How to find the remainders when ## 2^{50} ## and ## 41^{65} ## are?

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Here is the summary:In summary, when finding the remainders of ## 2^{50} ## and ## 41^{65} ## divided by ## 7 ##, we can use modular arithmetic to simplify the calculations. We first use the property that ## a^b\equiv (a \pmod m)^b \pmod m ## to simplify ## 2^{50} ## to ## 4 \pmod 7 ##. Then, we use the fact that ## a\equiv b \pmod m ## implies ## a^c\equiv b^c \pmod m ## to simplify ## 41^{65} ## to ## 6 \pmod 7 ##. Therefore, the remainders of
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Math100
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Homework Statement
Find the remainders when ## 2^{50} ## and ## 41^{65} ## are divided by ## 7 ##.
Relevant Equations
None.
Consider ## 2^{3}=8\equiv 1 \pmod 7 ##.
Then ## 2^{50}=2^{48}\cdot 2^{2}=(2^{3})^{16}\cdot 2^{2}\equiv 1^{16}\cdot 2^{2} \pmod 7\equiv 2^{2} \pmod 7\equiv 4 \pmod 7 ##.
Thus ## 2^{50}\equiv 4 \pmod 7 ##.
Now observe that ## 41\equiv 6 \pmod 7\equiv (-1) \pmod 7 ##.
Then ## 41^{65}\equiv (-1)^{65} \pmod 7\equiv (-1) \pmod 7\equiv 6 \pmod 7 ##.
Thus ## 41^{65}\equiv 6 \pmod 7 ##.
Therefore, the remainders when ## 2^{50} ## and ## 41^{65} ## are divided by ## 7 ## are ## 4 ## and ## 6 ##.
 
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Math100 said:
Homework Statement:: Find the remainders when ## 2^{50} ## and ## 41^{65} ## are divided by ## 7 ##.
Relevant Equations:: None.

Consider ## 2^{3}=8\equiv 1 \pmod 7 ##.
Then ## 2^{50}=2^{48}\cdot 2^{2}=(2^{3})^{16}\cdot 2^{2}\equiv 1^{16}\cdot 2^{2} \pmod 7\equiv 2^{2} \pmod 7\equiv 4 \pmod 7 ##.
Thus ## 2^{50}\equiv 4 \pmod 7 ##.
Now observe that ## 41\equiv 6 \pmod 7\equiv (-1) \pmod 7 ##.
Then ## 41^{65}\equiv (-1)^{65} \pmod 7\equiv (-1) \pmod 7\equiv 6 \pmod 7 ##.
Thus ## 41^{65}\equiv 6 \pmod 7 ##.
Therefore, the remainders when ## 2^{50} ## and ## 41^{65} ## are divided by ## 7 ## are ## 4 ## and ## 6 ##.
Perfect.
 
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Related to How to find the remainders when ## 2^{50} ## and ## 41^{65} ## are?

1. How do you find the remainder when 2^50 and 41^65 are divided?

To find the remainder when a number is divided by another number, you can use the modulo operator (%). The remainder is the number left over after the division is performed. For example, if you divide 10 by 3, the remainder is 1. So to find the remainder when 2^50 and 41^65 are divided, you would use the expressions 2^50 % 41 and 41^65 % 41, respectively.

2. Can you use a calculator to find the remainders?

Yes, most calculators have a built-in function for finding remainders. Look for the "mod" or "remainder" button on your calculator. However, if you are working with very large numbers like 2^50 and 41^65, it may be easier to use a computer program or a calculator that can handle larger numbers.

3. What is the significance of finding the remainders of these numbers?

Finding the remainders of 2^50 and 41^65 can be useful in various mathematical and scientific calculations. For example, in cryptography, remainders are used in modular arithmetic to encrypt and decrypt messages. In computer science, remainders are used in algorithms and data structures. In physics, remainders are used in calculations involving periodic phenomena.

4. Is there a specific method for finding the remainders of large numbers like 2^50 and 41^65?

Yes, there are various methods for finding remainders of large numbers. One method is to use the binomial theorem, which involves expanding the number into a sum of binomial coefficients. Another method is to use modular arithmetic, which involves finding the remainder after dividing by a specific modulus. There are also algorithms specifically designed for finding remainders of large numbers, such as the Barrett reduction algorithm.

5. Can you explain the concept of "cyclic remainders" and how it relates to 2^50 and 41^65?

Cyclic remainders refer to a pattern that occurs when finding the remainders of a number raised to different powers. For example, if you find the remainders of 2^1, 2^2, 2^3, and so on, you will notice that the remainders cycle through a pattern of 2, 4, 8, and so on. This is because 2 is a primitive root modulo 41, meaning that it generates all possible remainders when raised to different powers. Similarly, 41 is a primitive root modulo 2, so the remainders of 41^1, 41^2, 41^3, and so on will also cycle through a pattern. This concept can be applied to finding the remainders of 2^50 and 41^65, as their remainders will follow a cyclic pattern based on their primitive roots.

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