Mixing
If both the product and sum of four integers are even, which of the following could be the number of even...
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The answer says that only options 2 and 4 are right and option 0 is impossible.
But here is my solution which proves that option 0 is possible:
Let’s A=0, B=2, C=0, D=2;
So their sum is 0+2+0+2=4 (which is an even number)
And their product is 0*2*0*2 = 0 (which is also an even number)
So why choice 0 is incorrect?
Here is what the book tells me about it:
“Since these four integers have an even product, at least one of them must be even, so roman numeral I, 0, is impossible” . But I don't understand why
elementary-number-theory arithmetic
asked Feb 3 at 12:55
Mukhamedali ZhadigerovMukhamedali Zhadigerov
183
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|
show 11 more comments
$begingroup$
The answer says that only options 2 and 4 are right and option 0 is impossible.
But here is my solution which proves that option 0 is possible:
Let’s A=0, B=2, C=0, D=2;
So their sum is 0+2+0+2=4 (which is an even number)
And their product is 0*2*0*2 = 0 (which is also an even number)
So why choice 0 is incorrect?
Here is what the book tells me about it:
“Since these four integers have an even product, at least one of them must be even, so roman numeral I, 0, is impossible” . But I don't understand why
elementary-number-theory arithmetic
asked Feb 3 at 12:55
Mukhamedali ZhadigerovMukhamedali Zhadigerov
183
$endgroup$
1
$begingroup$
The product of odd numbers is always odd. Your example would appear to have $4$ even integers.
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– lulu
Feb 3 at 12:57
2
$begingroup$
In your example you have 4 even numbers, not $0$.
$endgroup$
– franzk
Feb 3 at 13:00
1
$begingroup$
Maybe I don't understand your question. You give an example with $4$ even digits but then ask about the case with $0$ even digits. It's clear that you can't have no even digits as the product of odd digits is odd.
$endgroup$
– lulu
Feb 3 at 13:00
1
$begingroup$
You can't have no even integers because the product of odd integers is odd. Your example with four even integers does not appear to be relevant.
$endgroup$
– lulu
Feb 3 at 13:04
1
$begingroup$
I think you have failed to grasp the question you were asked. You aren't being asked if it is possible for any of the integers to be $0$, you are asked to count the number of even integers and decide whether that count might be $0$.
$endgroup$
– lulu
Feb 3 at 13:14
|
show 11 more comments
$begingroup$
The answer says that only options 2 and 4 are right and option 0 is impossible.
But here is my solution which proves that option 0 is possible:
Let’s A=0, B=2, C=0, D=2;
So their sum is 0+2+0+2=4 (which is an even number)
And their product is 0*2*0*2 = 0 (which is also an even number)
So why choice 0 is incorrect?
Here is what the book tells me about it:
“Since these four integers have an even product, at least one of them must be even, so roman numeral I, 0, is impossible” . But I don't understand why
elementary-number-theory arithmetic
asked Feb 3 at 12:55
Mukhamedali ZhadigerovMukhamedali Zhadigerov
183
$endgroup$
The answer says that only options 2 and 4 are right and option 0 is impossible.
But here is my solution which proves that option 0 is possible:
Let’s A=0, B=2, C=0, D=2;
So their sum is 0+2+0+2=4 (which is an even number)
And their product is 0*2*0*2 = 0 (which is also an even number)
So why choice 0 is incorrect?
Here is what the book tells me about it:
“Since these four integers have an even product, at least one of them must be even, so roman numeral I, 0, is impossible” . But I don't understand why
elementary-number-theory arithmetic
elementary-number-theory arithmetic
asked Feb 3 at 12:55
Mukhamedali ZhadigerovMukhamedali Zhadigerov
183
asked Feb 3 at 12:55
Mukhamedali ZhadigerovMukhamedali Zhadigerov
183
edited Feb 3 at 13:05
Mukhamedali Zhadigerov
asked Feb 3 at 12:55
Mukhamedali ZhadigerovMukhamedali Zhadigerov
183
asked Feb 3 at 12:55
Mukhamedali ZhadigerovMukhamedali Zhadigerov
183
asked Feb 3 at 12:55
Mukhamedali ZhadigerovMukhamedali Zhadigerov
183
183
1
$begingroup$
The product of odd numbers is always odd. Your example would appear to have $4$ even integers.
$endgroup$
– lulu
Feb 3 at 12:57
2
$begingroup$
In your example you have 4 even numbers, not $0$.
