2 of 100: An Even/Odd Entanglement

Yes. First example came into my mind was $\frac{2}{4} \times 6 =3$ . You can construct infinitely many examples, like

$\Large \frac{\color{#3D99F6}{2^k a({\color{#D61F06}{2n+1}})}}{\color{#20A900}{2^{k+l}ab}} \ \times \ \color{#69047E}{2^l b} \ \ \color{#333333}{=} \ \ \color{#D61F06}{2n+1} \color{#333333}{,}$

Where $k>0$ , $l>0$ , $a \neq 0$ , $b \neq 0$ and $n$ are integers.

One could argue that the answer is “No”. We're in the domain of number theory, the study of natural numbers. Within that domain, you can't divide 6 by 4. (Alternatively, you can define the answer as 1, with a remainder of 2.)

After all, $\dfrac{2}{6} \times 4$ , is that an even or an odd number? That question doesn't make sense, does it?

Since they asked for "can," one example is sufficient to show that it can be done. And indeed, it can: $\frac{2}{4}$ • 6 = 3, which is an odd number.

Lets take an example. $\dfrac{2}{4} \times 6 = \dfrac{6}{2} = 3 =odd$

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$\frac{even1}{even2}$ * even3 can be an odd number.

If Even 1= (2^a)*Ka, Even 2= (2^b)Kb, Even 3= (2^c)Kc,

Then the expression could be shown as , $\frac{(2^a)Ka*(2^c)Kc}{(2^b)Kb}$ , which is equal to (2^(a+c-b))* $\frac{Ka*Kc}{Kb}$

If we can find any three number that satisfy the following:

a+c-b=0 and $\frac{Ka*Kc}{Kb}$ = an odd number,

then

$\frac{even1}{even2}$ * even3 can be an odd number.

E.g. a=1,b=3,c=2, Ka=15, Kb=21, Kc=7

Even1=2 * 15=30, Even2=8 * 21=168, Even3=4 * 7=28

$\frac{30}{168}$ * 28=5 which is an odd number

Let a be the numerator, b the denominator and c the even number that multiplies the fraction. If Q is the product of prime factors of a except 2, then a=2^k * Q. If the same way b=2^l * W and c=2^m * R. The fraction a c/b will be odd if and only if l=m+k and W divides perfectly (without remainder) Q R. (Sorry, i haven't learned to use latex yet.)

Any $\dfrac {\color{#D61F06}{\text{Even}}}{\color{#D61F06}{\text{Even}}}$ form can be converted into $\dfrac {\color{#D61F06}{\text{Odd}}}{\color{#D61F06}{\text{Even}}}$ or $\dfrac {\color{#D61F06}{\text{Even}}}{\color{#D61F06}{\text{Odd}}}$ or $\dfrac {\color{#D61F06}{\text{Odd}}}{\color{#D61F06}{\text{Odd}}}$ form. Obviously, we can't generalize for all. ;)

One example is:-

$\begin{aligned} \dfrac {\color{#D61F06}{\text{Even}}}{\color{#D61F06}{\text{Even}}} × \color{#3D99F6}{\text {Even}} & = \dfrac {\color{#D61F06}{\text{Odd}}}{\color{#3D99F6}{\text{Even}}} × \color{#3D99F6}{\text {Even}}\\ \\ & = \color{#D61F06}{\text {Odd}} × \dfrac {\color{#D61F06}{\text{Odd}}}{\color{#D61F06}{\text{Odd}}}\\ \\ & = \color{#20A900}{\text {Odd}} \end{aligned}$

Or

$\begin{aligned} \dfrac {10}{4} × 2 & = \dfrac {5}{2} × 2\\ \\ & = \color{#69047E}{5} \end{aligned}$

2/4 x 2 = 1

Good example: 2/4 x 2 = 1

if we can make any p/q=1/2, then 1/2 multiplied by 6,10,14,18,22...........makes the answer an odd number.

Even/Even×Even=Odd

If even/even results in a fraction, this product can result in an odd number.

(a/2a)x(even number) = odd number

[ a is an even integer ]

2/4 * 6 = 3.

The following example : $\frac{2}{4} \times 2 = 1$

proves that the answer is yes .

( $\frac{4}{8}$ )(10)=5

6/4 * 2 = 3; 6/4 is 1 1/2; 1 1/2 * 2 equals 3. First thing to pop up in my head.

Tree numbers that work are 6, 4 and 10. Such that: $\frac{6}{4}\times10=15$

Yes. There are infinite examples for which this is true.

50/4 x 2= 25

$\frac{10}{2}$ *10 = 25

An obvious solution set (popped into my head immediately) is $1 = (\frac{2}{2n})(n)$ for any even number $n$

Even number 2,4 and 6 will work

42/28*78=117

the three nos. will be (10/4)*6

$\frac{2}{4}$ x 6 =1

You usually think about this kind of riddles in terms of integers. But if you allow the intermediary solution to be a real number the solutions are easy to find and numerous: for example 10/8*4

(4/4)*4=1 ques doesnt mention even no. to be used cant be same

1 is and odd number.
2 X 1/2 = 1. Therefore, any even number (x) divided by its double (2x) and multiplied by 2 (an even number) will yield 1.

