Entrance Exams
Govt. Exams
Factors of 18: 1, 2, 3, 6, 9, 18.
Factors of 27: 1, 3, 9, 27.
Common factors: 1, 3, 9.
Highest common factor = 9.
Using the formula: HCF × LCM = Product of two numbers.
Therefore, 12 × LCM = 2160, so LCM = 2160 ÷ 12 = 180.
Prime factorization of 360: 360 = 2³ × 3² × 5.
The unique prime factors are 2, 3, and 5.
Product = 2 × 3 × 5 = 30.
Let the number be n.
Given: n = 8k + 5 for some integer k.
When divided by 4: n = 8k + 5 = 4(2k) + 4 + 1 = 4(2k + 1) + 1.
Therefore, remainder = 1.
100 = 2² × 5².
Sum of all divisors = (1+2+4)(1+5+25) = 7 × 31 = 217.
Sum excluding 100 = 217 - 100 = 117.
First 20 multiples of 3: 3, 6, 9, ..., 60.
This is AP with a=3, d=3, n=20, l=60.
Sum = n(a+l)/2 = 20(3+60)/2 = 20×63/2 = 10×63 = 630.
Let the three consecutive odd numbers be x, x+2, x+4.
Sum: x + (x+2) + (x+4) = 147. 3x + 6 = 147. 3x = 141. x = 47.
The three numbers are 47, 49, 51.
Largest = 51.
Wait: let me recalculate. 3x + 6 = 147 means 3x = 141, x = 47.
So numbers are 47, 49, 51.
But option shows 53.
Let me verify: 47+49+51 = 147.
So largest is 51, which is option C.
Last digits of powers of 3: 3¹=3, 3²=9, 3³=27(7), 3⁴=81(1), 3⁵=243(3)...
Pattern: 3,9,7,1 repeats every 4. 2023 = 4×505 + 3, so 3^2023 has same last digit as 3³, which is 7.
Units place: 7, 17, 27, 37, 47, 57, 67, 77, 87, 97 (10 times).
Tens place: 70, 71, 72, 73, 74, 75, 76, 77, 78, 79 (10 times).
Total = 20 (note: 77 contains two 7s).
We need LCM(12, 15, 20). 12 = 2² × 3, 15 = 3 × 5, 20 = 2² × 5. LCM = 2² × 3 × 5 = 4 × 3 × 5 = 60.