$endgroup$
– franzk
Feb 3 at 13:00
1
$begingroup$
Maybe I don't understand your question. You give an example with $4$ even digits but then ask about the case with $0$ even digits. It's clear that you can't have no even digits as the product of odd digits is odd.
$endgroup$
– lulu
Feb 3 at 13:00
1
$begingroup$
You can't have no even integers because the product of odd integers is odd. Your example with four even integers does not appear to be relevant.
$endgroup$
– lulu
Feb 3 at 13:04
1
$begingroup$
I think you have failed to grasp the question you were asked. You aren't being asked if it is possible for any of the integers to be $0$, you are asked to count the number of even integers and decide whether that count might be $0$.
$endgroup$
– lulu
Feb 3 at 13:14
|
show 11 more comments
1
$begingroup$
The product of odd numbers is always odd. Your example would appear to have $4$ even integers.
$endgroup$
– lulu
Feb 3 at 12:57
2
$begingroup$
In your example you have 4 even numbers, not $0$.
$endgroup$
– franzk
Feb 3 at 13:00
1
$begingroup$
Maybe I don't understand your question. You give an example with $4$ even digits but then ask about the case with $0$ even digits. It's clear that you can't have no even digits as the product of odd digits is odd.
$endgroup$
– lulu
Feb 3 at 13:00
1
$begingroup$
You can't have no even integers because the product of odd integers is odd. Your example with four even integers does not appear to be relevant.
$endgroup$
– lulu
Feb 3 at 13:04
1
$begingroup$
I think you have failed to grasp the question you were asked. You aren't being asked if it is possible for any of the integers to be $0$, you are asked to count the number of even integers and decide whether that count might be $0$.
$endgroup$
– lulu
Feb 3 at 13:14
1
1
$begingroup$
The product of odd numbers is always odd. Your example would appear to have $4$ even integers.
$endgroup$
– lulu
Feb 3 at 12:57
$begingroup$
The product of odd numbers is always odd. Your example would appear to have $4$ even integers.
$endgroup$
– lulu
Feb 3 at 12:57
2
2
$begingroup$
In your example you have 4 even numbers, not $0$.
$endgroup$
– franzk
Feb 3 at 13:00
$begingroup$
In your example you have 4 even numbers, not $0$.
$endgroup$
– franzk
Feb 3 at 13:00
1
1
$begingroup$
Maybe I don't understand your question. You give an example with $4$ even digits but then ask about the case with $0$ even digits. It's clear that you can't have no even digits as the product of odd digits is odd.
$endgroup$
– lulu
Feb 3 at 13:00
$begingroup$
Maybe I don't understand your question. You give an example with $4$ even digits but then ask about the case with $0$ even digits. It's clear that you can't have no even digits as the product of odd digits is odd.
$endgroup$
– lulu
Feb 3 at 13:00
1
1
$begingroup$
You can't have no even integers because the product of odd integers is odd. Your example with four even integers does not appear to be relevant.
$endgroup$
– lulu
Feb 3 at 13:04
$begingroup$
You can't have no even integers because the product of odd integers is odd. Your example with four even integers does not appear to be relevant.
$endgroup$
– lulu
Feb 3 at 13:04
1
1
$begingroup$
I think you have failed to grasp the question you were asked. You aren't being asked if it is possible for any of the integers to be $0$, you are asked to count the number of even integers and decide whether that count might be $0$.
$endgroup$
– lulu
Feb 3 at 13:14
$begingroup$
I think you have failed to grasp the question you were asked. You aren't being asked if it is possible for any of the integers to be $0$, you are asked to count the number of even integers and decide whether that count might be $0$.
$endgroup$
– lulu
Feb 3 at 13:14
|
show 11 more comments
2 Answers
2
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$begingroup$
The choice of 0 is incorrect, because if the product of integers is even, at least one must be even. (The product of (four) odd numbers is odd.) In your example, A=0, B=2, C=0, D=2, the four numbers are even (as is their sum and product). Also given in the comments is an example where two of the numbers are even (as is their sum and product), viz. A=0; B=1; C=0; D=3.
answered Feb 3 at 13:23
J. W. TannerJ. W. Tanner
4,9071420
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$begingroup$
Sorry, can't upvote because I have only 1 point :)
$endgroup$
– Mukhamedali Zhadigerov
Feb 3 at 13:26
add a comment |
$begingroup$
It might help to look at this algebraically. Let's say $a$, $b$, $c$, $d$ are any integers whatsoever.
