Solved Find The Remainders When 2 50 And 41 65 Are Divided Chegg (a) to find the remainders when 2^50 and 41^65 are divided by 7, we can use the property of modular arithmetic. for 2^50, we can use the concept of repeated squaring to simplify the calculation. Use congruence's to find the reminder when $2^ {50}$ and $41^ {65}$ are divided by 7. $2^ {50}$ $50= (7)^2 1$ $2^ {50}=2^ {7\cdot7 1}$ and i'm not sure where to go from here?.
Solved 7 Find The Remainders When A 1023457 Is Divided By Chegg Find the remainders when 2 50 and 41 65 are divided by 7. none. consider 2 3 = 8 ≡ 1 (mod 7). then 2 50 = 2 48 2 2 = (2 3) 16 2 2 ≡ 1 16 2 2 (mod 7) ≡ 2 2 (mod 7) ≡ 4 (mod 7). thus 2 50 ≡ 4 (mod 7). now observe that 41 ≡ 6 (mod 7) ≡ (1) (mod 7). then 41 65 ≡ (1) 65 (mod 7) ≡ (1) (mod 7) ≡ 6 (mod 7). thus 41 65 ≡ 6 (mod 7). There are 2 steps to solve this one. find the remainders when 2^50 and 41^65 are divided by 7. what is the remainder when the following sum is divided by 4? 1^5 2^5 3^5 99^5 100^5. not the question you’re looking for? post any question and get expert help quickly. In conclusion, the remainder when 2^50 is divided by 7 is 4, and the remainder when 41^65 is divided by 7 is 0. Find math textbook solutions? still have questions? find the remainder when 2^50 and 41^65 are divided by 7 get the answers you need, now!.
Solved 7 Find The Remainders When A 1023457 Is Divided By Chegg In conclusion, the remainder when 2^50 is divided by 7 is 4, and the remainder when 41^65 is divided by 7 is 0. Find math textbook solutions? still have questions? find the remainder when 2^50 and 41^65 are divided by 7 get the answers you need, now!. Thus the non negative remainder when 4165 is divided by 7 is 6. similarly, consider 2 50. we know that 2 1 ≡ 2 (mod 7) and 2 2 ≡ 4 (mod 7) and 2 3 ≡ 1 (mod 7) rise the power to 16 on both the sides, we get. (2 3) 16 ≡ 1 16 (mod 7) ⇒ 2 48 ≡ 1 (mod 7) multiply both the sides by 2 2, we have. ⇒ 2 48. 2 2 ≡ 1 . 2 2 (mod 7) ⇒ 2 50 ≡ 4 (mod 7). So the remainders when $2^ {50}$ and $41^ {65}$ are divided by $7$ are $\boxed {4}$ and $\boxed {6}$, respectively. now, let's find the remainder when the sum $1^5 2^5 3^5 \cdots 99^5 100^5$ is divided by 4. Find step by step advanced math solutions and your answer to the following textbook question: (a) find the remainders when $2^ {50}$ and $41^ {65}$ are divided by 7 . We first know 2 3 ≡ 1 (mod 7). then, we have: for 41 65: by observation, 1 5 2 5 3 5 4 5 ≡ 5 5 6 5 7 5 8 5 ≡ 9 5 10 5 11 5 12 5 ≡ (mod 4) we then only need to calculate 1 5 2 5 3 5 4 5 (mod 4) and times 25. thus 1 5 2 5 3 5 4 5 ≡ 1 4 × 2 3 9 × 27 0 ≡ 1 0 1 × 3 ≡ 0 (mod 4), which implies 0 × 25 ≡ 0 (mod 4).
Solved Exercise 1 Find The Following Remainders A 4566 Chegg Thus the non negative remainder when 4165 is divided by 7 is 6. similarly, consider 2 50. we know that 2 1 ≡ 2 (mod 7) and 2 2 ≡ 4 (mod 7) and 2 3 ≡ 1 (mod 7) rise the power to 16 on both the sides, we get. (2 3) 16 ≡ 1 16 (mod 7) ⇒ 2 48 ≡ 1 (mod 7) multiply both the sides by 2 2, we have. ⇒ 2 48. 2 2 ≡ 1 . 2 2 (mod 7) ⇒ 2 50 ≡ 4 (mod 7). So the remainders when $2^ {50}$ and $41^ {65}$ are divided by $7$ are $\boxed {4}$ and $\boxed {6}$, respectively. now, let's find the remainder when the sum $1^5 2^5 3^5 \cdots 99^5 100^5$ is divided by 4. Find step by step advanced math solutions and your answer to the following textbook question: (a) find the remainders when $2^ {50}$ and $41^ {65}$ are divided by 7 . We first know 2 3 ≡ 1 (mod 7). then, we have: for 41 65: by observation, 1 5 2 5 3 5 4 5 ≡ 5 5 6 5 7 5 8 5 ≡ 9 5 10 5 11 5 12 5 ≡ (mod 4) we then only need to calculate 1 5 2 5 3 5 4 5 (mod 4) and times 25. thus 1 5 2 5 3 5 4 5 ≡ 1 4 × 2 3 9 × 27 0 ≡ 1 0 1 × 3 ≡ 0 (mod 4), which implies 0 × 25 ≡ 0 (mod 4).
4 A Find The Remainders When 23 And 41 Are Divided By 7 Find step by step advanced math solutions and your answer to the following textbook question: (a) find the remainders when $2^ {50}$ and $41^ {65}$ are divided by 7 . We first know 2 3 ≡ 1 (mod 7). then, we have: for 41 65: by observation, 1 5 2 5 3 5 4 5 ≡ 5 5 6 5 7 5 8 5 ≡ 9 5 10 5 11 5 12 5 ≡ (mod 4) we then only need to calculate 1 5 2 5 3 5 4 5 (mod 4) and times 25. thus 1 5 2 5 3 5 4 5 ≡ 1 4 × 2 3 9 × 27 0 ≡ 1 0 1 × 3 ≡ 0 (mod 4), which implies 0 × 25 ≡ 0 (mod 4).
Solved 15 A Find An Integer Having The Remainders 1 2 5 5 Chegg
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