The product of the last two digits of (1919)^1919 is

Q: Wait, why do we use modulo 100?

The last two digits of any integer $N$ are simply $N \pmod{100}$. For example, $1234 \pmod{100} = 34$. It is the most efficient way to isolate the tail end of a number without calculating the whole thing, which would be impossible for a number as large as $(1919)^{1919}$.

Q: Why rewrite 1919 as 1920 - 1 instead of 1900 + 19?

We want to use the Binomial Theorem, which works best when one of the terms is easy to raise to a power. $1920$ is a multiple of $10$, so $1920^k$ will have many trailing zeros, making it vanish modulo 100. This is a classic JEE trick to simplify complex expansions!

Q: Why do we ignore terms with $1920^k$ for $k \ge 2$?

Because $1920 = 192 \times 10$. So $1920^2 = (192)^2 \times 100$. This is clearly a multiple of 100. Any power $k \ge 2$ will also be a multiple of 100. In modular arithmetic, multiples of the modulus are congruent to 0, so they effectively disappear from our calculation.

Question

The product of the last two digits of $(1919)^{1919}$ is

Sigma Percentile · Accepted Answer

### The Illusion of Complexity

Imagine you are staring at the expression $(1919)^{1919}$. It is a number so vast that if you tried to write it down, you would run out of paper, ink, and patience.

Yet, the JEE Advanced examiner asks for the product of its last two digits. At first glance, this feels like a trap designed to make you panic.

But here is the secret: in mathematics, whenever a problem asks for the 'last two digits,' it is a giant, neon sign pointing you toward modular arithmetic. We do not need the whole number; we only need the remainder when that number is divided by $100$.

We are looking for $(1919)^{1919} \pmod{100}$.

### The Binomial Key

How do we handle such a massive exponent? We need a tool that breaks down powers. The Binomial Theorem is our best friend here.

It tells us that $(x - a)^n$ can be expanded into a series of terms. But to use it effectively, we need to choose our $x$ and $a$ wisely.

If we write $1919$ as $(1920 - 1)$, we hit the jackpot. Why? Because $1920$ is a multiple of $10$.

When we raise $1920$ to any power $k$, we get $1920^k = (192 	imes 10)^k = 192^k 	imes 10^k$. For any $k \ge 2$, $10^k$ is at least $100$, meaning the entire term is a multiple of $100$.

In the world of modulo $100$, these terms are simply $0$. They vanish!

### The Vanishing Act

Let us apply the Binomial Theorem to $(1920 - 1)^{1919}$. The expansion looks like this:

$$ (1920 - 1)^{1919} = \binom{1919}{0} (1920)^{1919} - \dots + \binom{1919}{1918} (1920)^1 (-1)^{1918} + \binom{1919}{1919} (1920)^0 (-1)^{1919} $$

As we discussed, every term containing $1920^k$ where $k \ge 2$ becomes $0 \pmod{100}$. We are left with only the last two terms of the expansion.

Let us simplify them:

1. The second-to-last term: $\binom{1919}{1918} (1920)^1 (-1)^{1918}$. Since $\binom{n}{n-1} = n$, this is $1919 	imes 1920 	imes 1 = 1919 	imes 1920$.

2. The last term: $\binom{1919}{1919} (1920)^0 (-1)^{1919}$. Since $\binom{n}{n} = 1$ and $(-1)^{1919} = -1$, this is $1 	imes 1 	imes (-1) = -1$.

So, our expression simplifies to $(1919 	imes 1920) - 1 \pmod{100}$.

### The Final Stretch

Do not rush to multiply $1919$ by $1920$. Use the modulo trick again! Split $1919$ into $(1900 + 19)$.

Then our expression becomes:

$$ (1900 + 19) 	imes 1920 - 1 = (1900 	imes 1920) + (19 	imes 1920) - 1 $$

Since $1900$ is a multiple of $100$, the term $(1900 	imes 1920)$ is $0 \pmod{100}$. We are left with $(19 	imes 1920) - 1$.

Calculating $19 	imes 1920$ gives $36480$. Subtracting $1$ gives $36479$.

The last two digits are $79$. The question asks for the product of these digits: $7 	imes 9 = 63$.

We have tamed the beast! The final answer is **63**.

Animated Solution for Mathematics - Binomial Theorem: The product of the last two digits of $(1919)^{1919}$ is

Enter Numerical Value:

Visualized Solution

Objective: Last Two Digits

Rewriting the Base

Applying the Binomial Theorem

The Modulo $100$ Shortcut

Isolating the Significant Terms

Simplifying the Combinations

Breaking Down the Multiplication

Calculating the Final Product

Identifying the Last Two Digits

Final Answer: Product of Digits

The Sigma Insight: Binomial Expansion for Positive Integral Index

The Illusion of Complexity

The Binomial Key

The Vanishing Act

The Final Stretch

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Animated Solution for Mathematics - Binomial Theorem: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The product of the last two digits of (1919)1919 is