Fun math problem if you are bored.?

Let Mary's birth year be 10^3a + 10^2b + 10c + d

So her age on her birthday in 2016 wil be a + b + c + d

Let's approach this problem in terms of congruence modulo 9.

(In case you are not sure, the mathematics of congruence is the mathematics of remainders. Don't be put of by its seemingly trivial description. It is extremely important in number theory. In particular modulo 9 is very useful as the sum of the digits of a positive integer is equal to the remainder when that number is divided by 9)

Since 10^k ≡ 1 (mod 9) for all non-negative integers k

(Try it)

10^3a + 10^2b + 10c + d ≡ a + b + c + d (mod 9)

Also 2016 = (10^3 * 2) + (10^2 * 0) + (10 * 1) + 6 ≡ 2 + 0 + 1 + 6 ≡ 9 ≡ 0 (mod 9)

That is

2016 - (10^3a + 10^2b + 10c + d) ≡ a + b + c + d (mod 9)

0 - (a + b + c + d) ≡ a + b + c + d (mod 9)

ie

- (a + b + c + d) ≡ a + b + c + d (mod 9)

or

2(a + b + c + d) ≡ 0 (mod 9)

Since gcd(9, 2) = 1, we can safely divide both sides of the congruence by 2 without affecting the congruence. So

(a + b + c + d) ≡ 0 (mod 9)

This means that her age is a multiple of 9. In other words

a + b + c + d = 9n for some positive integer n.

Now

2016 - 2007 = 9

and 2 + 0 + 0 + 7 = 9 *1 = 9

So 2007 is a likely candidate.

But so is 1989. For

2016 - 1989 = 27

and

1 + 9 + 8 + 9 = 9*3 = 27

There are no more solutions, since any birth year before that would make Mary over 28 years old and the maximum sum of the digits you can have is 2 + 9 + 9 + 9 = 29 which, not only is not divisible by 9 but the year 2999 is somewhat later in the future.

Now you have to ask yourself: "Is 9 years of age too old for a doll?" Well, I don't know, but 27 certainly is, unless Mary is into collectable dolls, which we can assume to the contrary.

Good question, though.

Hello again.

Thanks for the answers. Best answer goes to Tomp for his mastery of congruence and the completeness of the answer.

Now if you are more into basic maths, here is how this could have been solved:

► Mary, being an human being, has a life expectancy of approximatively 100 years. So she is either born in this century (20AB) or the previous one (19AB).

► Considering this century:

   2 + 0 + A + B = 2016 – (2000 + 10A + B)

   2 + A + B = 16 – 10A – B

   11A + 2B = 14

Now 14 is even, and obviously 2B too. So 11A must also be even, implying the digit A must be even:

   A in { 0; 2; 4; 6; 8 }

Since any A greater than 1 would make B negative, the only possibility is

   A = 0

Leading to

   B = 7

With a birth year of 2007 and age of 9 in 2016.

► Considering the previous century:

   1 + 9 + A + B = 2016 – (1900 + 10A + B)

   10 + A + B = 116 – 10A – B

   11A + 2B = 106

Now 106 is even, and obviously 2B too. So 11A must also be even, implying the digit A must be even:

   A in { 0; 2; 4; 6; 8 }

If A=8 → 11A=88 and 106–88=18 → B=9

If A=6 → 11A=66 and 106–66=40 → B=20 ◄Impossible B is a digit

If A=4 → 11A=44 and 106–44=62 → B=31 ◄Impossible B is a digit

...and decreasing A will only increase B, leading to non-digit values of B.

So the only possibility is

   A = 8 and B=9

With a birth year of 1989 and age of 27 in 2016.

► Lastly you are told it is inappropriate to gift Mary a doll. Tomp's answer is thorough:

"Now you have to ask yourself: "Is 9 years of age too old for a doll?" Well, I don't know, but 27 certainly is, unless Mary is into collectable dolls, which we can assume to the contrary."

Thus the expected answer was 1989, reachable with basic maths and common sense.

Regards,

Dragon.Jade :-)

2007

1989 or 2007