UW Math Challenge of the Week Winner
April 19th, 2008 by erb
The UW Math Department offers a weekly challenge, with a small prize picked each week. A couple weeks ago, they picked me, much to my surprise :-).
While I only do these for fun, and the intent is for the prize to go to Undergrads, then Grad students, and finally Faculty, it was nice to be rewarded with a win. Last week’s problem, for which I won a Ben & Jerry’s gift certificate, was a Cryptarithm combined with an April Fool’s “Gotcha!” — Although I only caught the “gotcha” after doing the work and realizing the solution to the first part, the cryptarithm, was not unique.
Here’s this week’s challenge: Consider the following cryptarithm:
APRIL – FOOLS = DAY
Each letter A, P, R, I, L, F, O, S, D, and Y stands for a different digit (0-9).
After solving the cryptarithm, let UVW = |DAY – YAD| (so UVW is some 3-digit number, letting U and V be leading 0’s, if necessary).
Here’s the problem: what is UVW + WVU?
The real key is realizing that the APRIL FOOLS DAY cryptarithm is a RED HERRING. The problem requested is the solution to UVW + WVU, requiring knowledge of UVW and therefore DAY. But:
As I said, I first solved APRIL FOOLS DAY to find a solution, then realized afterwards that S and Y are interchangeable, meaning the solution I found wasn’t unique:
(I,L,S,Y)=(8,1,4,7): (10*8+1)–4=77, Y=7. DAY=367. UVW=396. UVW+WVU=1089.
But also:
(I,L,S,Y)=(8,1,7,4): (10*8+1)–7=74, Y=4. DAY=364. UVW=099. UVW+WVU=1089.
Then I “got” it. Albeit a little late. Here’s the better solution:
If D>Y : |DAY–YAD|=100D+10A+Y–100Y–10A–D = 99(D–Y)
In general: |DAY–YAD|=99|D–Y|
With single digits 0..9, let Ψ=|D–Y| < 10
99 Ψ = 90 Ψ + 9 Ψ = (90+9,180+18,270+27,…,720+72,810+81) = (99,198,198,297,396,495,594,693,792,891) = UVW
where
U = δ – 1
V = 9
W = 10 – δ
for δ = [1..9]
Therefore UVW + WVU = 101(U+W) + 20V = 101 (δ – 1 + 10 – δ) + 20 (9) = 101 (9) + 20 (9) = 121 (9) = 1089
… Independent of the actual values for D, A, Y.
The UW team presented a somewhat simpler derivation, obviously same conclusion. They mention there were actually six cryptarithm solutions.
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