\frac {x} {2x} \times 2 = 1

$\frac{2}{4}$ x2=1

Think with the smallest numbers you can imagine:

${\dfrac{2}{4}} \times {2} = 1$

This is true!

4/24 x 6 = 1

(2/8) * 4 = 1

$(4\div 2)\cdot 2 = 4\div (2\cdot 2) = 4\div 4 = 1$ .

Take the nos 6,4 and 14

If the final result is to be Odd, then the result of the division needs to be n.5 so that any doubling by the Even multiplier will produce an Odd integer; e.g.10/4*2=5

(2/4)*2 = 1

6/2=3 3x0=3

$\frac{6}{4}$ *2=3

$\frac{2}{4}$ 10

• (\frac{6/4}(2)

8/8= 1. 1*5=5

10 divided by 4 is 2.5 ~ 2.5 multiplied by 2 is 5 ~ 10, 4, and 2 are even, and 5 is odd.

You can always get the odd number of 1 if you multiply an even whole number by a multiple of it's reciprocal. ie. 2/4 x 2 = 1

((2 x3) x (2x5))/ (2 X2) Basically, have the 2's cancel and you good to go.

2/4 x 6 = 3

10/2 X 5 = 25

10/4 x 2=2.5 x 2=5

6/4 x 2 = 1.5 x 2 = 3

Yes (6/8)*4

$\frac{2}{20}$ *10=1

2 divided by 4 is one half. And half of 6 is 3

2/4 x 6 = 3

Example (42/84)*6=3

(6/16)x8=3

100/40=25x100=125

The first thing I tried was $\frac{4}{16}$ x 12 = 3 which worked wonderfully.

$\frac{42}{4}$ $X$ 2 = 21 i.e. an Odd Number This is an example can show that is possoble. There are many examples can fulfill this criteria.

(2/4)(6)= 3

Division can result in fractions which can yield odd numbers in many ways. For instance 2/4x6

An easy example is something like this: 2/8 x 4 = 1.

$\frac{2}{8}$ *12

N (2/8 *4)= N for N is any odd number.

I looked for even numbers that reduced to 3/2 then multiplied by 2.

Yes. 10/4 x 2 = 5 Just one example is sufficient to show this is true bc it asks "can... ever = an odd."

We have : $\frac{2 * blue}{2 * green} * 2 * purple = red$

that simplifies to : $\frac{2 * blue * purple}{green} = red$

So for every $green$ that is equals to $2 * blue$ and for every $purple$ that is odd, the triplet $(blue, green, purple)$ gives a $red$ odd value.

For example : $\frac{2*k}{4*k} * 2 * (2 * n + 1) = 2 * n + 1$

2/4 * 14 = 7 or 2k/4k * 2*(2n+1) = 2n + 1 = odd

I just asked my sis for three even numbers and then I picked yes

I myself am not really fond of solutions which just give a simple example, even though they are legitimate.
It is the same as solving the equation $x^2-5x+6=0$ just by stating that it has at most two solution and the solutions are $x=2$ and $x=3$ , which is easy in this particular case, but could you easily solve $x^2 + \pi x + e = 0$ in your head without using the quadratic formula?

As I said earlier, this solution is legitimate, but it leaves a question, "How did one come up with that solution?"
In this case the solutions were trivial to see by trial and error, but I don't like it as a problem solving strategy when there is an infinite amount of possible cases to consider in the worst case.

Here it is easy to state that $\dfrac28\cdot4=1$ which gives a solution to the problem, but what if there weren't any numbers satisfying the equation?

This is my solution to the problem:

$\Large \frac{\color{#3D99F6}{\text{Even}}}{\color{#20A900}{\text{Even}}} \ \times \ \color{#69047E}{\text{Even}} \ \ \color{#333333}{=} \ \ \color{#D61F06}{\text{Odd}}$

Now, we need to find the solutions to the equation $\dfrac AB\cdot C = d$ , where $A,B,C$ are even integers and $d$ is an odd integer.

But now, let $2^\alpha,2^\beta,2^\gamma$ be the greatest powers of $2$ dividing $A,B,C$ and let $a,b,c$ be (odd) integers such that $A = 2^\alpha a, B = 2^\beta b, C = 2^\gamma c$ .

Now, $2^{\alpha-\beta+\gamma} \cdot \dfrac ab\cdot c = d$ .

Since, $d$ is an odd integer, we must have $\alpha-\beta+\gamma = 0, b \mid ac$ .

