No. Wait, Maybe

Woah! This problem definitely tricked me when I first saw it! Then I wrote things out algebraically, and it was still tricky to figure out exactly what's going on. It's a wonderful question!

SOLUTION:

An even number is any number of the form $2x$ where $x$ is any whole number (integer). So the problem asks "Is it possible for $\frac{2x}{2y} \times 2z$ to equal an odd number?"

Now, let's use algebra:
$\frac{2x}{2y} \times 2z = \frac{x}{y} \times 2z\ = \frac{2xz}{y}$

This makes it much clearer what needs to happen: $y$ must have a factor of 2 , so that 2 can cancel out the 2 in the numerator of $\frac{2xz}{y}$ .

As an example $\fbox{x=3, y=6, z=5}$ will work. Our even numbers are then $2x = 6$ , $2y = 12$ , and $2z = 10$ .

$\frac{6}{12} \times 10 = 5$ and 5 is odd, so we're done. However...

GENERALIZATION:

To generalize, consider: $\frac{2xz}{y}\$ more closely. What does it take in order for the tuple of even numbers { $2x, 2y, 2z$ } to have the property that $\frac{2xz}{y} = \frac{2x}{2y} \times 2z =$ some odd integer?

1) Observe that in order for $\frac{2xz}{y}$ to be an integer, $y$ must be a factor of $2xz$ .
2) In order for $\frac{2xz}{y}$ to be odd, $y$ must be a factor that is itself divisible by the highest power of 2 that $2xz$ is divisible by.

The easiest way to see the necessity of both of these facts is to consider the prime factorization of all of the numbers ( $x, y,$ and $z$ ) involved, and to know that in order for the resulting product to be odd, all of the 2s must cancel.

FOLLOW-UP CHALLENGE:

If $x=12$ and $z=30$ . What value(s) of $y$ will make $\frac{2x}{2y} \times 2z$ odd?

In other words, can you make $\frac{24}{2y} \times 60$ odd?

First, we factor:
$x=12=2^2 \times 3$
$z=30=2 \times 3\times 5$

As per the argument above, $y$ must be a multiple of $2^{2+1+1} = 2^4 = 16$ that is also a factor of $2^4 \times 3^2 \times 5\ = 720$ . So, any of $\fbox{16, 48, 80, 144, or 720}$ will work.

Let's test $y=48$ .

$\frac{2x}{2y} \times 2z = \frac{24}{96} \times 60\ = 15$

Boom! Solved and generalized!

Let $a$ be an even number $\dfrac{a}{a^{2}}\times a=1$ and we know that $1$ is an odd number, just for $a \ne 0$

Taking the prime factorisations of an EVEN number is always of the form $2^n \times ODD$ where n is a positive integer > 1 (We know that no even primes exist other than two)

so taking our original proposal : $\boxed{1} \frac{EVEN}{EVEN} \times EVEN = ODD = \frac{2^a \times ODD}{2^b \times ODD} \times 2^c \times ODD$

We can see that if $\boxed{2} a + c - b = 0$ then all the powers of two will cancel out and we are left with $\boxed{1.1} \frac{ODD}{ODD} \times ODD$

We also know that $\boxed{3} \frac{ODD}{ODD}$ must always be ODD assuming there is an integer result and that $\boxed{4} ODD \times ODD$ must always be ODD

Assuming then that $\boxed{2}$ is true $from \boxed{1.1} \frac{ODD}{ODD} \times ODD = ODD \times ODD = ODD$

There fore so long as you can cancel the twos completely i.e. $\boxed{2}$ is true then $\frac{EVEN}{EVEN} \times EVEN = ODD$

Easy. Consider 6/4*2 = (6/2) = 3

If there is no restriction on use of numbers than considering 4/8×2=1 which is odd.

smallest even no. = 2,,, so, 2/4 x 6 = 3 is odd

(2/4) x 6 is the simple pattern.

Multiply each digit with 3 and get infinite answers.

