Question 466771
Question is a little vague, Im assuming based on the category that the weights need to be consecutive even/odd integers.
Lets call n the first and smallest weight.
The next consecutive even/odd integer would have to be (n+2).
1+2=3, 2+2=4, etc.
Thus we can say the next number is always 2 more than the previous number.
So the 4 weights would look like this,
n, n+2, n+4, n+6
Now set their sum equal to 40 and solve for n.
{{{n + (n+2) + (n+4) + (n+6)= 40}}}
Combine like terms
{{{4n + 12 = 40}}}
Subtract 12 on both sides
{{{4n = 28}}}
Divide by 4 on both sides
{{{n = 7}}}
Solution
7,9,11,13