If the fractional part of the number 2⁴⁰³/15 is k/15, then k is equal to : (A) 14 (B) 6 (C) 4 (D) 8

Q: Wait, why do we split $2^{403}$ into $2^3 \cdot 2^{400}$? Why not just $2^{403}$ directly?

Great question! We split the exponent to create a base that is easily divisible by $15$. We know $2^4 = 16$, and $16$ is just $1$ more than $15$. By isolating $2^4$, we turn the problem into $(15+1)^{100}$, which is incredibly easy to expand using the Binomial Theorem. If we didn't do this, we would be stuck trying to calculate a massive number!

Q: What happens to the $15\lambda$ part in the expansion? Do we just ignore it?

We don't ignore it, but we recognize its role. When we expand $(15+1)^{100}$, every term except the last one contains a factor of $15$. This means all those terms are perfectly divisible by $15$. When we divide the whole expression by $15$, these terms yield a remainder of $0$. So, they effectively disappear when we are only interested in the fractional part!

Q: Is it always possible to use the Binomial Theorem for these types of problems?

Yes, it is a classic JEE technique! Whenever you see a large power divided by a number, look for a power of the numerator that is close to a multiple of the denominator. This 'Binomial Trick' is a powerful tool in your arsenal for handling modular arithmetic and remainder problems.

Question

If the fractional part of the number $\frac{2^{403}}{15}$ is $\frac{k}{15}$, then k is equal to :

Options:

(A) 14

(B) 6

(C) 4

(D) 8

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are staring at the expression $\frac{2^{403}}{15}$. At first glance, it looks like a monster. Calculating $2^{403}$ is impossible by hand, and even a calculator would give up.

In the world of JEE mathematics, we don't use brute force; we use elegance. The problem asks for the fractional part of this number, which is $\frac{k}{15}$.

The fractional part is simply the remainder when the numerator is divided by the denominator. So, the entire challenge boils down to finding the remainder when $2^{403}$ is divided by $15$.

### The Magic of Base 16

To find this remainder, we need a clever observation. We are working with powers of $2$. Let us list them: $2^1=2$, $2^2=4$, $2^3=8$, and $2^4=16$.

Stop right there! Look at $16$. It is $15 + 1$. This is the golden key.

Whenever we divide by $15$, having a term like $16$ is a gift because $16 \equiv 1 \pmod{15}$. This means any power of $16$ will also leave a remainder of $1$ when divided by $15$.

### The Art of Splitting the Exponent

Now, we have $2^{403}$. We want to use our $2^4$ base. Is $403$ divisible by $4$? Not quite. But $400$ is!

So, we use the laws of indices to rewrite the expression:
$$2^{403} = 2^3 \cdot 2^{400} = 8 \cdot (2^4)^{100}$$

Now, substitute $2^4 = 16$: we get $8 \cdot (16)^{100}$. This is the turning point. We can write $16$ as $(15 + 1)$. So, our expression becomes:
$$8 \cdot (15 + 1)^{100}$$

### The Binomial Sledgehammer

Now, let us apply the Binomial Theorem to $(15 + 1)^{100}$. The expansion is:
$$\binom{100}{0} 15^0 + \binom{100}{1} 15^1 + \dots + \binom{100}{100} 15^{100}$$

Notice that every single term here contains a factor of $15$, except for the very first term, which is $\binom{100}{0} 1^0 = 1$. We can group all those terms containing $15$ into a single block, $15\lambda$, where $\lambda$ is some integer.

Thus, we have:
$$(15 + 1)^{100} = 15\lambda + 1$$

### The Final Cancellation

We are almost home. Substitute this back into our expression:
$$8 \cdot (15\lambda + 1) = 120\lambda + 8$$

Now, divide this by $15$:
$$\frac{120\lambda + 8}{15} = \frac{120\lambda}{15} + \frac{8}{15} = 8\lambda + \frac{8}{15}$$

Since $8\lambda$ is an integer, it has no fractional part. The entire fractional part of the original number is just $\frac{8}{15}$.

Comparing this to $\frac{k}{15}$, we find that **$k = 8$**. It is elegant, it is precise, and it is the kind of logical beauty that makes mathematics so thrilling. You didn't need to calculate the massive number; you just needed to understand its structure.

Animated Solution for Mathematics - Binomial Theorem: If the fractional part of the number $\frac{2 ^{403}}{15}$ is $\frac{k}{15}$ , then k is equal to :

Select Answer:

Visualized Solution

Understanding Fractional Parts

Finding a Suitable Base

Splitting the Exponent

Rewriting the Numerator

Substituting the Base

Applying Binomial Theorem

Simplifying the Expansion

Multiplying by the Coefficient

Dividing by the Divisor

Identifying the Fractional Part

The Sigma Insight: Binomial Expansion for Positive Integral Index

Analyzing the Setup

The Magic of Base 16

The Art of Splitting the Exponent

The Binomial Sledgehammer

The Final Cancellation

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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; } If the fractional part of the number 152403​ is 15k​, then k is equal to :