Then the integers $2a + 1$, $2b + 1$, $2c + 1$, and $2d + 1$ are all odd. Their sum is even: $2a + 2b + 2c + 2d + 4$. However, their product is $16abcd + 8abc + 8ab + $ yada, yada, yada, $+ 2c + 2d + 1$. You can have Wolfram Alpha do it for you if you don't feel like going through it yourself.
Now let's try $2a + 1$, $2b + 1$, $2c + 1$, and $2d$. There's only one even integer in here, but it's enough to make the product even. But now the sum is odd: $2a + 2b + 2c + 2d + 3$.
It's the same situation for $2a + 1$, $2b$, $2c$, and $2d$: even product but odd sum.
answered Feb 5 at 4:27
Robert SoupeRobert Soupe
11.5k21950
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2 Answers
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2 Answers
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$begingroup$
The choice of 0 is incorrect, because if the product of integers is even, at least one must be even. (The product of (four) odd numbers is odd.) In your example, A=0, B=2, C=0, D=2, the four numbers are even (as is their sum and product). Also given in the comments is an example where two of the numbers are even (as is their sum and product), viz. A=0; B=1; C=0; D=3.
answered Feb 3 at 13:23
J. W. TannerJ. W. Tanner
4,9071420
$endgroup$
$begingroup$
Sorry, can't upvote because I have only 1 point :)
$endgroup$
– Mukhamedali Zhadigerov
Feb 3 at 13:26
add a comment |
$begingroup$
The choice of 0 is incorrect, because if the product of integers is even, at least one must be even. (The product of (four) odd numbers is odd.) In your example, A=0, B=2, C=0, D=2, the four numbers are even (as is their sum and product). Also given in the comments is an example where two of the numbers are even (as is their sum and product), viz. A=0; B=1; C=0; D=3.
answered Feb 3 at 13:23
J. W. TannerJ. W. Tanner
4,9071420
$endgroup$
$begingroup$
Sorry, can't upvote because I have only 1 point :)
$endgroup$
– Mukhamedali Zhadigerov
Feb 3 at 13:26
add a comment |
$begingroup$
The choice of 0 is incorrect, because if the product of integers is even, at least one must be even. (The product of (four) odd numbers is odd.) In your example, A=0, B=2, C=0, D=2, the four numbers are even (as is their sum and product). Also given in the comments is an example where two of the numbers are even (as is their sum and product), viz. A=0; B=1; C=0; D=3.
answered Feb 3 at 13:23
J. W. TannerJ. W. Tanner
4,9071420
$endgroup$
The choice of 0 is incorrect, because if the product of integers is even, at least one must be even. (The product of (four) odd numbers is odd.) In your example, A=0, B=2, C=0, D=2, the four numbers are even (as is their sum and product). Also given in the comments is an example where two of the numbers are even (as is their sum and product), viz. A=0; B=1; C=0; D=3.
answered Feb 3 at 13:23
J. W. TannerJ. W. Tanner
4,9071420
answered Feb 3 at 13:23
J. W. TannerJ. W. Tanner
4,9071420
answered Feb 3 at 13:23
J. W. TannerJ. W. Tanner
4,9071420
answered Feb 3 at 13:23
J. W. TannerJ. W. Tanner
4,9071420
4,9071420
$begingroup$
Sorry, can't upvote because I have only 1 point :)
$endgroup$
– Mukhamedali Zhadigerov
Feb 3 at 13:26
add a comment |
$begingroup$
Sorry, can't upvote because I have only 1 point :)
$endgroup$
– Mukhamedali Zhadigerov
Feb 3 at 13:26
$begingroup$
Sorry, can't upvote because I have only 1 point :)
$endgroup$
– Mukhamedali Zhadigerov
Feb 3 at 13:26
$begingroup$
Sorry, can't upvote because I have only 1 point :)
$endgroup$
– Mukhamedali Zhadigerov
Feb 3 at 13:26
add a comment |
$begingroup$
It might help to look at this algebraically. Let's say $a$, $b$, $c$, $d$ are any integers whatsoever.
Then the integers $2a + 1$, $2b + 1$, $2c + 1$, and $2d + 1$ are all odd. Their sum is even: $2a + 2b + 2c + 2d + 4$. However, their product is $16abcd + 8abc + 8ab + $ yada, yada, yada, $+ 2c + 2d + 1$. You can have Wolfram Alpha do it for you if you don't feel like going through it yourself.
