Even Even × Even = Odd
Can an even number, divided by another even number, times another even number ever equal an odd number?
The three even numbers can be different numbers.
If "yes", what are three numbers that work? If "no", then why not?
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thank u 4 de hlp
how did you get that equation?that's nice.
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Just put some expressions that cancel each other!
What about (4/2) *6 =12 ? ?? and 8/4 *2 ???
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It doesn't say that outcome is always an odd number. It "can" be odd. It can be even.
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ok... Thanks ! I thought that in every case.
can i follow u?
can be a rational number (2/10)x6= 6/5 no ????
I thought of it as rewriting the equation as "even x even=odd x even = some number" which clearly is possible when you have 3 or more factors of a number. Eg 4 has 3 factors -1,2,4 and 2 x 2=1 x 4 or (2/4) x 2=1
Edit - having only 3 factors is sufficient only in this case. 2 must be a factor for the resulting number for the case in the problem to be true I.e. it must be an even number. Which is kind of obvious once stated in the odd × even =even x even way but I thought I should mention it otherwise as well.
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When k/p is integer, yes you are correct, however k/p can be any rational number.
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, 6 2 × 4 , is that an even or an odd number? That question doesn't make sense, does it?
But the problem never states that the two factors are either odd or even. Nor does it state that the factors are natural numbers or even integers. Yes, the numerator and denominator are both even, but the fraction itself is not constrained in any way that I can see by the problem statement.
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But the question “Is this number odd?” doesn't make sense for rational numbers (i.e. fractions). So, arguably, the problem domain is the set of natural numbers, or perhaps the set of integers, and in that case, the answer is a definite “no”.
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No one ever asked if the quotient was odd or even or neither. They asked if the product of the quotient of two even numbers and a third even number could be odd. It can. (See examples posted here.)
Math deals with fact. "Arguably" and "perhaps" are not fact. You can multiply two numbers that are neither even nor odd (e.g. 8/3 and 3/4) and get one that is (i.e. 2).
If you're really so determined to avoid non-integer fractions, just do some algebra (which is perfectly legal with integers) and transform it to (2*6)/4 before you execute the division. Poof, we've solved the problem without producing any fractions.
I think it does make sense. Just because your your example is not odd or even does not invalidate the question. There are odd solutions therefore the answer is yes - no ambiguity.
There is nothing in the question to suggest that we are limited to natural numbers, except that the three even numbers and the odd numbers are natural numbers. And division is certainly well-defined for any pair of natural numbers.
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I agree with "except that the three even numbers and the odd numbers are natural numbers" implies that we are limited to natural numbers, in which case division may be restricted to pairs of divisible numbers. Basically that's the trick of this trick question, seeing past that instinct.
(BTW your name seems really familiar, did you ever have a cat named after the owner of a very famous box?)
The question is not ambiguous. You have to think like a math teacher trying to trick his/her students. Most students will automatically think that the question refers to whole numbers, but that is not specified. So, you have to try different kind of numbers to see if it is ever possible to get an odd number in this type of situations. So, if you try even decimals, you could get an odd decimal. For example: 0.4/0.6 x 0.2=0.3... If the solution were obvious, the problem would not be as fun :-)
But you've made assumptions not stated in the question. The only things given are: 1) ((a/b)*c) = x 2) {a,b,c} ∊ 2N And the question asked is whether or not Ǝ x ∊ 2N-1
No constraints at all were placed on (a/b). If a constraint was placed on (a/b) such that it had to also be even then, absolutely, the answer would be "No"; but then you'd just reframe the question as "can two even numbers multiply to make an odd number", which is trivial.
Agreed. The question is ambiguous. I was assuming, maybe incorrectly, that the multiplication had to be between an even number and an even result of the fraction.
Exactly. I've answered "No" considering that, in A/B, A must be divisible by B. It is nonsense to ask whether a rational number is odd or even, and in Z associativity (a/b)c → a(c/b) doesn't hold. Just as a/0×0/b → a/b wouldn't hold for R.
An analogy in R would be: is there such x∊R that (√x)² = -x ? One might argue that in C every negative number with zero imaginary part is a solution, and that every such x is in R, but the context of problem implies that a square is always positive.
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Why are you defining division such that it can only produce integers? Most of math would disappear if we did that.
It didn't ask about rational numbers. It didn't ask about integers. It just asked about numbers.
"is there such x∊R that (√x)² = -x ?" There is no such x at all, real OR complex - except for zero.
It doesn't have to be an integer, I would think. Nonintegers like the one you mentioned are neither even nor odd, so they're not solutions. Your question is different from theirs because you asked "odd or even", which implies it's one or the other.
The question does make sense
Umm .. not sure what is happening in this question ... But, since every (a/b)Xc needs to be a number ... In order for this question to be valid (or be well-defined), we would need to define division (a/b), so that it remains integer for all integers a and b. one definition is what C++ or other programming languages as the integer portion (ceil) of the long devision. By that definition: 2/6x4 = ((2/6)x4) = (0x4)=0 so result is even. Similarly 2/4x6 =0, and it is impossible to have EVEN/EVEN*EVEN to be ODD.
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This isn't C++ math, just plain math. One half is still a number, even if it isn't a whole number.
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@Whitney Clark - How could you define a rational number odd ?? Maybe there is some special mathematics I am unaware of. That's just beyond me. Please enlighten me whether 1/2 is even or odd. This question is ill defined according to me.
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@Gaurav Sharma – One half is certainly not even or odd. But 2/4 = 1/2, so 2/4 x 6 = 6/2 = 3, which is odd. The result has to be odd; the intermediate steps need not be. They needn't even be integers! Why should they be?
