Here’s another. Once again,

- rot13 your answers.
- Show your work.
- First correct solution gets bragging rights.

PLIWA|PRISKWATWSTIWRNKRSPNL

No politics or religion.

Here’s another. Once again,

- rot13 your answers.
- Show your work.
- First correct solution gets bragging rights.

PLIWA|PRISKWATWSTIWRNKRSPNL

Comments are closed.

Is this in base 10? because something seems off

Scratch that it was my mistake

A base other than 10 is an interesting idea.

Base two is a little obvious, though. π

You could say that π

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

Murali wins.

Brag Brag

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

I did almost exactly what Jason did.

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.

I plan on doing this. For the record.

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.

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.

Naljnl, jurerβf 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.

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.