MATH 433
February 6, 2015
Quiz 3: Solutions
Problem 1. Determine the last digit of 433433 .
Solution: 3.
The last digit is the remainder under division by 10. We have 433 ≡ 3 mod 10. Then
4332 = 433 · 433 ≡ 3 · 3 = 9 (mod 10),
4333 = 4332 · 433 ≡ 9 · 3 = 27 ≡ 7 (mod 10),
4334 = 4333 · 433 ≡ 7 · 3 = 21 ≡ 1 (mod 10).
Now it follows that 433n+4 ≡ 433n mod 10 for all n ≥ 1. Therefore the last digits of the numbers
4331 , 4332 , 4333 , . . . , 433n , . . . form a periodic sequence with period 4. Since 433 ≡ 1 mod 4, the last digit of
433433 is the same as the last digit of 4331 , which is 3.
Problem 2. Find the inverse of the congruence class [8]29 .
Solution: [8]−1
29 = [11]29 .
To find the inverse of 8 modulo 29, we need to represent 1 as an integral linear combination of 8 and 29.
This can be done either by inspection or by the matrix method:
1 0 8
4 −1 3
4 −1 3
11 −3 1
1 0 8
→
→
→
→
.
0 1 29
−3 1 5
−3
1 5
−7
2 2
−7
2 2
From the first row of the last matrix we read off that 11 · 8 − 3 · 29 = 1. It follows that
[1]29 = [11 · 8 − 3 · 29]29 = [11 · 8]29 = [11]29 [8]29 ,
which means that [11]29 = [8]−1
29 .