SOLUTION: 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 bo

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): You can put this solution on YOUR website!

First suppose she is very young and was born after 2000.

The year she was born can be represented as 2000+B.
Her age in 2007 was 2007-(2000+B) = 7-B.
The sum of the digits of the year she was born is 2+B.
Her age is twice the sum of the digits:

So one possibility is that she was born in 2001. The sum of the digits of that year is 3, and her age is 2007-2001 = 6.

Now look for possibilities that she was born before 2000.

The year she was born (19AB) can be represented as 1900+10A+B.
Her age in 2007 was 2007-(1900+10A+B) = 107-10A-B.
The sum of the digits of the year she was born is 1+9+A+B = 10+A+B.
Her age is twice the sum of the digits:

A and B are non-negative single-digit integers. There are three solutions, easily found by inspection.

(1) A=7, B=1. She was born in 1971; her age is 2007-1971=36; the sum of the digits of her birth year is 1+9+7+1 = 18.

(2) A=6, B=5. She was born in 1965; her age is 2007-1965=42; the sum of the digits of her birth year is 1+9+6+5 = 21.

(3) A=5, B=9. She was born in 1959; her age is 2007-1959=48; the sum of the digits of her birth year is 1+9+5+9 = 24.

ANSWER: 4 possible years for her birth (2001, 1971, 1965, and 1959)

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