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

Answer 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.

Answer 2

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

Answer 3

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 .