Now, $ac = bd = k$ , for any odd integer $k$ , and we can freely choose $a$ and $b$ as any of the factors of $k$ , which now determine $c$ and $d$ .
Then we can freely choose any non-negative integers $\alpha$ and $\gamma$ , which now determine $\beta$ .

Hence, the original question can be rephrased to "Does there exist an odd integer which has at least one factor and does there exist any non-negative integers?".

Clearly, one divides every integer, especially the odd integers and axiomatically there exists at least one non-negative integer.

Hence, the solution is $\boxed{\text{Yes}}$ .

There are multiple solutions to this, but the one that I chose was: (2/8)×36= 9

It's very simple: you only have to take the tables of multiplication into consideration to know that even divided by even can be odd.

Eg : $\frac{10}{4}$ x2 = 5

2 multiply by 2 then divided by 4 = 1 Which is the odd number.

$\frac{10}{4}$ *2

Two divided by four equals one half. One half times two equals one.

10/4 x 2 = 5

It is possible when denominator is greater numerator in even/even,
2/8 * 4 =0.25 * 4 =1 4/6 * 2 =0.66 *2 =1.3

the first thing that came into my mind was 2/2 x 2 but that equals 2 so I thought of 2/4 x 2 which is 1

(4/16)x100=25

4/8x6 -> 0.5x6 -> 3

I set up an improper fraction whose value was 2.5 which multiplied by 2 is the odd integer 5

I just thought of the first 3 even numbers- 2/4 x 6

2/4 (or any equivalent to 1/2) multiplied by a multiple of 2 that is not a multiple of 4 works.

6 divide 4 times 2 equals 3

$\frac{12}{8}$ X 10 = 15

(2/4) × 2 = 1.... an odd #

Let's name these variables such that $\frac{a}{b}$ =c. If we can write a as m \cdot n, where m is any odd integer and n is any even integer; and we can write c as x \cdot y, where x is any odd integer and y is any even integer; then we can select b such that b=n \cdot y. Thus, the "even parts" cancel and leave us with an odd result.

2 divided by 4 multiplied by 6 gives the answer 3 which is odd

The simplest example is $\frac{2}{4}\times 2 = 1.$

(14/2) x 3 = 21

$\frac {2}{8}\cdot 4=1$

Any combination where the power of 2 in the prime factors of the green number equals the sum of the powers of 2 in the factors of the blue and green numbers will work. Question: Do the terms odd and even only apply to integers?

all even numbers have common divisor number ''2'' in order to create odd number from these numbers i can choose an odd number multiplied by 2 which equals an even number and i can choose another two number same even number like this: $\frac{2.odd}{2.even}$ . $even$ = $odd$

Rearrange to gain $\frac{Even}{Even}$ = $\frac{Odd}{Even}$ . Then choose any odd and even number for the RHS. We can double the numerator and denominator of the RHS fraction to gain an equal fraction, which also has an even numerator (and denominator), so satisfies the LHS of the equation, so the answer is $Yes$

YES consider: $\frac{2}{4}\times10=5$

Even number is in form of $2^{n}m$ where $n \in \mathbb{N}$ and $m$ any odd number, i.e. $m \in \{2x+1:x \in \mathbb{N}\}$

So, expression above could be written as $\frac{2^{n_1}m_1\cdot 2^{n_2}m_2}{2^{n_3}m_3}$ = $2^{n_1+n_2-n_3}\frac{m_1m_2}{m_3}$

So, for any combination of ${n_1, n_2, n_3}$ such that ${n_1+n_2-n_3}= 0$ , i.e. ${n_1+n_2=n_3}$ and ${m_1, m_2, m_3}$ such that ${m_3|m_1m_2}$ , the result will be an odd number.

e.g. ${n_1=1, n_2=2, n_3=3}$ and ${m_1=7, m_2=15, m_3=3}$ , the result is

$\frac{14\cdot 60}{24}$ = $\frac{(2^{1}\cdot7)\cdot (2^{2}\cdot 15)}{2^{3}\cdot 3}=35$

Well, it's an elementary question, 2*6/12=1 which is odd

Only intuition and logic

One example is that $\frac{12}{8}$ x 2 = 3. So, even/even x even becomes odd.

$\huge{\dfrac {2}{12} \times 18}$

The problem can be simplified to read $\frac{Even*Even}{Even}=Odd$ . Therefore the denominator of the left hand side of the equation must have the same number of 2s in its prime factorization as the numerator. This can easily be achieved if the denominator has more than one 2 in its prime factorization. An easy example of this would be setting the denominator as 4 and setting the two even numbers in the numerator as 2.

An even number may be of 2k or 4k type.

If the two even numbers in numerator is of form 2k , where k is odd &

If the even number in denominator is of form 4m , where m is odd, then-----

EVEN×EVEN÷EVEN= 2K×2Ķ÷4M =KĶ÷M =odd

Because, K,Ķ,M all are odd.

There's an example of the case.

18×22÷36=11 , that is odd.

The best example is 6/4*2..