Govt. Exams
Entrance Exams
Numbers divisible by 3: floor(100/3) = 33. Numbers divisible by both 3 and 5 (i.e., by 15): floor(100/15) = 6. Numbers divisible by 3 but not by 5 = 33 - 6 = 27. Wait, that's option B. Let me verify: 27 is correct.
Using the property: HCF × LCM = Product of two numbers. 6 × 60 = 12 × x. 360 = 12x. x = 30.
# Unit Digit of 7^2019
To find the unit digit of any power, we identify the cyclical pattern of unit digits for that base number.
Step 1: Find the Cyclical Pattern of Unit Digits for Powers of 7
The unit digit of powers of 7 repeats in a cycle. Let's calculate the first few powers:
The pattern repeats: 7, 9, 3, 1, 7, 9, 3, 1, ...
The cycle length is 4.
Step 2: Find the Position of 7^2019 in the Cycle
Divide the exponent 2019 by the cycle length 4 to find which position in the pattern it corresponds to:
Since the remainder is 3, we need the 3rd unit digit in our cycle pattern (7, 9, 3, 1).
The 3rd position corresponds to unit digit 3.
The unit digit of 7^2019 is 3.
Answer: (A) 3
We find the pattern of powers of 2 mod 7: 2^1≡2, 2^2≡4, 2^3≡1 (mod 7). The cycle repeats every 3 terms. Since 100 = 33×3 + 1, we have 2^100 ≡ 2^1 ≡ 2 (mod 7).
For n ≡ 3 (mod 7): possible numbers are 3, 10, 17, 24, 31, 38, 45, 52, 59... For n ≡ 5 (mod 11): possible numbers are 5, 16, 27, 38, 49, 60... Common number is 58. Check: 58 = 7(8) + 2... Let me recheck: 58/7 = 8 rem 2, not 3. Try 38: 38/7 = 5 rem 3 ✓, 38/11 = 3 rem 5 ✓. Answer is A=38.
20 = 2² × 5. Divisors are: 1, 2, 4, 5, 10, 20. Sum = 1 + 2 + 4 + 5 + 10 + 20 = 42.
Number of divisors = (3+1)(2+1)(1+1) = 4 × 3 × 2 = 24.
We know that GCD × LCM = Product of the two numbers. So 4 × LCM = 120. Therefore, LCM = 120/4 = 30.
144 = 2⁴ × 3². Odd divisors come only from 3² = (2+1) = 3 odd divisors: 1, 3, 9.
Check 22: 22÷9 = 2 rem 4 ✓, 22÷7 = 3 rem 1 ✗. Check 31: 31÷9 = 3 rem 4 ✓, 31÷7 = 4 rem 3 ✓. Answer is 31 (B, not A). Correction: Option B is correct.
