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Chapter 7: Problem 533

The least positive remainder when \(17^{30}\) is divided by 5 is (a) 2 (b) 4 (c) 3 (d) 1

Short Answer

Expert verified
The least positive remainder when \(17^{30}\) is divided by 5 is (b) 4.

Step by step solution

01

Find the remainder when the base is divided by 5

We want to find the remainder when 17 is divided by 5. We have: \[17 \equiv 2 \pmod 5\]
02

Raise the remainder to the power of 30

Now, we raise the remainder obtained in step 1 (which is 2) to the power of 30: \[2^{30}\] Note that we are not directly computing the value of \(2^{30}\), but rather, finding the remainder when it is divided by 5.
03

Determine the pattern of powers of 2 modulo 5

To find the remainder when \(2^{30}\) is divided by 5, we start by observing the powers of 2 modulo 5: \(2^1 \equiv 2 \pmod{5}\), \(2^2 \equiv 4 \pmod{5}\), \(2^3 \equiv 3 \pmod{5}\), \(2^4 \equiv 1 \pmod{5}\) We notice that there is a pattern that repeats every 4 powers, i.e., \((\)2, 4, 3, 1)$.
04

Find the remainder in the pattern of powers

Since we have discovered in step 3 that the powers of 2 modulo 5 repeat every 4 powers, we can find the remainder when \(2^{30}\) is divided by 4: \[30 \div 4 = 7 \text{ with a remainder of } 2\] Using the remainder (2), we know that it corresponds to the second number in the \((\)2, 4, 3, 1)$ pattern. Therefore: \[2^{30} \equiv 4 \pmod 5\]
05

Answer

Based on our calculations, the least positive remainder when \(17^{30}\) is divided by 5 is 4. Hence, the correct answer is: (b) 4

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Most popular questions from this chapter

In the expansion of \((x-y)^{10}\), (co-efficient of \(x^{7} y^{3}\) ) \(\left(\right.\) co-efficient of \(\left.x^{3} y^{7}\right)=\) (a) \({ }^{10} \mathrm{C}_{7}\) (b) \({ }^{2.10} \mathrm{C}_{7}\) (c) \({ }^{10} \mathrm{C}_{7}+{ }^{10} \mathrm{C}_{1}\) (d) 0

\(17^{\text {th }}\) term and \(18^{\text {th }}\) term are equal in expansion of \((2+x)^{40}\) then \(\mathrm{x}=\) (a) \((17 / 24)\) (b) \((17 / 12)\) (c) \((34 / 13)\) (d) \((34 / 15)\)

\((\sqrt{2}+1)^{5}+(\sqrt{2}-1)^{5}=\) (a) 58 (b) \(58 \sqrt{2}\) (c) \(-58\) (d) \(-58 \sqrt{2}\)

It \(\mathrm{A}\) and \(\mathrm{B}\) are coefficients of \(\mathrm{x}^{\mathrm{r}}\) and \(\mathrm{x}^{\mathrm{n}-\mathrm{r}}\) respectively in expansion of \((1+x)^{n}\) then \(=\) (a) \(\mathrm{A}+\mathrm{B}=\mathrm{n}\) (b) \(\mathrm{A}=\mathrm{B}\) (c) \(\mathrm{A}+\mathrm{B}=2^{\mathrm{n}}\) (d) \(A-B=2^{n}\)

\(\left[5^{(1 / 2)}+7^{(1 / 8)}\right]^{1024}\) has number of rational terms \(=\) (a) 0 (b) 129 (c) 229 (d) 178
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