Let the number (22)²⁰²² + (2022)²² leave the remainder α when divided by 3 and β when divided by 7.

Q: Why can't I just use a calculator or expand the binomial? This looks like a standard arithmetic problem.

The JEE Advanced exam isn't testing your ability to compute; it's testing your ability to simplify. Expanding $(22)^{2022}$ is physically impossible. Modular arithmetic is the elegant scalpel that cuts through the complexity, allowing you to find the remainder without ever knowing the actual value of the number.

Q: Is it mathematically legal to use negative remainders like $-1$? It feels like cheating.

It is not cheating; it is a mark of an expert! In modular arithmetic, $a \equiv b \pmod n$ means $a - b$ is a multiple of $n$. Since $6 - (-1) = 7$, which is a multiple of $7$, it is perfectly valid. Using negative remainders is a standard tool to make calculations trivial.

Q: What if the power was odd in the second term? Would the answer change?

Yes, it would! If the power were odd, $(-1)^{odd}$ would result in $-1$. In modular arithmetic, we prefer positive remainders, so you would add $7$ to that $-1$ to get $6$. Always check the parity of your exponent before finalizing your answer.

Question

Let the number $(22)^{2022} + (2022)^{22}$ leave the remainder $\alpha$ when divided by 3 and $\beta$ when divided by 7. Then $(\alpha^2 + \beta^2)$ is equal to

Options:

(A) 20

(B) 13

(C) 5

(D) 10

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The expression we are tasked with evaluating is:

$$E = (22)^{2022} + (2022)^{22}$$

Calculating this directly is impossible by hand. Instead, we employ **modular arithmetic** to reduce the magnitude of these terms by finding their remainders under specific divisors.

### Phase 1: The Modulo $3$ Conquest

Our first mission is to find $\alpha$, the remainder when $E$ is divided by $3$. We analyze the first term, $(22)^{2022}$.

Since $22 = 3 	imes 7 + 1$, we have $22 \equiv 1 \pmod 3$. Applying the property $a \equiv b \pmod n \Rightarrow a^k \equiv b^k \pmod n$, we get:

$$(22)^{2022} \equiv 1^{2022} \equiv 1 \pmod 3$$

Next, we examine the second term, $(2022)^{22}$. Using the divisibility rule for $3$ (sum of digits $2+0+2+2 = 6$), we see that $2022$ is divisible by $3$.

Thus, $2022 \equiv 0 \pmod 3$, which implies:

$$(2022)^{22} \equiv 0^{22} \equiv 0 \pmod 3$$

Combining these results, we find the remainder $\alpha$:

$$\alpha = 1 + 0 = 1$$

### Phase 2: The Modulo $7$ Strategy

Now, we hunt for $\beta$, the remainder when $E$ is divided by $7$. We start again with the base $22$.

Since $22 = 7 	imes 3 + 1$, we have $22 \equiv 1 \pmod 7$. Therefore:

$$(22)^{2022} \equiv 1^{2022} \equiv 1 \pmod 7$$

For the second term, $(2022)^{22}$, we divide $2022$ by $7$:

$$2022 = 7 	imes 288 + 6$$

This gives $2022 \equiv 6 \pmod 7$. To simplify calculations, we use the negative remainder $6 \equiv -1 \pmod 7$.

Raising this to the power of $22$:

$$(-1)^{22} = 1$$

Since $22$ is an even exponent, the negative sign vanishes. Thus, the remainder $\beta$ is:

$$\beta = 1 + 1 = 2$$

### Final Calculation

We have successfully determined the values $\alpha = 1$ and $\beta = 2$. The problem requires us to compute $\alpha^2 + \beta^2$.

Substituting our findings:

$$\alpha^2 + \beta^2 = 1^2 + 2^2 = 1 + 4 = 5$$

The final result is **$5$**. Through the power of modular arithmetic, we have tamed the monster with elegance and precision.

Animated Solution for Mathematics - Binomial Theorem: Let the number $(22)^{2022} + (2022)^{22}$ leave the remainder $α$ when divided by 3 and $β$ when divided by 7. Then $(α^{2} + β^{2})$ is equal to

Select Answer:

Visualized Solution

Problem Breakdown

Finding $α$ : First Term Modulo $3$

Applying Power to First Term

Finding $α$ : Second Term Modulo $3$

Applying Power to Second Term

Calculating $α$

Finding $β$ : First Term Modulo $7$

Applying Power for Modulo $7$

Finding $β$ : Second Term Modulo $7$

Applying Power to Negative Remainder

Calculating $β$

Final Calculation: $α^{2} + β^{2}$

The Sigma Insight: Binomial Expansion for Positive Integral Index

Analyzing the Setup

Phase 1

Phase 2

Final Calculation

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Select Answer:

Visualized Solution

Problem Breakdown

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α: First Term Modulo .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 3

Applying Power to First Term

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α: Second Term Modulo .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 3

Applying Power to Second Term

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

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β: First Term Modulo .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 7

Applying Power for Modulo .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 7

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β: Second Term Modulo .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 7

Applying Power to Negative Remainder

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

Final Calculation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α2+β2