<s>[INST] For $a+b+c+d=16$, are there always four integers $x, y, z, w$ that satisfy some conditions including $10\mid x+2y+3z+4w$? 
For $a, b, c, d \in \Bbb{N}, a+b+c+d=16$, do four integers $x, y, z, w$ always exist that satisfy the following three conditions?

*

*$\ 0\leq x\leq a, \ 0\leq y\leq b, \ 0\leq z\leq c, \ 0\leq w\leq d.$

*$\ x+y+z+w=10.$

*$\ 10\mid x+2y+3z+4w.$

This was from the Math Competition in Korea, which ended today.
My attempt:
\begin{align}
\text{if} \; x\geq 10 \text{ or } &y\geq 10 \text{ or } z\geq 10 \text{ or } w\geq 10: \\
x\geq10; \; & (x, y, z, w) = (10, 0, 0, 0) \\
y\geq10; \; & (x, y, z, w) = (0, 10, 0, 0) \\
z\geq10; \; & (x, y, z, w) = (0, 0, 10, 0) \\
w\geq10; \; & (x, y, z, w) = (0, 0, 0, 10) \\
\ \\
\end{align}
Then I tried all kinds of $(x, y, z, w)$s which satisfy conditions 2 and 3, and the results were like this...
\begin{align}
& (10, 0, 0, 0) \\
& (1, 8, 1, 0), (2, 6, 2, 0), (2, 7, 0, 1), (3, 4, 3, 0), (3, 5, 1, 1), (4, 2, 4, 0), (4, 3, 2,1), (4, 4, 0, 2), (5, 0, 5, 0), (5, 1, 3, 1), (5, 2, 1, 2), (6, 0, 2, 2), (6, 1, 0, 3), (0, 10, 0, 0) \\
& (0, 1, 8, 1), (0, 2, 6, 2), (0, 3, 4, 3), (0, 4, 2, 4), (0, 5, 0, 5), (1, 0, 7, 2), (1, 1, 5, 3), (1, 2, 3, 4), (1, 3, 1, 5), (2, 0, 4, 4), (2, 1, 2, 5), (2, 2, 0, 6), (3, 0, 1, 6), (0, 0, 10, 0) \\
& (0, 0, 0, 10)
\end{align}
What else should I do here?
 [/INST] zwim found by brute force that
no such $x,y,z,w$ exist for $(a,b,c,d)=(1,7,7,1)$.
Let me prove that fact by hand. Suppose integer $x,y,z,w$ satisfy condition 1 and 2 for $(a,b,c,d)=(1,7,7,1)$.
$x+2y+3z+4w=3(x+y+z+w)-2x-y+w\ge30-2\times1-7+0=21.$
$x+2y+3z+4w=2(x+y+z+w)-x+z+2w\le20-0+7+2\times1=29.$
The two inequalites above imply $x,y,z,w$ do not satisfy the condition 3.
</s>