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Chapter 7: Problem 530

When \(5^{20}\) is divided by 48 then remainder is (a) 2 (b) 0 (c) 1 (d) 5

Short Answer

Expert verified
The remainder when \(5^{20}\) is divided by 48 is 1. The correct answer is (c) 1.

Step by step solution

01

Write the expression in congruence form

We want to find the remainder when dividing \(5^{20}\) by 48. This can be written equivalently as finding the value of \(5^{20} \mod 48\).
02

Express 48 as the product of its factors

To make our calculations easier, we can express 48 as the product of its prime factors, which are 2 and 3. We have: \(48 = 2^4 \cdot 3\).
03

Use the Chinese Remainder Theorem

The Chinese Remainder Theorem (CRT) states that if the moduli are pairwise relatively prime, we can find the remainder modulo their product (in this case, 48) by finding the remainders modulo each of the factors (2^4 and 3). So, we want to find the values of \(5^{20} \mod 2^4\) and \(5^{20} \mod 3\).
04

Calculate the remainders modulo each of the factors

Notice that \(5^{20} \equiv 1 \pmod{2^4}\) because any odd number raised to an even exponent is odd and hence odd number will have remainder 1 when divided by 2^4. For the second congruence, we will use Fermat's Little Theorem. Because 5 and 3 are relatively prime, we know that \(5^{(3-1)} \equiv 1 \pmod{3}\), or \(5^2 \equiv 1 \pmod{3}\). This means 😃 \[5^{20} = (5^2)^{10} \equiv 1^{10} \equiv 1 \pmod{3}\] Now we have that \(5^{20} \equiv 1 \pmod{2^4}\) and \(5^{20} \equiv 1 \pmod{3}\).
05

Reconstruct the remainder modulo 48

We are now to see the remainder of \(5^{20} \pmod{48}\). Since both congruences have the same remainder (1), we can conclude that \(5^{20} \equiv 1 \pmod{48}\). Therefore, the remainder when \(5^{20}\) is divided by 48 is 1. The correct answer is (c) 1.

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Most popular questions from this chapter

\({ }^{\mathrm{n}} \mathrm{C}_{1}+2 \cdot{ }^{\mathrm{n}} \mathrm{C}_{2}+3 \cdot{ }^{\mathrm{n}} \mathrm{C}_{3}+\ldots+\mathrm{n} \cdot{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}}=\) (a) n \(\cdot 2^{n-1}\) (b) \((\mathrm{n}-1) 2^{\mathrm{n}-1}\) (c) \((\mathrm{n}+1) 2^{\mathrm{n}-1}\) (d) \((\mathrm{n}-1) 2^{\mathrm{n}}\)

\(\mathrm{R}=(3+\sqrt{5})^{2 \mathrm{n}}\) and \(\mathrm{f}=\mathrm{R}-[\mathrm{R}]\), Where [] is an integer part function then \(\mathrm{R}(1-\mathrm{f})=\) (a) \(2^{2 \mathrm{n}}\) (b) \(4^{2 \mathrm{n}}\) (c) \(8^{2 \mathrm{n}}\) (d) \(1^{2 \mathrm{n}}\)

If the ratio of co-efficient of three consecutive terms in expansion of \((1+x)^{n}\) is \(1: 7: 42\) then \(n=\) (a) 35 (b) 45 (c) 55 (d) 65

The Co-efficient of \(\mathrm{x}^{\mathrm{n}}\) in the expansion of \((1+\mathrm{x})(1-\mathrm{x})^{\mathrm{n}}\) is (a) \((-1)^{\mathrm{n}-1}(\mathrm{n}-1)^{2}\) (b) \((-1)^{\mathrm{n}}(1-\mathrm{n})\) (c) \(n-1\) (d) \((-1)^{\mathrm{n}-1} \mathrm{n}\)

In the binomial expansion of \((a-b)^{n}, n \geq 0\), the sum of \(5^{\text {th }}\) and \(6^{\text {th }}\) terms is zero then \((\mathrm{a} / \mathrm{b})=\) (a) \([5 /(\mathrm{n}-4)]\) (b) \([6 /(\mathrm{n}-5)]\) (c) \([(n-5) / 6]\) (d) \([(n-4) / 5]\)
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