Question 1192305: On her birthday in 2007, Rachel's age is equal to twice the sum of the digits of the
year in which she was born. How many possible years are there in which she could
have been born?
Found 2 solutions by htmentor, greenestamps:
Answer by htmentor(1343)   (Show Source):
You can put this solution on YOUR website! First, let's consider the possibility that Rachel was born in the 2000's. Then we can express the year she was born as 2000 + x. Thus, her age A = 2007 - (2000 + x)
If twice the sum of the digits = her age, we can write A = 2(2 + x).
Thus, we have two equations A = 4 + 2x and A = 7 - x.
Multiplying the 2nd equation by 2, and adding (1) and (2) we get:
3A = 18 -> A = 6.
Now, we consider if Rachel is born in the year 19xy.
Then we can express the year she was born as 1900 + 10x + y, and her age would be:
A = 2007 - (1900 + 10x + y)
And the equation for twice the sum of the digits is:
2(1 + 9 + x + y) = A
Simplifying we have 107 - 10x - y = A and 20 + 2x + 2y = A
The expression for x in terms of A, which we get by multiplying (1) by 2 and adding, is: A = 78 - 6x
Since x can take on the values [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], we have these possible ages: [78, 72, 66, 60, 54, 48, 42, 36, 30, 24].
This gives possible birth years [1929, 1935, 1941, 1947, 1953, 1959, 1965, 1971, 1977, 1983].
Of these, only 1959, 1965 and 1971 have the correct x value.
So, the answer is 1959, 1965, 1971 and 2001.
Answer by greenestamps(13200)  (Show Source):
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