- LPU MCA Semester 1
- PEA515
- Unit 1
PEA515 · Unit 1: Number system and averages
Unit digit calculation and remainder theorem
30 practice questions · in the MTE and ETE syllabus
Key points
- Reduce the exponent modulo the cycle length
- Remainder 0 means the last digit of the cycle
- Remainders of sums and products can be taken term by term
- Look for a power that leaves remainder 1
The unit digit of a power follows a repeating cycle that depends only on the unit digit of the base.
| Base ends in | Cycle of unit digits | Length |
|---|---|---|
| 0, 1, 5, 6 | the same digit | 1 |
| 4, 9 | 4, 6 and 9, 1 | 2 |
| 2 | 2, 4, 8, 6 | 4 |
| 3 | 3, 9, 7, 1 | 4 |
| 7 | 7, 9, 3, 1 | 4 |
| 8 | 8, 4, 2, 6 | 4 |
Divide the exponent by the cycle length; a remainder of 0 means the last digit of the cycle. Example: 7^45, and 45 = 4 x 11 + 1, so the unit digit is 7.
Remainders: the remainder of a sum or product equals the sum or product of the remainders (reduced again). Example: remainder of 2^10 divided by 7: 2^3 = 8 leaves 1, and 10 = 3 x 3 + 1, so 2^10 leaves 2^1 = 2.
Exam trap: when the exponent leaves remainder 0, use the fourth term of the cycle, not the first.
Try two questions
Find the last digit of 32 x 20.
- 0
- 2
- 1
- 3
Show the answer
A. Multiply only the unit digits 2, 0 step by step, keeping the last digit: 0.
Why not B: Adds the unit digits instead of multiplying them.
Why not C: Slips by one when carrying while multiplying the units.
Why not D: Arithmetic slip in the final step.
What is the unit digit of 7^91?
- 7
- 4
- 3
- 2
Show the answer
C. Last digit 7 repeats in a cycle; power 91 sits at position 3 of that cycle: unit digit 3.
Why not A: Multiplies the last digit by the power instead of raising it.
Why not B: Arithmetic slip in the final step.
Why not D: Arithmetic slip in the final step.