3. How many zeros at the end does the number 2022! have? (n! is a product of all positive integers from 1 to n. 2! = 1 x 2, 3! = 1 x 2 x 3, etc.) Can't use the calculator or any other electronic help.
3. How many zeros at the end does the number 2022! have? (n! is a product of all positive integers from 1 to n. 2! = 1 x 2, 3! = 1 x 2 x 3, etc.) Can't use the calculator or any other electronic help.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![3. How many zeros at the end does the number 2022! have? (n! is a product of all positive integers from 1 to n. 2! = 1 x
2, 3! = 1 x 2 x 3, etc.) Can't use the calculator or any other electronic help.
4. MOON
Using recursive definition of a number, compute the value of MOON27 in decimal representation. You can use the table:
0123456789 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
0 1 23456789 A BCDE F GHIKL M NOP QR
where the first row contains decimal values and the second row contains digits base 27. Note that some letters have
been skipped on purpose. Please use values from the table for your computation.
Show every step of your computation:
M 27 = [1digit]
MO 27 =[2digits]
MOO 27 =[3digits]
and the final result
MOON 27 = [4digits]
5. In the lecture we found a private key d for the following RSA values:
P = 7, and q = 13. We computed N = 91 and o = 72. Then for e = 41 we computed d = e-1 mod 72. Simplify this problem
and instead of o = (p-1) (g-1) take o = LCM((p-1), (q-1)) and compute d modulo this p. Now use this new private key to
decrypt our ciphertext 48. Show your work.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F05dd3cb4-971a-4726-ae57-476c6d039f50%2F8ade5191-950c-485b-8fe8-14b71a519ac8%2F2rfdpqn_processed.png&w=3840&q=75)
Transcribed Image Text:3. How many zeros at the end does the number 2022! have? (n! is a product of all positive integers from 1 to n. 2! = 1 x
2, 3! = 1 x 2 x 3, etc.) Can't use the calculator or any other electronic help.
4. MOON
Using recursive definition of a number, compute the value of MOON27 in decimal representation. You can use the table:
0123456789 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
0 1 23456789 A BCDE F GHIKL M NOP QR
where the first row contains decimal values and the second row contains digits base 27. Note that some letters have
been skipped on purpose. Please use values from the table for your computation.
Show every step of your computation:
M 27 = [1digit]
MO 27 =[2digits]
MOO 27 =[3digits]
and the final result
MOON 27 = [4digits]
5. In the lecture we found a private key d for the following RSA values:
P = 7, and q = 13. We computed N = 91 and o = 72. Then for e = 41 we computed d = e-1 mod 72. Simplify this problem
and instead of o = (p-1) (g-1) take o = LCM((p-1), (q-1)) and compute d modulo this p. Now use this new private key to
decrypt our ciphertext 48. Show your work.
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