Division!

Here’s another. Once again,

• First correct solution gets bragging rights.
```     PLI
WA|PRISK
WA
TWS
TIW
RNK
RSP
NL
```

Mike Schilling

Mike has been a software engineer far longer than he would like to admit. He has strong opinions on baseball, software, science fiction, comedy, contract bridge, and European history, any of which he's willing to share with almost no prompting whatsoever.

1. Is this in base 10? because something seems off

• Scratch that it was my mistake

• Mike Schilling

A base other than 10 is an interesting idea.

```      ABA
ABA|AABAB
ABA
AAB
ABA
A
```
• Jason Tank

Base two is a little obvious, though. π

• Mike Schilling

You could say that π

2. Solution:

C = 1
C = 1 —- (1)

be ryfr JN qvivqrq ol C pnaabg or JN

Gurersber
J > 5 — (2)
E < 9 — (3)
Va snpg,
E < J — (4)
Nyfb, tvira (1)
X- Y = 1 — (5)
F = 0 fvapr A – F = A — (6)

Tvira (6) naq (1)
A < 5

fvapr JF – VJ = EA

J – 1 – V = E — (7)

fvapr V k N = ?C
[V, N] = [3, 7] — (8)

vs V = 7, gura J – E = 8
Gura J = 9 naq E = 1
ubjrire, fvapr (1)
V = 3 — (9)
N = 7 — (10)

Fvapr (9) naq (7)
J – E = 4 — (11)
Fvapr CEV – JN -1 = GJ naq (11)
G = 5 — (12)
Gung yrnirf gur ahzoref
2, 4, 6, 8, 9

Tvira (5),

X = 9 naq — (13)
Y = 8 — (14)

Tvira (11),
J = 6 — (15) naq
E = 2 — (16)

naq gung yrnirf
A = 4 — (17)

• There is a typo in one of the lines

Fvapr CEV β JN -1 = GJ naq (11)

should be

Fvapr CEV β JN -10 = GJ naq (11)

but everything else remains the same

• Mike Schilling

Murali wins.

3. Jason Tank

I think Murali and I took similar paths.

Svefg bss, C vf boivbhfyl bar, fvapr JN gvzrf C vf vgfrys.

V gvzrf N rdhnyf fbzrguvat jvgu gur ynfg qvtvg bs bar. Gur bayl ahzoref gung unir fhpu n cebqhpg ner frira naq guerr, fb gung’f jung V naq N ner.

V – N rdhnyf J. J vf rvgure sbhe be fvk. Jr yngre frr gung J – V rdhnyf E. J qbrfa’g obeebj sebz gur G gb vgf yrsg, fb V vf yrff guna J. Gurersber, V zhfg or guerr, N zhfg or frira, juvpu va ghea zrnaf J vf fvk, naq E zhfg or gjb.

Ng gur obggbz, A – F rdhnyf A. A qbrfa’g obeebj sebz gur E gb gur yrsg, fb F zhfg or mreb. F – J lvryqf A, fb A vf sbhe.

(Ng guvf cbvag, V’z pregnva gur 0-9 jbeq zhfg or fcevag-jnyx, ohg V’yy xrrc ybbxvat zngurzngvpnyyl.)

X – C lvryqf Y. Fvapr C vf bar, X zhfg or Y + bar. Gur bayl ahzoref yrsg gung svyy guvf erdhverzrag ner rvtug naq avar, fb Y vf rvtug naq X vf avar.

Y gvzrf JN lvryqf GVJ, fb G unf gb or svir, bhe ynfg erznvavat ahzore.

• I don’t think that there are that many different ways to solve the problem. But let’s see how others get there

• Mike Schilling

I did almost exactly what Jason did.

• kenB

Well, since I’m at work and thus in a hurry, and the competition was already over, I used my customary strategy:

C – abguvat = abguvat, fb C = 1

X – 1 = Y, fb jbeq pbagnvaf “YX”

AX – FC = AY, fb F = 0 be 9 — ohg tvira gung C vf gur frpbaq yrggre, F vf cebonoyl gur svefg. (Be nygreangryl, sbe F gb or 9, X jbhyq unir gb or 0, juvpu jbhyq znxr Y unir gb or 9, ohg V jrag jvgu gur jbeq nccebnpu svefg).