If you make sure a smaller even number is in the numerator and a larger one in the denominator, you'll end up with a fraction. Multiply the fraction as necessary to create a whole odd number.

ex. $\frac {2}{4}$ x 2 = 1

2÷4×2=1. 1 is odd whereas 2 and 4 are both even numbers.

2 x 2 / 4 = 1. Just Need to find a denominator to cancel out the power of 2 in the numerator, such that the final quotient is a whole number

Yes. $\frac{2}{8}4=1$ . Just arrange the number of factors of 2 appropriately.

6 * 2 /4 = 3, 10 * 2/4, 10*6/12, the examples are endless

*let the two even numbers in numerator be 2 and the even in denominator be 4 * $\frac{2}{4}$ x 2 = 1

6/4*2=3 fluff fluff fluff

If the first two evens divide to give an integer then this integer will be even and the overall answer will be even but you could get a fraction ex. (6÷4)x2 = 3

Simply tricky..

One example I got is (6/8*12) = 9.

(2/4)*6=3. So I took an even integer, divided it by double of the integer, and multiplied the numerator by the sum of double of the integer and the difference between double the integer and the integer. In code speak: Int x; Int y = 2 * x; Int z = y + (y - x);

(x / y) * z;

I just thought about it for a few seconds, then wrote 2/4*6=3. Sometimes all you need is a little common sense.

Any fraction with a decimal equivalent of 0.5 multiplied by a whole even integer equivalent to two times any odd number will satisfy the conditions to produce the desired result. THere are many solution sets, though. Any fraction with a decimal equivalent of .25 multiplied by an even integer that is a multiple of 4 and an odd number will also work, e.g., $\frac {2}{8}*$ {12, 20, 28, 36, 44, 52, just to name a few} (and the answer will be that number divided by 4). Zandra Vinegar's explanation is the best, as it uses a formula to derive the solution.

6 x 4 / 8 = 3, in general (odd x even1) x (even2) / ( even3) = odd ,where even3 = even1 x even2

(4/8)*2 = 1, which is odd by definition.

Take any fraction who when multiplied by 2, gives an odd number. Multiply the top and bottom of the fraction by two, the fraction remains the same but it is now even over even. Now multiply your fraction by 2, and you will have the odd number you planned to get at the start.

Assuming its possible.

2k+1 = (2n/2z)*2y

2k+1 = 2ny/z

k+ 1/2 = ny/z

We find that,

If ny/z = (2w+ 1)/2, then you will always get odd numbers as result.

n,y = odd , z =1 ... Infinite possibilities

2 divided by 4 multiplied by 10 = 5

2 / 4 * 2 = 0.5 * 2 = 1

Don't think too much and take a=2,b=4,c=2, then ac/b = 1.

2/4 * 6 = 3

it can be done using algebra

(2x/2y)*2z

or, (2xz)/y

so if we make y a multiple of 2 we will get the solution as odd

let y= (multiple of 2) and x = z = ( any odd number)

hence substituting values of x,y,z in (2x/2y)*2z we get (odd)

so ans is YES.

6/4 x 2 = 3

An example is 2/4 * 6 which is equal to 3

10/4=2.5. Multiply that by 2 and you get 5

2/4 (6) gvies 3

I just tried with the first three non negative even numbers 2, 4 and 6.

=> [2(EVEN) / 4(EVEN)] * 6(EVEN) = 3(ODD)

6/4*2=3 6,4,2 are even nos

if we consider the formula to be (a/b)*c=x then how can we get an odd number? well the easiest way consider this is to think would be what can we multiply with an even number to make it odd? if an integer is multiplied by an even number you get an even number so we must be looking for a fraction so thinking about possible fractions 1/2 will work if multiplied by an even number that's not a multiple of 4 but is a multiple of 2 1/3 will work if multiplied by an even number that's not a multiple of 6 but is a multiple of 3 1/4 will work if multiplied by an even number that's not a multiple of 8 but is a multiple of 4 we can add whole numbers to the fraction without changing this behaviour basically if we multiply a fraction that is a multiple of the denominator we get a integer and of that set of numbers if they are twice as large and another member in the set we will generate a multiple of two

My initial thoughts were: Half of every other even number starting with 2 is always odd. Just make the denominator double the numerator and you get 1/2. Half of 2, 6, 10, 14, 18, 22....... all are odd.

