To find how many numbers between 50 and 150 are divisible by 7, we identify the smallest and largest multiples of 7 in this range, then count them using the arithmetic sequence formula.
Step 1: Find the smallest multiple of 7 greater than 50
Divide 50 by 7:
The next whole number is 8, so the smallest multiple is:
Step 2: Find the largest multiple of 7 less than 150
Divide 150 by 7:
The largest whole number is 21, so the largest multiple is:
Step 3: Identify the arithmetic sequence
The multiples of 7 from 56 to 147 form an A.P. with:
- First term: \(a_1 = 56\)
- Common difference: \(d = 7\)
- Last term: \(a_n = 147\)
Step 4: Use the A.P. formula to find the count
For an A.P., \(a_n = a_1 + (n-1)d\)
Answer: There are 14 numbers between 50 and 150 divisible by 7 (Option A)
Let numbers be 3k and 5k. Since gcd(3,5)=1, LCM = 3k×5k/1 = 15k. Given LCM = 150, so 15k = 150, k = 10. Numbers are 30 and 50.
Using the property: GCD(a,b) × LCM(a,b) = a × b. Therefore: 12 × 144 = 36 × b. So 1728 = 36b, which gives b = 48.
Let number be 10a + b. Reversed number is 10b + a. Given: (10b + a) - (10a + b) = 27, so 9b - 9a = 27, thus b - a = 3. Also a + b = 9. Solving: b = 6, a = 3. Original number = 36.
Let three consecutive odd numbers be x, x+2, x+4. Their sum: x + (x+2) + (x+4) = 51. So 3x + 6 = 51, thus 3x = 45, and x = 15.
Prime factorizations: 24 = 2³ × 3, 36 = 2² × 3², 60 = 2² × 3 × 5. LCM = 2³ × 3² × 5 = 8 × 9 × 5 = 360.
If one number is 60 and product is 2160, then other number = 2160 ÷ 60 = 36. Sum = 60 + 36 = 96. Wait, let me verify GCD(60, 36) = 12. Yes, 12 is correct. Sum = 96.
Numbers divisible by both 4 and 6 are divisible by LCM(4,6) = 12. Numbers from 1 to 500 divisible by 12: ⌊500/12⌋ = 41.666..., so 41 numbers. Actually, ⌊500÷12⌋ = 41, but we need to check: 12 × 41 = 492. So there are 41 numbers. Let me recalculate: 500 ÷ 12 = 41.666, so answer is 41. Wait, the options suggest 42. Let me verify: counting from 12, 24, 36...492. That's 492/12 = 41. The correct count is 41, but closest option is 42.
Factoring: x² - 5x + 6 = (x - 2)(x - 3) = 0. Therefore x = 2 or x = 3.
[When a number is divisible by two coprime numbers, it must be divisible by their product.]
We are told that a number is divisible by both 9 and 11, meaning both divide evenly into this number with no remainder.
Two numbers are coprime if their greatest common divisor (GCD) is 1. Since 9 = 3² and 11 is prime, they share no common factors.
When a number is divisible by two coprime numbers, it must be divisible by their product (by the Fundamental Theorem of Arithmetic).
The answer is (B) 99.
