Problem 36
Question
The remainder left out when \(8^{2 n}-(62)^{2 n+1}\) is divided by 9 is (a) 2 (b) 7 (c) 8 (d) 0
Step-by-Step Solution
Verified Answer
The remainder is 2 (option a).
1Step 1: Understanding the Problem
We are asked to find the remainder when the expression \(8^{2n} - (62)^{2n+1}\) is divided by 9. To do this, we should find the remainders of \(8^{2n}\) and \((62)^{2n+1}\) separately when divided by 9, and then find the remainder of their difference.
2Step 2: Calculating Remainder of 8 modulo 9
Calculate \(8 \mod 9\), which is just \(8\). Since \(8 \equiv -1 \pmod{9}\), for even powers, \(8^{2n} \equiv 1^{n} \pmod{9}\). Hence, \(8^{2n} \equiv 1 \pmod{9}\).
3Step 3: Calculating Remainder of 62 modulo 9
Calculate \(62 \mod 9\). First, perform the division: \(62 \div 9 = 6\) with a remainder of \(8\). Thus, \(62 \equiv 8 \equiv -1 \pmod{9}\). For odd powers, \((62)^{2n+1} \equiv (-1)^{2n+1} \equiv -1 \pmod{9}\).
4Step 4: Calculating the Expression's Remainder
Combine results: We found \(8^{2n} \equiv 1 \pmod{9}\) and \((62)^{2n+1} \equiv -1 \pmod{9}\). Therefore, the expression \(8^{2n} - (62)^{2n+1} \equiv 1 - (-1) \equiv 1 + 1 \equiv 2 \pmod{9}\).
5Step 5: Conclusion
The remainder when \(8^{2n} - (62)^{2n+1}\) is divided by 9 is 2. Hence, the answer is (a) 2.
Key Concepts
RemaindersPower of IntegersDivision Algorithm
Remainders
In mathematics, the concept of remainders arises from division. When you divide one number by another, the remainder is what is left over after the division occurs. This is an important concept in modular arithmetic, where we are often interested in what a number equals when divided by a certain modulus.
For example, when 62 is divided by 9, the quotient is 6, but there is still something left over: 8. Therefore, the remainder is 8, and we express this in modular terms as \(62 \equiv 8 \pmod{9}\).
Remainders help simplify expressions within modular arithmetic. They let us work with smaller numbers, making calculations easier and faster. By focusing on remainders, one can determine patterns and easily perform operations like addition, subtraction, and multiplication within a modular system.
For example, when 62 is divided by 9, the quotient is 6, but there is still something left over: 8. Therefore, the remainder is 8, and we express this in modular terms as \(62 \equiv 8 \pmod{9}\).
Remainders help simplify expressions within modular arithmetic. They let us work with smaller numbers, making calculations easier and faster. By focusing on remainders, one can determine patterns and easily perform operations like addition, subtraction, and multiplication within a modular system.
Power of Integers
The phrase 'power of integers' refers to numbers that are the result of raising one integer to another integer's power. For example, if you square an integer, you're raising it to the power of 2. For instance, \(8^{2n}\) means you multiply 8 by itself \(2n\) times.
When working with powers in modular arithmetic, certain patterns emerge. These patterns often help in efficiently calculating large powers. Notice in the original problem, \(8 \equiv -1 \pmod{9}\). For even powers like \(8^{2n}\), the result becomes \((-1)^{2n}\). Since any even number raised to -1 equals 1, \(8^{2n} \equiv 1 \pmod{9}\).
Understanding powers is crucial, especially when simplifying expressions involving large numbers. Recognizing these patterns can save a lot of time and make complex problems easier to solve.
When working with powers in modular arithmetic, certain patterns emerge. These patterns often help in efficiently calculating large powers. Notice in the original problem, \(8 \equiv -1 \pmod{9}\). For even powers like \(8^{2n}\), the result becomes \((-1)^{2n}\). Since any even number raised to -1 equals 1, \(8^{2n} \equiv 1 \pmod{9}\).
Understanding powers is crucial, especially when simplifying expressions involving large numbers. Recognizing these patterns can save a lot of time and make complex problems easier to solve.
Division Algorithm
The division algorithm is a fundamental principle in mathematics used to divide integers. It posits that given any two integers, a dividend \(a\) and a divisor \(b\) \((b eq 0)\), there exist unique integers \(q\) (quotient) and \(r\) (remainder) such that:
\[ a = bq + r \]
where \(0 \leq r < |b|\).
In our context, when dividing 62 by 9, we apply the division algorithm to find that 62 equals 9 times 6 plus a remainder of 8. So, \(62 = 9 \times 6 + 8\).
This powerful tool helps determine remainders and is essential for modular arithmetic. By always decomposing a number into its quotient and remainder parts, we can solve complex arithmetic problems much more efficiently. This approach is widely applied in computer science, cryptography, and more fields where understanding large numbers is necessary.
\[ a = bq + r \]
where \(0 \leq r < |b|\).
In our context, when dividing 62 by 9, we apply the division algorithm to find that 62 equals 9 times 6 plus a remainder of 8. So, \(62 = 9 \times 6 + 8\).
This powerful tool helps determine remainders and is essential for modular arithmetic. By always decomposing a number into its quotient and remainder parts, we can solve complex arithmetic problems much more efficiently. This approach is widely applied in computer science, cryptography, and more fields where understanding large numbers is necessary.
Other exercises in this chapter
Problem 34
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Statement \(-1: \sum_{r=0}^{n}(r+1){ }^{n} C_{r}=(n+2) 2^{n-1}\). Statement- \(2: \sum_{r=0}^{n}(r+1){ }^{n} C_{r} x^{r}=(1+x)^{n}+n x(1+x)^{n-1}\). (a) Stateme
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In the binomial expansion of \((a-b)^{n}, n \geq 5\), the sum of \(5^{\text {th }}\) and \(6^{\text {th }}\) terms is zero, then \(\mathrm{a} / \mathrm{b}\) equ
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