Given (x/y) multiplied by z. For all even (or odd) integers x and y that are greater than 0, If y = 2x, Then x/y = 1/2. If z is even and z/2 = an odd integer, Then 1/2 of z is ALWAYS odd!

Yes you can get an odd number. Your 3 numbers can be any numbers (represented by x,y, and z here) that satisfy these conditions: x (any even number), y (where y = 2x), and z (where z is even and z/2 = any odd number)

I had a fleeting moment of foolish assumption which went along the lines of "Well if both numbers are multiples of 2 then dividing one by the other must be a multiple of 2" which is nonsense. I realised that you can make "Even/Even" be any fraction you like, and came up with $\frac{18}{20} \times 10 = 9$

To quote it very simply, 2/6x9 = 3 Which is odd. Problem solved!

Simple. Let A=2,B=4 then C=6 then (2/4)*6=3 Hence we got odd number

Divisibility is something I am fascinated with and know much of its secrets

take first even number=6, second even number=8 & third even number=4 (6/8)*4=3= odd

If the dividends are $2^n \times$ an-odd-number and $2^m \times$ other-odd-number, then it works for any divisor that is $2^{n+m}$

For example, if the dividends are 12 and 10, then it works if the divisor is 8.

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

The dividends:

$12 = 2^2 \times 3$

$10 = 2^1 \times 5$

The divisor:

$8 = 2^3 = 2^{2+1}$

Why it works that way?

Simply because odd $\times$ odd $=$ odd. So if you can cancel out all of the even factors (the 2's), you will get an odd number on the right hand side.

Yes! For example: I have an odd number $m$ , and an even number $n$ , that three numbers are: $n$ , $2^kn$ , and $2^km$

(2/4)x6 Since both an odd and even number times an even number is even, I thought the quotient must not be odd or even. That left a fraction.

2,4,6...(2÷4)×6=3..many such ways..

2/4 x 2 = 1 or generally 2/4 x even = odd

(2/4 =.5 .5 x 2=1),

(2/4 =.5 .5 x 6=3),

(2/4=.5 .5 x 10=5),

(4/8=.5 .5 x 2=1),

(4/8=.5 .5x 6=3),

(6/12 = .5 .5x14=7)

(8/16=.5 .5x 18=9)

See the pattern? Any two even numbers where the dividend is twice the divisor will equal .5

.5 times every alternating even number (2, 6, 10, 14, 18......)* will yield an odd product

     * (4,8, 12, 16.....) will yield an even product

Any time you divide an even number that is a prime*2 by 4 and multiply by 2 you will get the prime back. You can also do this with any odd number. So there are infinitely many numbers this works for.

Any two even numbers when divided reduce to a fraction of 1/x where x is an even number can then be multiplied by x to get 1. 1 is an odd number. For example, 2/8 = 1/4 then multiply by 4 to get 1.

I just thought of 2/4 * 2 (or any other way of getting a half)

Umm......I guess we just need to remove the 2's from all the terms to make sure that only the odd term remains.Therefore 2,4,6 and other pairs where the denominator has equal number of 2's as the numerator will work.

Let's take 3 even nos like 12,8,6 then 12/8*6=9 proved....!!!!

let 'a' be an odd number and 'b' be an even number let 'c' = a x b, let 'd' = b x b then both 'c' and 'd' will always be even

so we have 3 even numbers b, c and d if evaluate (c / d) x b then

(c / d) x b = a , which is an odd number

Really nice problem!

We can look at it this way

An even number multiplied by an even number is always even

And an even number multiplied by an odd number is even.

$Even \times Even=Even$

$Even \times Odd=Even$

Equating the two equations,

$Even \times Even=Even \times Odd$

$\frac{Even}{Even} \times Even=Odd$

This shows that it is possible.

$NB$ : This is only true for even numbers that can be expressed as the product of an odd and an even number and at the same time be expressed as the product of two even numbers.

As is shown in the combined equation