Ybbxvat ng erznvavat yrggref, bayl gjb ibjryf, fb cebonoyl FCE…; tvira YX naq n J, cebonoyl JNYX. Rnfl sebz gurer gb trg “Fcevagjnyx”.

For the next one, you should mess with my head by having the answer not be a word.

4. I plan on doing this. For the record.

5. Bxnl, urer’f jurer V nz. V xabj gung Crr vf Bar naq Rff vf Mreb. V nyfb xabj gung qbhoyr-lbh vf rvgure fvk be sbhe.

Ohg V unira’g unq zhpu zber gvzr guna gung.

Zber fbba. V unir n CCI gbavtug.

6. C K JN = JN
Y K JN = GVJ
V K JN = EFC

Fb C vf gur tvzzvr.

Tvira gung JN K V raqf va 1, JN K V vf rvgure fbzr inevnag bs frira K guerr be avar K avar. Fvapr jr xabj gung N =/= V, jr xabj gung N be V vf frira naq gur bgure vf guerr. Naq tvira gung guvegrra zvahf frira vf fvk naq naq frira zvahf guerr vf sbhe, jr xabj gung J vf rvgure fvk be sbhe.

Naljnl, jurer’f Mreb?
Yrg’f ryvzvangr. Jr xabj gung abar bs gur gbc ahzoref ner mreb naq abar bs gur fvqr ahzoref ner 0 fb gung’f svir evtug gurer. (N *ZVTUG* unir orra bar, ohg V zvahf N vf abg V). Jr nyfb xabj gung G naq E naq A nera’g mreb rvgure. Gung yrnirf obgu F be X.

Naq gurer ner ab jbeqf gung ortva jvgu XC.

AY + EFC = EAX

Juvpu gryyf hf gung AY + FC vf yrff guna 100 (naq tvira gung FC rdhnyf *BAR*, jr xabj gung AY pbhyq or 98 ba qbja). Naq tvira gung jr xabj gung F vf mreb, jr xabj gung Y + C rdhnyf X

Fb Y + 1 = X

Urpx, sbe gung znggre, EA + VJ vf yrff guna 100 gbb.

Fb J + A = 10 naq jr xabj gung J vf rvgure fvk be sbhe fb A vf rvgure sbhe be fvk (naq pnaabg or nabgure ahzore)

Fb jr xabj gung Y vf abj rira orpnhfr jr xabj gung N vf bqq naq Y gvzrf N vf rira. Fb X vf bqq gbb. (Naq, sbe gung znggre, vf rvgure 5 be 9).

Fb jr xabj gung Y vf rvgure 2 be 8.

V nz gverq bs nyy bs gurfr “rvguref”.

Tvira gung jr xabj A + J erfhygf va n pneevrq bar, jr xabj gung E + V + 1 vf J (naq gurer’f ab pneevrq bar). Fb jr xabj gung E naq V pnaabg or 1 (tvira gung C vf 1) fb jr xabj gung E + V pnaabg = 3… fb J vf FVK! (bu, naq jr nyfb xabj gung V vf 7 be 3 naq vs abguvat vf pneevrq, jr xabj vg’f 3…)

Juvpu znxrf A rdhny sbhe.

Juvpu znxrf N rdhny frira.

Fb E + V unir gb rdhny svir, naq V vf guerr, jr xabj gung E vf 2.

Juvpu tvirf zr FCEVA-JN–

Tvira gung Y + 1 vf X, gurer’f bayl ebbz sbe gjb ahzoref arkg gb rnpu bgure ng gur raq, jr’ir tbg FCEVA-JNYX.

Naq gurer’f n G yrsg.

Fcevagjnyx.

Jurj.

V srry qhzo.

• Mike Schilling

Naljnl, jurerβf Mreb?

A – F = A, naq gurer’f ab pneel, orpnhfr gur E gb gur yrsg – E vf mreb.

• I wasn’t certain about the no carry. I did stare at that for a while, though.

• Mike Schilling

Anyway, nicely done.

By the way, one technique that can be helpful when you have a lot of “either M or N” is to make a table, and cross off columns when you can, something like (numbers and letters made up)

X: 3 5
Y: 5 3
Z: 8 2 (X – Y)
W:1 7 (X + Z)

“But I already know that A is 1, so the second column must be right.”

• Dude. I’ll do that next time.

And I will try to not rely on my knowledge of English Words.