Problem 591

The layer sum of $wxyz$ equals 2014.

This means that the sum of the integer with digits $wxyz$, the integer with digits $xyz$, the integer with digits $yz$, and the integer $z$ is 2014.

Note that the integer with digits $wxyz$ equals $1000w+100x+10y+z$, the integer with digits $xyz$ equals $100x+10y+z$, and the integer with digits $yz$ equals $10y+z$.

Therefore, we have $$(1000w+100x+10y+z)+(100x+10y+z)+(10y+z)+z = 2014$$ or $$1000w+200x+30y+4z = 2014 \quad (*)$$ Each of $w,x,y,z$ is a single digit and $w \neq 0$.

Now $w$ cannot be 3 or greater, or the left side of $(*)$ would be at least 3000, which is too large. Thus, $w=1$ or $w=2$.

If $w=2$, then $2000+200x+30y+4z=2014$ and so $200x+30y+4z=14$ or $100x+15y+2z=7$. This would mean that $x=y=0$ (since otherwise the terms $100x+15y$ would contribute more than 7), which gives $2z=7$ which has no integer solutions. Thus, $w \neq 2$.

Therefore, $w=1$.

This gives $1000+200x+30y+4z=2014$ and so $200x+30y+4z=1014$ or $100x+15y+2z=507$.

Since $0 \leq y \leq 9$ and $0 \leq z \leq 9$, then $0 \leq 15y + 2z \leq 15(9)+2(9)=153$.

Since $100x$ is a multiple of 100 and $0 \leq 15y+2z \leq 153$, then $100x=400$ or $100x=500$ so $15y+2z=507 - 400 = 107$ or $15y+2z= 507 - 500 = 7$. From above, we saw that $15y+2z$ cannot equal 7, so $15y+2z=107$, which means that $100x=400$ or $x=4$.

Thus, $15y+2z=107$.

Since $2z$ is even, then $15y$ must be odd to make $15y+2z$ odd.

The odd multiples of 15 less than 107 are $15, 45, 75, 105$.

Since $0 \leq 2z \leq 18$, then we must have $15y=105$ or $y=7$. This gives $2z=2$ or $z=1$.

Therefore, the integer $wxyz$ is $1471$. (Checking, $1471+471+71+1 = 2014$.)

Finally, $w+x+y+z=1+4+7+1=13$.