-My tens and units digits are consecutive integers whose product is 2/3 of my hundreds digit.
-My units digit is 1/3 of my hundreds digit.
-The sum of my digits id equal to the sum of the first five counting numbers.
What year am I?
2 * 3 = 6 = (2/3) 9
Assuming the year has a thousands digit, it must be 1 to sum to 15 = 1 + 2 + 3 + 4 + 5.
1923
To find the answer, we'll break down the given information and look for a pattern. Let's start step by step:
1. "My tens and units digits are consecutive integers whose product is 2/3 of my hundreds digit."
Let's assume the tens digit is 'x' and the units digit is 'x+1'. According to this information, the product of 'x' and 'x+1' is equal to 2/3 of the hundreds digit. Mathematically, we can write this as: x * (x+1) = (2/3) * (hundreds digit).
2. "My units digit is 1/3 of my hundreds digit."
This tells us that the units digit is equal to 1/3 of the hundreds digit. Mathematically, we can write this as: x+1 = (1/3) * (hundreds digit).
3. "The sum of my digits is equal to the sum of the first five counting numbers."
The sum of the digits is x + (x+1) + hundreds digit. According to this information, this sum is equal to 1 + 2 + 3 + 4 + 5 = 15.
Now, let's solve these equations simultaneously:
From equation 2, we can rewrite it as: hundreds digit = 3 * (x+1).
Substituting this value of hundreds digit into equation 1:
x * (x+1) = (2/3) * 3 * (x+1).
x^2 + x = 2(x+1).
x^2 + x = 2x + 2.
x^2 - x - 2 = 0.
Factoring this quadratic equation:
(x - 2)(x + 1) = 0.
This gives two possible values for x: x = 2 or x = -1.
Since we are dealing with a year, the tens and units digits cannot be negative. Therefore, we'll consider x = 2.
From equation 2, we can find the value of the hundreds digit:
hundreds digit = 3 * (x + 1) = 3 * (2 + 1) = 9.
The year will be given by the hundreds, tens, and units digits:
Year = hundreds digit * 100 + tens digit * 10 + units digit = 9 * 100 + 2 * 10 + (2 + 1) = 902 + 2 + 1 = 905.
Therefore, the year is 905.