MS_Kekas wrote: Four positive integers $a, b, c, d$ satisfy the condition: $a < b < c < d$. For what smallest possible value of $d$ could the following condition be true: the arithmetic mean of numbers $a, b, c$ is twice smaller than the arithmetic mean of numbers $a, b, c, d$? We have $\frac{a+b+c+d}4=2\frac{a+b+c}3$ and so $d=\frac53(a+b+c)\ge \frac 53(1+2+3)=10$ And indeed $\boxed{d=10}$ fits for example with $(1,2,3,10)$