I have an example here which not support the statement above : (56/8) x 2 = 14. I'm confused by the answer given too.
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That's even. It asked for odd.
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My LHS are all even number as the questions stated but I couldn't get an odd number. May I know what's the trick for this question?
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@Lightning Jj – The fraction doesn't have to be a whole number. No reason was given that it should be.
Since they asked for "can," one example is sufficient to show that it can be done. And indeed, it can: 4 2 • 6 = 3, which is an odd number.
yes, you are right
I stand corrected. Well done!
took same exactly same example!
Yes, even I took the same approach. It is much quicker and easier.
I used the opposite approach. 1.5 x 2 = 3. so choose any even divisor divisible by 4, and multiply it by 1.5, and you get the solution. 6/4, 12/8, 18/12, etc.
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I did the same thing. I assumed integers were required in the steps.
I did 6/4×2
I made the mistake of assuming that resulting integers were required. If the numerator is allowed to be less than the denominator, then the answer is quite obvious. Otherwise, if the numerator HAS to be greater than the denominator, then the answer is 'no'. Prove me wrong? :)
Lets take an example. 4 2 × 6 = 2 6 = 3 = o d d
e v e n 2 e v e n 1 * 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 , ( 2 b ) K b ( 2 a ) K a ∗ ( 2 c ) K c , which is equal to (2^(a+c-b))* K b K a ∗ K c
If we can find any three number that satisfy the following:
a+c-b=0 and K b K a ∗ K c = an odd number,
then
e v e n 2 e v e n 1 * 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
1 6 8 3 0 * 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 Even Even form can be converted into Even Odd or Odd Even or Odd Odd form. Obviously, we can't generalize for all. ;)
One example is:-
Even Even × Even = Even Odd × Even = Odd × Odd Odd = Odd
Or
4 1 0 × 2 = 2 5 × 2 = 5
(12/2)*4=24(even)
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Lol, im not telling that we can generalize that its odd for all, but there are many such ratios which give this phenomenon. We can't generalize that its even for all. The example is one such scenario.
It asked for odd.
Good example: 2/4 x 2 = 1
nice job :D
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
nice solution
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 ]
Not if the even number you're multiplying by is a multiple of 4. for example (2/4)*8=4.
The following example : 4 2 × 2 = 1
proves that the answer is yes .
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: 4 6 × 1 0 = 1 5
Yes. There are infinite examples for which this is true.
An obvious solution set (popped into my head immediately) is 1 = ( 2 n 2 ) ( n ) for any even number n
Even number 2,4 and 6 will work
the three nos. will be (10/4)*6
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
Think with the smallest numbers you can imagine:
4 2 × 2 = 1
This is true!
( 4 ÷ 2 ) ⋅ 2 = 4 ÷ ( 2 ⋅ 2 ) = 4 ÷ 4 = 1 .
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
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 divided by 4 is one half. And half of 6 is 3
Example (42/84)*6=3
The first thing I tried was 1 6 4 x 12 = 3 which worked wonderfully.
4 4 2 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.
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.
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 : 2 ∗ g r e e n 2 ∗ b l u e ∗ 2 ∗ p u r p l e = r e d
that simplifies to : g r e e n 2 ∗ b l u e ∗ p u r p l e = r e d
So for every g r e e n that is equals to 2 ∗ b l u e and for every p u r p l e that is odd, the triplet ( b l u e , g r e e n , p u r p l e ) gives a r e d odd value.
For example : 4 ∗ k 2 ∗ 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
−
5
x
+
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
+
π
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 8 2 ⋅ 4 = 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:
Even Even × Even = Odd
Now, we need to find the solutions to the equation B A ⋅ C = d , where A , B , C are even integers and d is an odd integer.
But now, let 2 α , 2 β , 2 γ be the greatest powers of 2 dividing A , B , C and let a , b , c be (odd) integers such that A = 2 α a , B = 2 β b , C = 2 γ c .
Now, 2 α − β + γ ⋅ b a ⋅ c = d .
Since, d is an odd integer, we must have α − β + γ = 0 , b ∣ a c .
Now,
a
c
=
b
d
=
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
α
and
γ
, which now determine
β
.
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 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.
2 multiply by 2 then divided by 4 = 1 Which is the odd number.
Two divided by four equals one half. One half times two equals one.
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
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.
Let's name these variables such that b a =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 4 2 × 2 = 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: 2 . e v e n 2 . o d d . e v e n = o d d
Rearrange to gain E v e n E v e n = E v e n O d d . 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 Y e s
YES consider: 4 2 × 1 0 = 5
Even number is in form of 2 n m where n ∈ N and m any odd number, i.e. m ∈ { 2 x + 1 : x ∈ N }
So, expression above could be written as 2 n 3 m 3 2 n 1 m 1 ⋅ 2 n 2 m 2 = 2 n 1 + n 2 − n 3 m 3 m 1 m 2
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 1 m 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 = 1 5 , m 3 = 3 , the result is
2 4 1 4 ⋅ 6 0 = 2 3 ⋅ 3 ( 2 1 ⋅ 7 ) ⋅ ( 2 2 ⋅ 1 5 ) = 3 5
Well, it's an elementary question, 2*6/12=1 which is odd
Only intuition and logic
One example is that 8 1 2 x 2 = 3. So, even/even x even becomes odd.
The problem can be simplified to read E v e n E v e n ∗ E v e n = O d d . 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.
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Yes. First example came into my mind was 4 2 × 6 = 3 . You can construct infinitely many examples, like
2 k + l a b 2 k a ( 2 n + 1 ) × 2 l b = 2 n + 1 ,
Where k > 0 , l > 0 , a = 0 , b = 0 and n are integers.