# Let $x$ be the year that is $P$ years before $2017$. If the sum of the digits of $x$ is equal to $P$, how many such $P$ exist?

I have this question that I found in a past paper. I’ve solved it, but would there be a more efficient way to solve this? Here is the question:

Let $$x$$ be the year that is $$P$$ years before $$2017$$. If the sum of the digits of $$x$$ is equal to $$P$$, how many such $$P$$ exist?

I have derived a few things, most of them pretty obvious:

1. $$P$$ and $$x$$ are integers.
2. $$P$$ must be less than or equal to 28, because the year with the greatest sum of digits is 1999, with a digit sum of 28
3. Because of 2., $$x$$ has a range of 1 to 28.
4. $$x$$ can only be a 4-digit number.

Now, all I did was to start from 1 and go all the way back to 28 years, substituting values for $$x$$ and $$P$$. I feel like this is very inefficient. Is there a better way to do this?

• One thing that probably helps is the observation that $x \equiv P (\text{mod} \ 9),$ so $x + P = 2017 \to 2x \equiv 1 (\text{mod}\ 9) \to x, P \equiv 5 (\text{mod} \ 9).$ This should reduce the number of required checks by about a factor of $9$. Jun 22 at 7:22

## 2 Answers

Let $$x=1000a_1+100a_2+10a_3+a_4$$

$$x=2017-P$$

$$1000a_1+100a_2+10a_3+a_4=2017-(a_1+a_2+a_3+a_4)$$

$$1001a_1+101a_2+11a_3+2a_4=2017$$

Since $$a_1, a_2, a_3, a_4\in \{0,1,2,3,4,5,6,7,8,9\}$$, we conclude that $$a_1\leq2$$.

Taking $$a_1=2$$, means that $$101a_2+11a_3+2a_4=15$$. This further gives $$a_2=0$$ so that $$11a_3+2a_4=15$$.

Now, $$11a_3+2a_4=15$$ has only one possible solution $$a_3=1, a_4=2$$.

Taking $$a_1=1$$, means that $$101a_2+11a_3+2a_4=1016$$. This further gives $$a_2=9$$ (otherwise $$11a_3+2a_4>117$$) so that $$11a_3+2a_4=107$$.

Now, $$11a_3+2a_4=107$$ has only one possible solution $$a_3=9, a_4=4$$. Note that if $$a_3<9$$, we must have $$a_4>18$$ which is not possible.

Thus, $$x\in\{2012,1994\}$$ and $$P\in\{5,23\}$$.

Consequently, there are only $$2$$ possible values of $$P$$.

• What about 1994? That works. Jun 22 at 7:49
• You are right. I have edited and completed the solution. Jun 22 at 8:06

You've identified the possible years as being in the form $$19ab$$ or $$20ab$$.

If the year is $$19ab$$ then that has a value of $$1900 + 10a +b$$.

You can create two equations:

First $$P=2017-(1900+10a+b) \Rightarrow P=117-10a-b$$

Then $$P=1 + 9 + a + b \Rightarrow P=10+a+b$$

Combining those gives $$10+a+b=117-10a-b$$

$$2b=107-11a$$

$$b=\frac{107-11a}2$$

For this to work, $$a$$ must be odd.

$$a=9 \Rightarrow b=4$$

$$a=7 \Rightarrow b=15$$ - too large

$$a=5, 3, 1$$ will fail similarly.

$$a=9 \Rightarrow b=4$$

If the year is $$20ab$$ then that has a value of $$2000 + 10a +b$$.

You can create two equations:

First $$P=2017-(2000+10a+b) \Rightarrow P=17-10a-b$$

Then $$P=2 + 0 + a + b \Rightarrow P=2+a+b$$

Combining those gives $$2+a+b=17-10a-b$$

$$2b=15-11a$$

$$b=\frac{15-11a}2$$

For this to work, $$a$$ must be odd.

$$a=1 \Rightarrow b=2$$

$$a=3 \Rightarrow b<0$$ , similarly for $$a=5, 7, 9$$.

That gives you two possible answers: 1994 and 2012

• Thank you for your answer! I feel like this answer is the simplest and most easy to understand, so I have marked it correct. Jun 23 at 3:29