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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?

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    $\begingroup$ 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 $9$. $\endgroup$ Jun 22 at 7:22
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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 $P$.

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  • $\begingroup$ What about 1994? That works. $\endgroup$
    – Tyrcnex
    Jun 22 at 7:49
  • $\begingroup$ You are right. I have edited and completed the solution. $\endgroup$
    – Ashiq
    Jun 22 at 8:06
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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

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

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