1
$\begingroup$

Numeric puzzle $$\text{Happy} - \text{new} - \text{year} = 2018$$ where each letter corresponds to a single digit. How many solutions does this equation have?

$\endgroup$
  • $\begingroup$ Do you want an "elegant" solution, or a solution based on "brute force" ? $\endgroup$ – Peter Dec 25 '17 at 15:27
  • $\begingroup$ any right solution) $\endgroup$ – Arbron Dec 25 '17 at 15:35
  • $\begingroup$ For a start, note that $H=1$, $y$ has to be pretty large and $a$ has to be pretty small. There are only a few $(a,y)$ cases to consider, so pick one and see how many ways there are to fill in the other numbers. You will probably see a pattern as you go along. $\endgroup$ – Ross Millikan Dec 25 '17 at 15:38
  • $\begingroup$ @RossMillikan, what rules out the possibility $H=0$? $\endgroup$ – Barry Cipra Dec 25 '17 at 15:45
  • 1
    $\begingroup$ @BarryCipra: In puzzles like this we do not allow leading zeros. At least that was the rule I have seen followed. $\endgroup$ – Ross Millikan Dec 25 '17 at 16:01
1
$\begingroup$

First letter H=1, then second letter a=0 or 2. If a=0 then y=8 and if a=2 then y=9. So only five variants for "Happy" exist: 10558, 10668, 10778, 10998, 12009. For each of them exist two variants: 10558 - 234 - 8306 = 2018,

10558 - 236 - 8304 = 2018,

10668 - 243 - 8407 = 2018,

10668 - 247 - 8403 = 2018,

10778 - 254 - 8506 = 2018,

10778 - 256 - 8504 = 2018,

10998 - 274 - 8706 = 2018,

10998 - 276 - 8704 = 2018,

12009 - 364 - 9627 = 2018,

12009 - 367 - 9624 = 2018

There are 10 solutions for the puzzle.

$\endgroup$
1
$\begingroup$

With PARI/GP , I found these $10$ solutions :

? q=0;forvec(z=vector(8,j,[0,9]),if(length(Set(z))==8,[a,e,h,n,p,r,w,y]=z;if(h*n
*y>0,if(h*10000+a*1000+p*100+p*10+y-n*100-10*e-w-y*1000-e*100-a*10-r==2018,q=q+1
;print(q,"   ",z)))))
1   [0, 3, 1, 2, 5, 4, 6, 8]
2   [0, 3, 1, 2, 5, 6, 4, 8]
3   [0, 4, 1, 2, 6, 3, 7, 8]
4   [0, 4, 1, 2, 6, 7, 3, 8]
5   [0, 5, 1, 2, 7, 4, 6, 8]
6   [0, 5, 1, 2, 7, 6, 4, 8]
7   [0, 7, 1, 2, 9, 4, 6, 8]
8   [0, 7, 1, 2, 9, 6, 4, 8]
9   [2, 6, 1, 3, 0, 4, 7, 9]
10   [2, 6, 1, 3, 0, 7, 4, 9]
?
$\endgroup$
0
$\begingroup$

This is just to record a possibility overlooked by the other answers. If you allow $H=0$, then there is at least one more solution:

$$09117-000-7099=2018$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.