When Ben added $2$ of $4$ together repeatedly, he got $4$ consecutive numbers. Show that either $3$ or $1$ of his $4$ numbers were even

Question

Ben had $4$ numbers: $a, b, c,$ and $d$.

$a+c=e$

$a+d=f$

$b+c=g$

$b+d=h$

$e,f,g,h$ are consecutive. Prove that either $1$ or $3$ of $a,b,c,d$ were even

What is the source of this question? It has come up a couple times recently, as in http://math.stackexchange.com/questions/1800038/dr-math-and-his-family-question-how-to-solve-without-trial-and-error — lulu, May 26 '16 at 10:22

score 1 · Answer 1 · answered May 26 '16 at 09:53

1

Hint: Try to derive that $d = c +1$ and $b = a+2$. Use that to get the conclusion.

answered May 26 '16 at 09:53

Marc

and I can use this in reverse? – SuperNinja741 Does Gaming May 26 '16 at 09:55
1

I'm sorry, I don't understand what you are saying. – Marc May 26 '16 at 09:57

score 1 · Accepted Answer · answered May 26 '16 at 10:41

1

Let sum all this expressions: $$ (a+c)+(a+d)+(b+c)+(b+d)=e+f+g+h $$ $$ 2(a+b+c+d)=(n+1)+(n+2)+(n+3)+(n+4) $$ $$ a+b+c+d=2n+5 $$

So, the sum of this numbers is odd. Therefore, 1 or 3 ot them is odd and 3 or 1, respectevly, ot them is even.

answered May 26 '16 at 10:41

bigant146

How did you get from the above line to bottom? – SuperNinja741 Does Gaming May 26 '16 at 11:21
$e$, $f$, $g$, $h$ are consecutive, so they are permutation of $n+1$, $n+2$, $n+3$, $n+4$ for some n – bigant146 May 26 '16 at 11:25

score 0 · Answer 3 · answered May 26 '16 at 10:33

c~d=1 or 2 or 3 ,

a~b=1 or 2 or 3 ,

and c~d not equal to a~b since efgh are consecutive.

So, if c~d=1 , then 1 one them is even and other odd.

and a~b can't be 1 or 2 since c~d =1 (c and d become consecutive).

Hence a~d =3 ,then 1 one them is even and other odd.

Hence 2 of them are even , 2 of them odd , Thus "1 or 3 of a,b,c,d were even" can't be possible in all cases.

Note- '~' for difference between the two values i.e magnitude value .

3 Answers3