Key Concept: Finding Last Two Digits using Modular Arithmetic and Binomial Theorem

To determine the last two digits of any integer $N$, we are essentially looking for the remainder when $N$ is divided by 100. This is expressed in modular arithmetic as $N \pmod{100}$. For problems involving large powers, the Binomial Theorem is an indispensable tool, especially when the base of the power can be conveniently expressed as $(100k \pm x)$ or $(10k \pm x)$, where $x$ is a small integer.

The Binomial Theorem states that for any non-negative integer $n$: $(a+b)^n = \sum_{k=0}^{n} {^nC_k} a^{n-k} b^k = {^nC_0} a^n b^0 + {^nC_1} a^{n-1} b^1 + \ldots + {^nC_{n-1}} a^1 b^{n-1} + {^nC_n} a^0 b^n$ When working with modulo 100, if one term in the binomial expansion is a multiple of 100 (or its square/higher power is), many terms in the expansion become $0 \pmod{100}$, simplifying the calculation significantly. Specifically, if $a$ is a multiple of 10 (e.g., $10k$), then $a^2 = (10k)^2 = 100k^2$ which is a multiple of 100. This means any term containing $a^m$ where $m \ge 2$ will be $0 \pmod{100}$.

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Step-by-Step Solution

1. Goal: Express the problem in terms of modular arithmetic. Our objective is to find the last two digits of $(1919)^{1919}$. This means we need to calculate $(1919)^{1919} \pmod{100}$.

2. Rewrite the Base for Binomial Expansion. To effectively use the Binomial Theorem for modulo 100, we need to express the base, 1919, in a form like $(100k \pm x)$ or $(10k \pm x)$. We observe that 1919 is close to 1920, which is a multiple of 10 and, more crucially, $20^2 = 400$ is a multiple of 100. We can write $1919 = 1920 - 1$. Why this choice? By choosing $a = 1920$, we ensure that $a \equiv 20 \pmod{100}$, and $a^2 \equiv 20^2 \equiv 400 \equiv 0 \pmod{100}$. This property will dramatically simplify our binomial expansion as terms with powers of $a$ greater than or equal to 2 will vanish modulo 100.

   So, the expression becomes $(1920 - 1)^{1919}$.

3. Apply the Binomial Theorem. Let $a = 1920$, $b = -1$, and $n = 1919$. Applying the Binomial Theorem: $(1920 - 1)^{1919} = \sum_{k=0}^{1919} {^{1919}C_k} (1920)^{1919-k} (-1)^k$ Expanding the last few terms (which are the only ones that will matter modulo 100): $= \ldots + {^{1919}C_{1917}}(1920)^2(-1)^{1917} + {^{1919}C_{1918}}(1920)^1(-1)^{1918} + {^{1919}C_{1919}}(1920)^0(-1)^{1919}$

4. Identify Terms Relevant to Last Two Digits (Modulo 100). We are interested in the value modulo 100. Let's analyze the terms based on the power of 1920:

   • Recall that $1920 \equiv 20 \pmod{100}$.
   • Therefore, $1920^2 \equiv 20^2 \pmod{100} \equiv 400 \pmod{100} \equiv 0 \pmod{100}$.
   • This implies that any term containing $(1920)^k$ where $k \ge 2$ will be a multiple of 100. For example, ${^{1919}C_{1917}}(1920)^2(-1)^{1917} \equiv 0 \pmod{100}$.
   • Why is this important? Because these terms contribute nothing to the last two digits of the overall sum. We only need to consider terms where the power of 1920 is less than 2 (i.e., $1920^1$ and $1920^0$).

   Thus, the only terms that are not necessarily $0 \pmod{100}$ are the last two terms of the expansion: $(1919)^{1919} \equiv {^{1919}C_{1918}}(1920)^1(-1)^{1918} + {^{1919}C_{1919}}(1920)^0(-1)^{1919} \pmod{100}$

5. Evaluate the Relevant Terms. We use the properties of binomial coefficients: ${^nC_{n-1}} = {^nC_1} = n$ and ${^nC_n} = {^nC_0} = 1$.

   • Second to last term: ${^{1919}C_{1918}}(1920)^1(-1)^{1918}$

     • ${^{1919}C_{1918}} = {^{1919}C_1} = 1919$.
     • $(-1)^{1918} = 1$ (since 1918 is an even number).
     • This term evaluates to $1919 \times 1920 \times 1 = 1919 \times 1920$.

   • Last term: ${^{1919}C_{1919}}(1920)^0(-1)^{1919}$

     • ${^{1919}C_{1919}} = 1$.
     • $(1920)^0 = 1$.
     • $(-1)^{1919} = -1$ (since 1919 is an odd number).
     • This term evaluates to $1 \times 1 \times (-1) = -1$.

   Combining these, we get: $(1919)^{1919} \equiv (1919 \times 1920 - 1) \pmod{100}$

6. Calculate the Result Modulo 100. Now we need to find the last two digits of $(1919 \times 1920 - 1)$. We can simplify the multiplication by taking remainders modulo 100 first:

   • $1919 \equiv 19 \pmod{100}$
   • $1920 \equiv 20 \pmod{100}$

   Substitute these into the expression: $1919 \times 1920 - 1 \equiv (19 \pmod{100} \times 20 \pmod{100} - 1) \pmod{100}$ $\equiv (19 \times 20 - 1) \pmod{100}$ $\equiv (380 - 1) \pmod{100}$ $\equiv 379 \pmod{100}$ The remainder of 379 when divided by 100 is 79. Therefore, the last two digits of $(1919)^{1919}$ are $79$.

7. Find the Product of the Last Two Digits. The last two digits are 7 and 9. Their product is $7 \times 9 = 63$.

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Tips and Common Mistakes

• Tip 1: Choose the Right Expansion Base: When finding the last two digits of $X^N$, always try to express $X$ as $(100k \pm y)$ or $(10k \pm y)$. The most effective form for modulo 100 is often $(10k \pm y)$ where $y$ is small, particularly if $10k$ is a multiple of $20$ (since $20^2=400 \equiv 0 \pmod{100}$). In this case, $1920-1$ was perfect because $1920$ is a multiple of $20$.
• Tip 2: Simplify with Modular Arithmetic at Each Step: Do not wait until the end to take the modulo. Simplify intermediate products and sums modulo 100 to keep numbers manageable. For instance, $1919 \times 1920 \pmod{100}$ is much easier to calculate as $(19 \times 20) \pmod{100} = 380 \pmod{100} = 80 \pmod{100}$.
• Common Mistake: Ignoring Parity of Exponents: In expansions like $(a-b)^n$, the term $(-b)^k$ alternates between positive and negative. Always pay close attention to the parity (even or odd) of the exponent $k$ for $(-1)^k$. An odd exponent yields $-1$, while an even exponent yields $1$.
• Common Mistake: Single Digit Last Digits: If a calculation yields a single digit (e.g., $5 \pmod{100}$), remember that the last two digits are $05$, not just $5$. The problem asks for the product of the last two digits, so this distinction is crucial (e.g., for $05$, the digits are $0$ and $5$, product is $0$).

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Summary and Key Takeaway

The product of the last two digits of $(1919)^{1919}$ is $\mathbf{63}$. This problem perfectly illustrates the synergy between the Binomial Theorem and modular arithmetic. By strategically rewriting the base to exploit the property that $20^2 \equiv 0 \pmod{100}$, we can reduce a seemingly complex calculation involving large powers to a simple evaluation of just two terms from the binomial expansion. This technique is fundamental for efficiently solving problems related to finding the last few digits of large numbers.