How many 2-digit numbers are divisible by 3 and what is their sum?

There are 30 two-digit numbers divisible by 3, and their sum is 1665.

Here's a detailed breakdown:

Finding the Two-Digit Numbers Divisible by 3

• The smallest two-digit number divisible by 3 is 12 (3 x 4).
• The largest two-digit number divisible by 3 is 99 (3 x 33).
• The two-digit numbers divisible by 3 form an arithmetic sequence: 12, 15, 18,... 99.
• To find how many numbers are in this sequence, we can use the formula for the nth term of an arithmetic sequence: an = a1 + (n - 1)d.
  • Where:
    • an = the last term (99)
    • a1 = the first term (12)
    • n = the number of terms (what we want to find)
    • d = the common difference (3)
• Therefore: 99 = 12 + (n - 1) * 3
  • Solving for n:
    • 99 - 12 = (n - 1) * 3
    • 87 = (n - 1) * 3
    • 87 / 3 = n - 1
    • 29 = n - 1
    • n = 30

Thus, there are 30 two-digit numbers divisible by 3.

Finding the Sum of the Two-Digit Numbers Divisible by 3

• We can find the sum of an arithmetic sequence using the formula: S = (n/2) * (a1 + an)

  • Where:
    • S is the sum
    • n is the number of terms
    • a1 is the first term
    • an is the last term

• Applying the values: S = (30/2) * (12 + 99)

• S = 15 * 111

• S = 1665

Therefore, the sum of all two-digit numbers divisible by 3 is 1665.

As stated in the reference, "Let it contain n terms. Then 12 + n - 1 x 3 = 99 or n = 30. Required sum = 30/2 x 12+99 = 1665." This result confirms our calculations.