d \pmod{6}$ from mod 7 pairs: $5,1,3$ Find $n$ mod 12 (since lcm(4,6)=12) for each pair: - $(1,5)$: $n=5$ - $(1,1)$: $n=1$ - $(1,3)$: $n=9$ - $(0,5)$: $n=5$ - $(0,1)$: $n=1$ - $(0,3)$: $n=3$ - $(3,5)$: $n=11$ - $(3,1)$: $n=7$ - $(3,3)$: $n=3$ - $(2,5)$: $n=10$ - $(2,1)$: $n=2$ - $(2,3)$: $n=6$ Only keep pairs where $c$ and $d$ come from valid mod 5 and mod 7 pairs: - Mod 5 $n$ residues: $1,0,3,2$ - Mod 7 $n$ residues: $5,1,3$ Valid $n$ mod 12 are: - $(1,5) \to 5$ - $(0,1) \to 1$ - $(3,3) \to 3$ - $(2,1) \to 2$ 13. **Count valid $(m,n)$ pairs mod 12:** - Valid $m$ mod 12: $\{0,1,2,6\}$ (4 values) - Valid $n$ mod 12: $\{1,2,3,5\}$ (4 values) Total valid pairs mod 12: $4 \times 4 = 16$ 14. **Total possible pairs mod 12:** - $m$ mod 12 has 12 possible residues - $n$ mod 12 has 12 possible residues - Total pairs mod 12: $12 \times 12 = 144$ 15. **Probability:** Since $m$ and $n$ are chosen uniformly from $1$ to $600$, and $600$ is divisible by $12$, the residues mod 12 are uniformly distributed. Therefore, the probability is: $$\frac{16}{144} = \frac{1}{9}$$ **Final answer:** $\boxed{\frac{1}{9}}$