50th root of a number x is 12 and 50th root of another number y is 18. Then the remainder obtained o

Q: Why can't we just calculate $12^{50}$ and $18^{50}$ directly?

That is a great question! In competitive exams like JEE, you are never expected to compute massive numbers like $12^{50}$. The goal is to use algebraic properties—in this case, the Binomial Theorem and modular arithmetic—to simplify the expression before any calculation is needed. We are looking for the 'soul' of the number, not its full decimal expansion.

Q: Why did we choose to write $12$ as $(15 - 3)$ and $18$ as $(15 + 3)$?

We chose $15$ because it is a multiple of $5$. Since our divisor is $25 = 5^2$, any term in the binomial expansion that contains $15^k$ where $k \ge 2$ will automatically be a multiple of $15^2 = 225$. Since $225$ is a multiple of $25$, those terms become $0$ in modulo $25$. This is the 'trick' that collapses the entire expression!

Q: Wait, we got a remainder of $-2$. Can a remainder be negative?

In formal mathematics, a remainder is typically defined as a non-negative integer. If you arrive at a negative result like $-2 \pmod{25}$, you simply add the divisor ($25$) to it to bring it into the standard positive range: $-2 + 25 = 23$. So, $23$ is your correct, positive remainder.

Question

$50^{	ext{th}}$ root of a number $x$ is 12 and $50^{	ext{th}}$ root of another number $y$ is 18. Then the remainder obtained on dividing $(x + y)$ by 25 is ______.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To find the remainder of $x + y$ divided by $25$, where $x = 12^{50}$ and $y = 18^{50}$, we must evaluate:
$$(12^{50} + 18^{50}) \pmod{25}$$

The modulus $25$ is equal to $5^2$. This suggests we should express the bases $12$ and $18$ in terms of a multiple of $5$.

We observe that $12 = 15 - 3$ and $18 = 15 + 3$. Since $15$ is a multiple of $5$, its square $15^2 = 225$ is a multiple of $25$.

### The Binomial Dance

We rewrite the expression using the binomial expansion:
$$x + y = (15 - 3)^{50} + (15 + 3)^{50}$$

Applying the Binomial Theorem, we note that for an even power $n$, the odd-powered terms of the expansion cancel out. This leaves us with:
$$x + y = 2 \cdot \left[ \binom{50}{0} 15^{50} + \binom{50}{2} 15^{48} \cdot 3^2 + \dots + \binom{50}{48} 15^2 \cdot 3^{48} \right] + 2 \cdot \binom{50}{50} 3^{50}$$

### The Modulo Magic

Every term inside the bracket contains a factor of $15^k$ where $k \ge 2$. Since $15^2 = 225$, which is divisible by $25$, all these terms are congruent to $0 \pmod{25}$.

The expression simplifies significantly:
$$x + y \equiv 2 \cdot 3^{50} \pmod{25}$$

### The Final Ascent

To calculate $3^{50} \pmod{25}$, we use the property $3^5 = 243$. Since $243 = 250 - 7$, we have:
$$3^5 \equiv -7 \pmod{25}$$

Squaring both sides yields:
$$3^{10} \equiv (-7)^2 = 49 \equiv -1 \pmod{25}$$

Raising this result to the power of $5$:
$$3^{50} = (3^{10})^5 \equiv (-1)^5 = -1 \pmod{25}$$

Finally, we multiply by the factor of $2$:
$$2 \cdot (-1) = -2 \equiv 23 \pmod{25}$$

The remainder of $x + y$ divided by $25$ is **23**.

Animated Solution for Mathematics - Binomial Theorem: $5 0^{th}$ root of a number $x$ is 12 and $5 0^{th}$ root of another number $y$ is 18. Then the remainder obtained on dividing $(x + y)$ by 25 is ______.

Enter Numerical Value:

Visualized Solution

Defining $x$ and $y$

The Target Expression

Strategic Decomposition

Binomial Expansion Identity

Applying the Expansion

Modulo $25$ Analysis

Simplifying the Remainder

Calculating $3^{10} (mod 25)$

Final Computation

Conclusion

The Sigma Insight: Binomial Expansion for Positive Integral Index

Analyzing the Setup

The Binomial Dance

The Modulo Magic

The Final Ascent

Similar Questions

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Enter Numerical Value:

Visualized Solution

Defining .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y

The Target Expression

Strategic Decomposition

Binomial Expansion Identity

Applying the Expansion

Modulo .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 25 Analysis

Simplifying the Remainder

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 310(mod25)

Final Computation

Conclusion

The Sigma Insight: Binomial Expansion for Positive Integral Index

Analyzing the Setup

The Binomial Dance

The Modulo Magic

The Final Ascent

Similar Questions

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