Paper I · 9 Logical Reasoning & Analytical Ability · 9.1 Number system and simplification
9.1.2 Remainders, unit digits and factors
Unit digits repeat in short cycles, remainders can be worked on small pieces, and the number of factors comes straight from the prime factorisation. (Factorials, factor counting and remainder theorems each have a fuller note in this unit: MAB-NS-09, 10 and 11.)
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Core facts
- Unit-digit cycles:
| Last digit | Cycle | Length |
|---|---|---|
| 0, 1, 5, 6 | same digit always | 1 |
| 4 | 4, 6 (odd power 4, even power 6) | 2 |
| 9 | 9, 1 (odd power 9, even power 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 |
- Rule: divide the power by 4 and use the remainder; a remainder of 0 means the 4th term.
- Dividend = divisor × quotient + remainder; the remainder is always less than the divisor.
- If N = p^a × q^b × r^c (p, q, r primes): number of factors = (a + 1)(b + 1)(c + 1).
- Trailing zeros of a product = the smaller of the counts of 2s and 5s.
- aⁿ + bⁿ is divisible by (a + b) when n is odd; aⁿ − bⁿ is divisible by (a − b) always, and by (a + b) when n is even.
Example 1. Find the unit digit of 8^47. Solution. 47 ÷ 4 leaves 3; the 3rd term of 8, 4, 2, 6 is 2. Answer: 2
Example 2. Find the unit digit of 4^63. Solution. The power is odd, so the unit digit is 4. Answer: 4
Example 3. How many factors does 200 have? Solution. 200 = 2³ × 5². Factors = 4 × 3. Answer: 12
Explained
Type 1: unit digit of a product or sum. Reduce each base to its last digit and each power modulo 4, then multiply or add the unit digits.
Example 4. Unit digit of 2^37 × 7^54 × 9^21. (a) 2 (b) 4 (c) 6 (d) 8 Solution. 2^37: 37 ÷ 4 leaves 1 → 2. 7^54: 54 ÷ 4 leaves 2 → 9. 9^21: odd power → 9. 2 × 9 × 9 = 162. Answer: (a) 2
Example 5. Unit digit of 3^50 + 7^50. Solution. 50 ÷ 4 leaves 2: 3^50 ends in 9, 7^50 ends in 9. 9 + 9 = 18. Answer: 8
Example 6. Unit digit of (1,237)^89 × (564)^22. Solution. Only last digits matter: 7^89 (89 ÷ 4 leaves 1) → 7; 4^22 (even) → 6. 7 × 6 = 42. Answer: 2
Type 2: remainder of a power. Find a small power that leaves 1 (or −1), then only the leftover power matters.
Example 7. Remainder when 3^41 is divided by 8. Solution. 3² = 9 leaves 1. 3^41 = (3²)^20 × 3, so the remainder is 3. Answer: 3
Example 8. Remainder when 5^63 is divided by 6. Solution. 5 leaves −1 on 6; (−1)^63 = −1, that is 6 − 1. Answer: 5
Type 3: remainder carried to a factor of the divisor.
Example 9. A number leaves 64 when divided by 255. What does it leave when divided by 15? Solution. 15 divides 255, so the answer is 64 ÷ 15's remainder. Answer: 4
- This works only when the new divisor is a factor of the old one.
Type 4: odd, even and total factors. Odd factors ignore the power of 2; even factors = total − odd.
Example 10. How many factors of 1,800 are even? (a) 9 (b) 18 (c) 27 (d) 36 Solution. 1,800 = 2³ × 3² × 5². Total = 4 × 3 × 3 = 36. Odd = 3 × 3 = 9. Even = 36 − 9. Answer: (c) 27
Speed tips.
- Write the four-term cycle once at the top of the rough sheet; most unit-digit questions then take one line.
- A factor count for a non-square is always even; an odd count means the number is a perfect square.
Common traps
- ✗ "8^48 ends in 8 because 48 ÷ 4 leaves 0, so take the first term." ✓ Remainder 0 means the 4th term: 8^48 ends in 6.
- ✗ "Number of factors of 2³ × 5² = 3 × 2 = 6." ✓ Add 1 to each exponent: 4 × 3 = 12.
- ✗ "N ÷ 255 leaves 64, so N ÷ 15 leaves 64." ✓ The remainder must be less than 15: 4.
- ✗ "Even factors of 1,800 = 36 ÷ 2 = 18." ✓ Even = total − odd = 36 − 9 = 27.
- ✗ "aⁿ + bⁿ is always divisible by a + b." ✓ Only when n is odd.
Quick revision
- Cycles: 2, 3, 7, 8 → length 4; 4, 9 → length 2; 0, 1, 5, 6 → fixed.
- Power ÷ 4 remainder 0 means the 4th term of the cycle.
- Factors of p^a q^b r^c = (a+1)(b+1)(c+1).
- Odd factors: drop the power of 2. Even = total − odd.
- Only perfect squares have an odd number of factors.
- For remainders of powers, find a power leaving 1 or −1.
- Remainder carries over only to a factor of the divisor.
- Successive division: build the number from the last quotient outwards.
- aⁿ + bⁿ ÷ (a + b) when n is odd.
- 1! + 2! + … + n! (n ≥ 4) ends in 3.
Go deeper: Deeper (beyond what APPSC Paper I usually asks)
Deeper Advanced
Example 11. Sum of all factors of 180. Solution. 180 = 2² × 3² × 5. Sum = (1 + 2 + 4)(1 + 3 + 9)(1 + 5) = 7 × 13 × 6. Answer: 546
Example 12. A number leaves 31 when divided by a certain divisor; twice the number leaves 9 with the same divisor. Find the divisor. Solution. Twice the number would leave 62 if 62 were below the divisor. It leaves 9, so the divisor is 62 − 9 = 53 (and 53 > 31, so the first remainder is allowed). Answer: 53
Example 13. A number divided by 5 leaves 2; the quotient divided by 7 leaves 4. What is the remainder when the number is divided by 35? Solution. Work from the inside. Smallest quotient = 4, so the number = 5 × 4 + 2 = 22. Every such number is 35k + 22. Answer: 22
Example 14. Find the last two digits of 3^2026. (a) 09 (b) 29 (c) 49 (d) 69 Solution. 3^20 = 3,486,784,401 ends in 01. 2026 = 20 × 101 + 6, so the last two digits are those of 3^6 = 729. Answer: (b) 29
- Useful "ends in 01" powers: 3^20, 7^4 (2,401) and 9^10.
- Unit digit of a sum of factorials: from 5! onwards each ends in 0, so 1! + 2! + … + n! (n ≥ 4) ends in the unit digit of 1 + 2 + 6 + 24 = 33, that is 3.
- Trailing zeros of n! = ⌊n/5⌋ + ⌊n/25⌋ + ⌊n/125⌋ + … (full method in MAB-NS-09).
Practise this topic. Questions on Numerical ability, with answers explained.
All Number system and simplification notes · APPSC Paper I syllabus · APPSC Paper I Mock 1 with answers · Start free
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