Now let's try $2a + 1$, $2b + 1$, $2c + 1$, and $2d$. There's only one even integer in here, but it's enough to make the product even. But now the sum is odd: $2a + 2b + 2c + 2d + 3$.
It's the same situation for $2a + 1$, $2b$, $2c$, and $2d$: even product but odd sum.
answered Feb 5 at 4:27
Robert SoupeRobert Soupe
11.5k21950
$endgroup$
add a comment |
$begingroup$
It might help to look at this algebraically. Let's say $a$, $b$, $c$, $d$ are any integers whatsoever.
Then the integers $2a + 1$, $2b + 1$, $2c + 1$, and $2d + 1$ are all odd. Their sum is even: $2a + 2b + 2c + 2d + 4$. However, their product is $16abcd + 8abc + 8ab + $ yada, yada, yada, $+ 2c + 2d + 1$. You can have Wolfram Alpha do it for you if you don't feel like going through it yourself.
Now let's try $2a + 1$, $2b + 1$, $2c + 1$, and $2d$. There's only one even integer in here, but it's enough to make the product even. But now the sum is odd: $2a + 2b + 2c + 2d + 3$.
It's the same situation for $2a + 1$, $2b$, $2c$, and $2d$: even product but odd sum.
answered Feb 5 at 4:27
Robert SoupeRobert Soupe
11.5k21950
$endgroup$
add a comment |
$begingroup$
It might help to look at this algebraically. Let's say $a$, $b$, $c$, $d$ are any integers whatsoever.
Then the integers $2a + 1$, $2b + 1$, $2c + 1$, and $2d + 1$ are all odd. Their sum is even: $2a + 2b + 2c + 2d + 4$. However, their product is $16abcd + 8abc + 8ab + $ yada, yada, yada, $+ 2c + 2d + 1$. You can have Wolfram Alpha do it for you if you don't feel like going through it yourself.
Now let's try $2a + 1$, $2b + 1$, $2c + 1$, and $2d$. There's only one even integer in here, but it's enough to make the product even. But now the sum is odd: $2a + 2b + 2c + 2d + 3$.
It's the same situation for $2a + 1$, $2b$, $2c$, and $2d$: even product but odd sum.
answered Feb 5 at 4:27
Robert SoupeRobert Soupe
11.5k21950
$endgroup$
It might help to look at this algebraically. Let's say $a$, $b$, $c$, $d$ are any integers whatsoever.
Then the integers $2a + 1$, $2b + 1$, $2c + 1$, and $2d + 1$ are all odd. Their sum is even: $2a + 2b + 2c + 2d + 4$. However, their product is $16abcd + 8abc + 8ab + $ yada, yada, yada, $+ 2c + 2d + 1$. You can have Wolfram Alpha do it for you if you don't feel like going through it yourself.
Now let's try $2a + 1$, $2b + 1$, $2c + 1$, and $2d$. There's only one even integer in here, but it's enough to make the product even. But now the sum is odd: $2a + 2b + 2c + 2d + 3$.
It's the same situation for $2a + 1$, $2b$, $2c$, and $2d$: even product but odd sum.
answered Feb 5 at 4:27
Robert SoupeRobert Soupe
11.5k21950
answered Feb 5 at 4:27
Robert SoupeRobert Soupe
11.5k21950
answered Feb 5 at 4:27
Robert SoupeRobert Soupe
11.5k21950
answered Feb 5 at 4:27
Robert SoupeRobert Soupe
11.5k21950
11.5k21950
add a comment |
add a comment |
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$begingroup$
The product of odd numbers is always odd. Your example would appear to have $4$ even integers.
$endgroup$
– lulu
Feb 3 at 12:57
2
$begingroup$
In your example you have 4 even numbers, not $0$.
$endgroup$
– franzk
Feb 3 at 13:00
1
$begingroup$
Maybe I don't understand your question. You give an example with $4$ even digits but then ask about the case with $0$ even digits. It's clear that you can't have no even digits as the product of odd digits is odd.
$endgroup$
– lulu
Feb 3 at 13:00
1
$begingroup$
You can't have no even integers because the product of odd integers is odd. Your example with four even integers does not appear to be relevant.
$endgroup$
– lulu
Feb 3 at 13:04
1
$begingroup$
I think you have failed to grasp the question you were asked. You aren't being asked if it is possible for any of the integers to be $0$, you are asked to count the number of even integers and decide whether that count might be $0$.
$endgroup$
– lulu
Feb 3 at 13:14