Jax bought exactly enough trees to plant eight equal rows. Then one tree died and couldn't be planted, but he still had enough trees left to plant exactly nine equal rows. After that, three trees were stolen, but he still had enough trees left to plant exactly ten equal rows. If he bought the least number of trees satisfying these three conditions, how many trees did he buy?
Jax bought exactly enough trees to plant eight equal rows. Then one tree died and couldn't be planted, but he still had enough trees left to plant exactly nine equal rows. After that, three trees were stolen, but he still had enough trees left to plant exactly ten equal rows. If he bought the least number of trees satisfying these three conditions, how many trees did he buy?
I can't think of a formula to solve this, so I did it by brute force.
First, I mentalized all the multiples of 8, and subtracted 1 each time to see if the result is a multiple of 9.
I thought of 8, 16, 24 ... 56, 64 When I got to 64 I saw that 64 – 1 = 63, which is a multiple of 9.
Then 3 less than that 63 is 60, which is a multiple of 10. So the answer to the problem is 64.
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Using Chinese Remainder theorem + Modular multiplicative Inverse, we have:
N mod 8/2 ==0 ==N mod 4 ==0
N mod 9 ==1 ==N mod 9 ==1
N mod 10/2==2==N mod 5 ==2 [The moduli: 4, 9, 5 should be relatively prime to each other]
LCM[4, 5, 9] ==180n + 172 [which is the remainder of the CRT + MMI], where n==0, 1, 2, 3.....etc.
N ==180 x 1 + 172 ==352 - trees that Jax bought.
Note: With 3 trees stolen, the CRT does not give integer solution. So, I changed 3 to 2 trees stolen, since the question was seen before with 2 trees stolen!
I think the Chinese Remainder Theorem lost something in translation.
"If he bought the least number of trees satisfying these three conditions..."
Using the Slobbovian Penumbral Postulate, we find that 64 is less than 352.
Furthermore, the number 352 doesn't even satisfy the three conditions.
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+2440 Ron’s (as guest #1) solution presentation demonstrates an excellent method for solving small number simultaneous modular equations. It’s intuitive and quick; especially when one of the equivalencies is zero. ... And it works when the modulo numbers are not co-prime.
GA
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+2440
Here is a formal solution, using Euler's totient function –calculated from non-prime numbers.
There are often (not always) solutions to simultaneous modular equations with non-co-prime modulo numbers, when one of the co-primes modulos is equivalent to zero.
The first product of zero (0) is included as a presentation continuity formality.
\(\begin{array}{rcll} n &\equiv& {\color{red}0} \pmod {{\color{green}8}} \\ n &\equiv& {\color{red}1} \pmod {{\color{green}9}} \\ n &\equiv& {\color{red}4} \pmod {{\color{green}10}} \\ \text{Set } m &=& 8\cdot 9\cdot 10 = 720 \\ \end{array}\)
\({ \begin{array}{l} n = {\color{red}0} \cdot {\color{green}9\cdot 10} \cdot \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ [ { (\color{green}9 \cdot 10) }^{\varphi({\color{green}8}) -1 } \pmod {{\color{green}8}} ] }_{=\text{modulo inverse }(9\cdot 10) \mod 8 } }_{=(9\cdot 10)^{4-1} \mod {8}} }_{=(9\cdot 10)^{3} \mod {8}} }_{=(90\pmod{8})^{3} \mod {8}} }_{=(2)^{3} \mod {8}} }_{= 0} + {\color{red}1} \cdot {\color{green}8\cdot 10} \cdot \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ [ { (\color{green}8\cdot 10) }^{\varphi({\color{green}9}) -1} \pmod {{\color{green}9}} ] }_{=\text{modulo inverse } (8\cdot 10) \mod {9}} }_{=(8\cdot 10)^{6-1} \mod {9}} }_{=(8\cdot 10)^{5} \mod {9}} }_{=(80\pmod{9})^{5} \mod {9}} }_{=(8)^{5} \mod {9}} }_{=8} + {\color{red}4} \cdot {\color{green}8\cdot 9} \cdot \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ [ { (\color{green}8\cdot 9) }^{\varphi({\color{green}10}) -1 } \pmod {{\color{green}10}} ] }_{=\text{modulo inverse } (8\cdot 9) \mod 10 } }_{=(8\cdot 9)^{4-1} \mod { 10}} }_{=(8\cdot 9)^{3} \mod {10}} }_{=(72\pmod{10})^{3} \mod {10}} }_{=(2)^{3} \mod {10}} }_{=8}\\ \\ n = {\color{red}0} \cdot {\color{green}9\cdot 10} \cdot [0] + {\color{red}1} \cdot {\color{green}8\cdot 10} \cdot [8] + {\color{red}4} \cdot {\color{green}8\cdot 9} \cdot [8] \\ n = 0+ 640 + 2304 \\ n = 2944 \\ n \pmod {m} \rightarrow 2944 \pmod {720} \equiv 64 \\ \\ n = 64 + k\cdot 360\qquad k \in Z\\ \hspace {10em} \mathbf{n_{min}} \mathbf{=} \mathbf{64} \end{array} } \)
GA
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