14. The sign and unit place digit of (−9)99 are, respectively:
a. Negative, 9
b. Negative, 1
c. Positive, 9
d. Not possible to determine without calculating the full number
The sign and unit place digit of (−9)^99 are negative and 9, respectively, making option (a) correct. Since 99 is odd, the power preserves the negative sign from the base. The unit digit follows the pattern of 9^odd powers, which cycles as 9, 1 but lands on 9 for odd exponents.
Step-by-Step Solution
To solve, first determine the sign: (−9)^99 = −(9^99) because odd exponents keep the negative sign. Next, find the unit digit of 9^99. The unit digits of 9^n cycle every 2: 9^1 ends in 9, 9^2 in 1, 9^3 in 9, repeating. For odd n=99, unit digit is 9. Thus, the number is negative ending in 9.
Option Analysis
| Option | Description | Correct? | Reason |
|---|---|---|---|
| a. Negative, 9 | Sign negative, unit digit 9 | ✅ Yes | Matches odd exponent: negative sign, 9^odd unit=9 |
| b. Negative, 1 | Sign negative, unit digit 1 | ❌ No | Unit digit 1 occurs only for even exponents |
| c. Positive, 9 | Sign positive, unit digit 9 | ❌ No | Sign stays negative for odd power |
| d. Not possible to determine without calculating the full number | Cannot determine | ❌ No | Unit digit found via cyclicity (mod 10), no full calculation needed |
Unit Digit Cyclicity Rule
Unit digits of powers depend only on the base’s unit digit modulo 10. For base ending in 9:
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Cycle length 2: odd exponents → 9; even → 1.
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Proof: 9 ≡ −1 mod 10, so 9^n ≡ (−1)^n mod 10. For odd n, ≡ −1 ≡ 9 mod 10.
Negative base doesn’t alter the digit pattern; only sign changes. Ideal for CSIR NET Part A quantitative aptitude.
