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
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