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

To prove that \(2b + 4c + 3d + a\) is a multiple of 5, given that \(2a + 4b + 3c + d\) is a multiple of 5, we can proceed as follows:

### Step 1: Express the given condition
We are given that:

\[
2a + 4b + 3c + d \equiv 0 \pmod{5}
\]

This means that the expression \(2a + 4b + 3c + d\) is congruent to 0 modulo 5.

### Step 2: Analyze the target expression
We need to show that \(2b + 4c + 3d + a\) is also a multiple of 5, i.e.,

\[
2b + 4c + 3d + a \equiv 0 \pmod{5}
\]

### Step 3: Perform a substitution using congruence
To prove this, let's consider how we might use the given condition. Notice that we can express \(a\) from the given congruence in terms of the other variables modulo 5.

Given:

\[
2a + 4b + 3c + d \equiv 0 \pmod{5}
\]

This can be rewritten as:

\[
2a \equiv - (4b + 3c + d) \pmod{5}
\]

Since \( -1 \equiv 4 \pmod{5} \), we have:

\[
2a \equiv 4(4b + 3c + d) \pmod{5}
\]

Now multiply both sides by the multiplicative inverse of 2 modulo 5, which is 3, to isolate \(a\):

\[
a \equiv 3 \times 4(4b + 3c + d) \pmod{5}
\]

Simplifying:

\[
a \equiv 2(4b + 3c + d) \pmod{5}
\]

### Step 4: Substitute \(a\) in the target expression
Now, substitute this expression for \(a\) into the expression we need to prove:

\[
2b + 4c + 3d + a \equiv 2b + 4c + 3d + 2(4b + 3c + d) \pmod{5}
\]

Expanding this:

\[
2b + 4c + 3d + 8b + 6c + 2d \pmod{5}
\]

Combine like terms:

\[
(2b + 8b) + (4c + 6c) + (3d + 2d) \equiv 10b + 10c + 5d \pmod{5}
\]

Since \(10b\), \(10c\), and \(5d\) are all multiples of 5, we have:

\[
10b + 10c + 5d \equiv 0 \pmod{5}
\]

This shows that \(2b + 4c + 3d + a\) is indeed a multiple of 5.

### Conclusion
We have proven that if \(2a + 4b + 3c + d\) is a multiple of 5, then \(2b + 4c + 3d + a\) is also a multiple of 5 for all integers \(a\), \(b\), \(c\), and \(d\).

Would you like more details on any specific step, or do you have any questions?

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**Related Questions:**

1. What is the significance of modular arithmetic in proofs like this?
2. How can we determine the multiplicative inverse of an integer modulo \(n\)?
3. Why is it necessary to consider the problem modulo 5?
4. Can we generalize this proof for any modulus other than 5?
5. How do modular congruences help in solving equations with multiple variables?
6. Could this problem be solved using a different approach, such as direct computation?
7. What are some common strategies for proving properties of integer expressions modulo \(n\)?
8. How would the proof change if the coefficients were different (e.g., 3a + 5b + c + d)?

**Tip:** When working with modular arithmetic, understanding the properties of congruences, like how they behave under addition and multiplication, is crucial for constructing proofs.

Discover how to prove that the expression 2a + 4b + 3c + d is congruent to 0 modulo 5. This detailed mathematical proof demonstrates the application of modular arithmetic and congruences in solving equations with multiple variables. Suitable for advanced mathematics students.

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