Not sure where to post this question or what type of question it is
- Thread starter bondi
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Please read the post entitled "Read Before Posting."If 66x=10 and 56x=1 what does 61x=
And then give us the problem EXACTLY as it is given to you. What you have posted makes no sense.
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HallsofIvy
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IF this were "normal multipication" then the first equation would tell us that x= 10/66= 5/33. But 56x= 56(5/33)= 280/33, not 1 so this is not "normal multiplication but, presumably, some other operation. Unfortunately, given only what is written here, there are an infinite number of such operations.If 66x=10 and 56x=1 what does 61x=
Hello, bondi!
I assume this problem involves Modulo Arithmetic.
I can't recreate all my steps, but here is my solution.
\(\displaystyle \text{From }\:\begin{Bmatrix}66x \,\equiv\,10\text{ (mod b)} \\ 56x \,\equiv\,1\text{ (mod b)} \end{Bmatrix}\;\text{ I found that: }\,b =13\)
\(\displaystyle \text{Then: }\:66x \equiv 10\text{ (mod 13)} \quad\Rightarrow\quad x \equiv 10\text{ (mod 13)} \)
\(\displaystyle \text{Hence: }\:61x \:=\:61(10) \:=\:610 \)
\(\displaystyle \text{Therefore: }\:61x \:\equiv\:12\text{ (mod 13)}\)
I assume this problem involves Modulo Arithmetic.
I can't recreate all my steps, but here is my solution.
\(\displaystyle \text{From }\:\begin{Bmatrix}66x \,\equiv\,10\text{ (mod b)} \\ 56x \,\equiv\,1\text{ (mod b)} \end{Bmatrix}\;\text{ I found that: }\,b =13\)
\(\displaystyle \text{Then: }\:66x \equiv 10\text{ (mod 13)} \quad\Rightarrow\quad x \equiv 10\text{ (mod 13)} \)
\(\displaystyle \text{Hence: }\:61x \:=\:61(10) \:=\:610 \)
\(\displaystyle \text{Therefore: }\:61x \:\equiv\:12\text{ (mod 13)}\)