Last-Two-Digits Test for 4

Divisibility rules are a staple of the HSC Mathematics Extension 1 syllabus, particularly in the Proofs topic where you learn to construct rigorous arguments. In this problem, we explore why a number is divisible by 4 exactly when its last two digits are – a fact you’ve likely used without proof. By working through the biconditional structure and modular arithmetic, you’ll strengthen your ability to write clear, logical proofs that link algebraic representation to modular congruences.

Problem Statement

Prove that a number is divisible by 4 if and only if the last two digits form a number divisible by 4.

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Hints

Attempt the proof independently first. Focus on the key theorem, algebraic transformation, or contradiction setup that links the hypothesis to the target conclusion.

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Solutions

Biconditional Proof (If and Only If)

This requires proving both directions:

Forward Direction ($\implies$): If $N$ is divisible by 4, then its last two digits form a number divisible by 4.

Reverse Direction ($\impliedby$): If the last two digits form a number divisible by 4, then $N$ is divisible by 4.

Setup: Express any integer $N$ as:

$$N = 100A + L$$

where $A$ is the number formed by all digits except the last two, and $L$ is the two-digit number formed by the last two digits ($0 \leq L \leq 99$).

This representation is useful because it isolates the final two digits in $L$ and bundles everything else into $A$. Since $100$ is a multiple of $4$, the contribution of the $100A$ term modulo $4$ is zero, making the analysis very clean.

Forward Direction Proof:

We assume that $N$ is already a multiple of $4$ and show that $L$ must be too. Assume $4 \mid N$, so $N \equiv 0 \pmod{4}$. Substituting the digit representation into this congruence gives:

$$\begin{aligned} 100A + L &\equiv 0 \pmod{4} \end{aligned}$$

Since $100 = 4 \times 25$, we have $100A \equiv 0 \pmod{4}$. Because $100$ is a multiple of $4$, any whole multiple of $100$ is certainly divisible by $4$. This term disappears modulo $4$, leaving us with just $L$:

$$\begin{aligned} 0 + L &\equiv 0 \pmod{4} \\ L &\equiv 0 \pmod{4} \end{aligned}$$

Thus $4 \mid L$.

Reverse Direction Proof:

Now we prove the converse. Assuming the last two digits $L$ are divisible by $4$, we need to show that the whole number $N$ is divisible by $4$. Assume $4 \mid L$, so $L \equiv 0 \pmod{4}$.

Since $100 = 4 \times 25$, we have $100A \equiv 0 \pmod{4}$. Again, the $100A$ term contributes nothing modulo $4$, so the entire number $N$ is congruent to $L$, which we assumed to be $0 \pmod{4}$:

$$\begin{aligned} N = 100A + L &\equiv 0 + 0 \pmod{4} \\ N &\equiv 0 \pmod{4} \end{aligned}$$

Thus $4 \mid N$.

Conclusion:

Since both directions are proven, we conclude:

$$N \text{ is divisible by } 4 \iff \text{last two digits of } N \text{ divisible by } 4$$

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Takeaways

• Iff Proof Structure: Must prove both ($\implies$) and ($\impliedby$) directions independently
• Digit Representation: Writing $N = 100A + L$ separates last two digits for modular analysis
• Key Observation: Since $100 \equiv 0 \pmod{4}$, divisibility of $N$ by 4 depends only on last two digits
• Generalization: Same technique proves divisibility rules for powers of 2 (e.g., rule for 8 uses last three digits)

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Further Readings

If you found this proof interesting, be sure to check out these relevant HSC booklets to sharpen your reasoning skills: HSC Proofs, HSC Vectors, HSC Mechanics