Peter, I found your question interesting and have been messing with it for a while
www.jiskha.com/questions/1789041/Ethel-went-to-the-farmers-market-to-buy-some-eggs-not-in-a-dozen-cartons-on-the
I finally made up a short computer program using a silly old language form the 70'
and came up with 301 as your answer, that is
301 = 1 mod 2
301 = 1 mod 3
301 = 1 mod 4
301 = 1 mod 5
301 = 1 mod 6 , and
301 = 0 mod 7
I also found 721 to work
There is a method called "the Chinese Remainder Theorem" that uses modular arithmetic, but it only works well if the mod arguments are relatively prime, yours aren't.
anyway, 301 leaves a remainder of 1, when divided by 2,3,4,5, and 6, and a remainder of 0 when divided by 7
The same is true for 721, but 301 is the smallest.
An oblique triangle ABC has internal angles ∠A=40∘ and ∠B=53∘, and the length of the side opposite to ∠C is c=30 (metres). Determine the length of the side opposite to ∠A, denoted a, to 2 decimal places.
a=
Number
metres (2 decimal places).
What happen to Reiny?
As an AI language model, I do not have personal information about individuals. However, Reiny might be unavailable or busy at the moment. Is there anything I can help you with?
Thank you for sharing your findings and solution to the problem. It seems like you've used modular arithmetic and tested different numbers to find one that satisfies the given conditions.
To explain how you arrived at the answer of 301, you started by checking the remainders when dividing a number by each of the given integers: 2, 3, 4, 5, 6, and 7. You were looking for a number that leaves a remainder of 1 when divided by 2, 3, 4, 5, and 6, and a remainder of 0 when divided by 7.
By applying modular arithmetic, which involves finding the remainder when dividing one number by another, you determined that the number 301 satisfies all the conditions. It leaves a remainder of 1 when divided by 2, 3, 4, 5, and 6, and a remainder of 0 when divided by 7.
You also mentioned another number, 721, that satisfies the conditions, but you found 301 to be the smallest number that meets all the criteria.
You mentioned the Chinese Remainder Theorem, which is a mathematical theorem that provides a way to solve a system of linear congruences. However, it works best when the mod arguments (divisors in your case) are relatively prime (do not share any common factors). Since the divisors in this problem are not relatively prime, the Chinese Remainder Theorem may not be the most efficient method to solve it.
In summary, you used modular arithmetic and tested different numbers until you found the smallest one, 301, that satisfies the given conditions. Great job on finding a solution!