Counting odds?
You are given two integers J and K .
At most, how many of the following statements can be simultaneously true?
- J + K is an odd number.
- J − K is an odd number.
- J × K is an odd number.
- J ÷ K is an odd number.
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4 solutions
J / K is only guaranteed to be odd if J and K are odd! If J and K are even, it can be both odd and even. 8 / 2 = 4 , 6 / 2 = 3
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It isn't guaranteed to be odd in this scenario. For instance, 1 / 3 = 3 1 which is a non-integer and so neither odd nor even.
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Yes, it's true. But, the problem maker is asking the best scenario.
What if J:odd, K:even, J/K would be odd.
I disagree with the last one. J/K is odd only if both J and K are odd.
If both are even J/K is even
I disagree with the third. If J is 1 and K is 3, than 3 answers are correct.
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By my calculation:
J + K = 4 (even)
J − K = − 2 (even)
J × K = 3 (odd)
J ÷ K = 3 1 (neither)
What values did you come up with?
According to me.... The answer should have been 3... Because.... For example.... Let J be 10 & K be 3... Then J*K = 30 {even} J-K = 7 {odd} J + K = 13 {odd} J / K = 3.33... {odd}
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3.33... is not odd. It is a non-integer, therefore neither odd nor even.
Examples:
J = 4 , K = 2 gives 0 true statements.
J = 6 , K = 2 gives 1 true statement.
J = 2 , K = 1 gives 2 true statements.
J + K is odd iff J − K is odd. Therefore, 3 or 4 true would require both these statements to be true. However, these first two statements require that J is odd and K is even or vice versa. Which would make J × K even. If J is odd and K is even, then J ÷ K is necessarily a non-integer. If J is even and K is odd, J ÷ K is either a non-integer or even. Hence both the last two statements are false. Therefore, we can't have 3 or 4 true statements.
There is only one correct response and that is '2'. It is a matter of English.
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The problem has changed since I solved it yesterday. Previously there was a set of checkboxes - basically "Which of these is a possible number of true statements in the list?". For some reason they've now been changed to radiobuttons, with the question simply asking for the maximum number that can be true.
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Yes, we modified the problem because it seems like people were getting confused with the setup (only 9% correct), and using "most" caught the essence of the problem sufficiently (also, solve rate now up to 31%).
What about if I take j=6 and K=3
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6 + 3 = 9, 6 - 3 = 3 (both odd) 6 * 3 = 18, 6 / 3 = 2 (both even) so two out of the four are odd in that case.
An integer (pronounced IN-tuh-jer) is a whole number (not a fractional number) that can be positive, negative, or zero. (grab from google) so if J =1 and K=0 would't all 4 work
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1 * 0 = 0, an even number. 1 / 0 is an invalid operation and does not give a number at all.
So you just have the first two true, not the other two.
All it says is that the first two numbers are integers. couldn't you use 1 and 2 which would give 3, -1, and .5? That is three solutions
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If you use J = 1 and K = 2 , then the values are 3 , − 1 , 2 and 2 1 . Only the first two of these are odd. A non-integer is neither odd nor even.
3,2 3-2=1 3+2=1 3*2=6 3/2 gives 1.5 that considers (odd)? Thats 3 answers
If J and K are both even, all sums can be even -> 0
If J and K are both even, J/K can be odd (e.g. 6/2) -> 1
If J is even and K is odd (or vice versa), J + K and J - K are odd, but J*K and J/K are even (or non-integer) -> 2 is possible, but 3 or 4 are not
I might be misunderstanding the question but... Even/even can produce an odd number when one is divided by the other. Even/odd can produce an odd sum or subtraction. Odd/odd produces an odd number on multiplication. So for all pairs of numbers at least one of the items CAN be true. In which case 0 is not a valid answer
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It CAN be the case that 0 are true: Use 8 and 2 for example. Results in 10, 6, 16, 4
The question is subject to misinterpretation because it does not specify that you need to have the same J & K values for all the conditions. If we change the values for each, all 4 are possible.
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Yes you are right
I fully agree that the question is worded poorly
what ... why would you do that :I ...
For me ÷ stand for integer divison , that is a % b = (a - (a mod b))/b A / stands for division. So please rephrase the 3rd line and write that "is integer and odd" and add that / is the"real" division and not integer division.
The question has been changed! Maurice's solution was a (correct) answer to a PREVIOUS version of the question, namely: which answerS are possible (multiple answers could be checked, in this version!)
Assume one of them is of the form 2n and other is of 2m+1. So adding or subtracting them would give us a dangling (extra) 1. Multiplying them gives a number 2n(m+1) which is always even. And dividing them doesn't give us an integer.
J + K is odd only if J and K have opposite parity.
J − K is odd only if J and K have opposite parity.
J K is odd only if J and K are both odd.
J / K is odd only if J and K have the same parity.
Hence, at most two of the four given statements can be true.