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5ugxr

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Suppose a, b, c, and d are integers with the property that 2a + 4b + 3c + d is a multiple of 5. Prove that a + 2b + 4c + 3d is a multiple of 5. [Note: Your proof must work for all integers a, b, c, and d, not just one or several specific choices of these values.]

I'm not sure how I can get started on this problem.
If 2a + 4b + 3c + d is a multiple of 5, we can have 2a + 4b + 3c + d = 5k where k is an integer

For a + 2b + 4c + 3d can I assume it is a multiple of 5 and then try to prove otherwise? Im not too familar with how to make proofs
Suppose a + 2b + 4c + 3d = 5j where j is an integer

Then,
2a+4b+3c+d=5k
a+2b+4c+3d=5j

If I add the equations I get
3a+6b+7c+4d=5(k+j)

If I subtract the equations I get
a+2b-c-2d=5(k-j)

Can I get a hint if im in the right direction or what I should be doing instead
 
5ugxr said:
For a + 2b + 4c + 3d can I assume it is a multiple of 5 and then try to prove otherwise?
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I don't see what use would that be.

Hint #1 : 2a + 4b + 3c + d = 2a + 4b - 2c - 4d mod 5
 
blamocur said:
I don't see what use would that be.

Hint #1 : 2a + 4b + 3c + d = 2a + 4b - 2c - 4d mod 5
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I didnt learn mod yet, but I think I can understand some... You are saying that we can subtract 5b or 5c since 5 x anything is a multiple of 5 so we can just subtract it from the expression... Thank you, I got it now :)
 
blamocur said:
Do you mean that you know how to complete the proof?
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Is there anything additional that I have to do to complete my proof? I'm new to proofs so I'm not sure what else I need to include.
I was able to get 2a+4b+3c+d to a+2b+4c+3d by multiplying by 3 to have 6a+12b+9c+3d, then subtracting 5a, 10b, and 5c to get
a+2b+4c+3d.
 
5ugxr said:
Is there anything additional that I have to do to complete my proof? I'm new to proofs so I'm not sure what else I need to include.
I was able to get 2a+4b+3c+d to a+2b+4c+3d by multiplying by 3 to have 6a+12b+9c+3d, then subtracting 5a, 10b, and 5c to get
a+2b+4c+3d.
Click to expand...
What you need next is to actually write the proof. You have the idea, but it is not a proof until you put it into an orderly form so that each step can be checked.

Start with, "Suppose a, b, c, and d are integers with the property that 2a + 4b + 3c + d is a multiple of 5. Then ..."

Once you've written a first draft of the proof itself, we can probably make suggestions.
 
Dr.Peterson said:
What you need next is to actually write the proof. You have the idea, but it is not a proof until you put it into an orderly form so that each step can be checked.

Start with, "Suppose a, b, c, and d are integers with the property that 2a + 4b + 3c + d is a multiple of 5. Then ..."

Once you've written a first draft of the proof itself, we can probably make suggestions.
Click to expand...

Suppose a, b, c, and d are integers with the property that 2a + 4b + 3c + d is a multiple of 5. Then, any multiple of 2a + 4b + 3c + d must be a multiple of 5, since 2a + 4b + 3c + d is initially a multiple of 5.

Therefore, we can multiply 2a + 4b + 3c + d by 3, to get 6a + 12b + 9c + 3d, which is a multiple of 5.

We can note that any multiple of 5 multiplied by a, b, c, or d, is a multiple of 5. For example, 5a, and 10b must be multiples of 5.

We can split our expression into parts: 6a + 12b + 9c + 3d => 5a + a + 10b + 2b + 5c + 4c + 3d. We can set this equal to 5k, where k is an integer.

Therefore, we have 5a + a + 10b + 2b + 5c + 4c + 3d = 5k.
We can isolate the multiples of 5 on one side:
a + 2b + 4c + 3d = 5k -5a-10b-5c
a + 2b + 4c + 3d = 5(k-a-2b-c)

Since a,b,c, and k are integers, and are multiplied by 5, then 5(k-a-2b-c) is always some multiple of 5.

This means that a + 2b + 4c + 3d must be a multiple of 5
 
5ugxr said:
Suppose a, b, c, and d are integers with the property that 2a + 4b + 3c + d is a multiple of 5. Then, any multiple of 2a + 4b + 3c + d must be a multiple of 5, since 2a + 4b + 3c + d is initially a multiple of 5.

Therefore, we can multiply 2a + 4b + 3c + d by 3, to get 6a + 12b + 9c + 3d, which is a multiple of 5.

We can note that any multiple of 5 multiplied by a, b, c, or d, is a multiple of 5. For example, 5a, and 10b must be multiples of 5.

We can split our expression into parts: 6a + 12b + 9c + 3d => 5a + a + 10b + 2b + 5c + 4c + 3d. We can set this equal to 5k, where k is an integer.

Therefore, we have 5a + a + 10b + 2b + 5c + 4c + 3d = 5k.
We can isolate the multiples of 5 on one side:
a + 2b + 4c + 3d = 5k -5a-10b-5c
a + 2b + 4c + 3d = 5(k-a-2b-c)

Since a,b,c, and k are integers, and are multiplied by 5, then 5(k-a-2b-c) is always some multiple of 5.

This means that a + 2b + 4c + 3d must be a multiple of 5
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That's a very good first draft. I think everything you need is there.

I could make some suggestions, but since it's so good, I'm going to pass it back to you: Read through it, looking for anything that you find hard to follow, or perhaps a little too wordy, and rewrite it as a second draft. (Proofs are essentially persuasive essays, so writing a second draft works just the same way.) You're likely to see some of the same things I see that could be clarified, and that will be good practice for you. After that, any ideas I or others have will probably just be stylistic details.

I'll just mention one specific: "6a + 12b + 9c + 3d => 5a + a + 10b + 2b + 5c + 4c + 3d" doesn't mean what you want it to mean. They are equal, not an implication!
 
5ugxr said:
Is there anything additional that I have to do to complete my proof? I'm new to proofs so I'm not sure what else I need to include.
I was able to get 2a+4b+3c+d to a+2b+4c+3d by multiplying by 3 to have 6a+12b+9c+3d, then subtracting 5a, 10b, and 5c to get
a+2b+4c+3d.
Click to expand...
This is a better proof than what I had in mind, shorter and more elegant!
 
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