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_ _ _ _ _

CRABS -  0

EERIE - 0

SLEEP -  2

SLEPT - 2

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SPEAR

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SLABS if 2 then SL_ _ _ due to _ _ A B S being 0 in CRABS

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if SLABS is not 2  but Spear is then S ? E ? ? as there is no way for there to be 1 in _ _ _ E P in SLEEP and 1 in _ _ _ P T in SLEPT and 1 in _ _ _ A R in SPEAR at the same time.  This means there are 2 in S L E ? ?  and 2 in S P E ? ?  thus 2 in S ? E ? ?

_ _ _ _ _

CRABS -  0

EERIE - 0

SLEEP -  2

SLEPT - 2

SPEAR -  3

SLABS -  1

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_ E _ _ _ because:
1) Assume ---EP=1 then since SLEEP=1 it means SLE--=1 and since SLEPT = 2 it means --PT=1
2) Assume ---PT=1 similar to above it follows that ---EP=1
3) From #1&#2 we have ---EP=1 <=> ---PT=1 and either of those imply SLE--=1. However if ---EP=1 and ---PT=1 then the only two possibilities are ---ET=2 or ---PP=2.
4) Assume ad absurdum ---EP=1. Then SLE--=1 as above, therefore SPE-- can be at most 2 and at worst 0 (since there is only one letter change from SLE--=1). From SPEAR=3 it follows that ---AR must be at least 1. However this contradicts both possibilities in #3 above (---ET=2 or ---PP=2). Therefore ---EP is not 1, and by #3 ---PT is not 1.
5) Assume ---EP=2 then SLE--=0 therefore ---PT=2 which is a contradiction. Similarly ---PT cannot be 2.
6) From #4&#5 we find that ---EP=0 and ---PT=0 (not 1, not 2, cannot be greater than 2 letters). Therefore SLE--=2 from SLEEP=2.
7) SPEAR=3 so SPE-- is at least 1 since ---AR can be at most 2.
8) Assume -L---=1, then S-E--=1 from #6 and since SLABS=1 it follows that S-ABS=0 therefore --E--=1.
9) Assume -L---=0, then S-E--=2 from #6, therefore --E--=1
10) Regardless of -L--- being 1 or 0 from #8&#9 it follows that --E--=1

STEAM

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The above is - -E - -, copy paste error in the first sentence, correct in last sentence.

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_ _ E _ _

CRABS -  0

EERIE - 0

SLEEP -  2

SLEPT - 2

SPEAR -  3

SLABS -  1

STEAM -  3

+5 araver (I think   lol)

sheaf

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S---- because reasons

1) SLABS=1 but CRABS=0 so SL---=1
2) Assume S----=0 then from #1 it means -L---=1. But STEAM=3 and since obviously ST---=0 we get --EAM=3. But SPEAR=3 and S----=0 and ----R=0 (because ----M=1 from --EAM=3) therefore -P-EA-=3 which means -P---=1 which contradicts -L---=1.

S _ E _ _

CRABS -  0

EERIE - 0

SLEEP -  2

SLEPT - 2

SPEAR -  3

SLABS -  1

STEAM -  3

SHEAF -  2

+10 ARAVER

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CHARM

if 2 then RM are last.  Not C or A from CRABS - 0 and not H from SHEAF - 2; the established S and E. And from SPEAR - 3 the second letter would need to be P since we now know AR cannot be last: then SPERM

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SPERM

CRABS -  0

EERIE - 0

SLEEP -  2

SLEPT - 2

SPEAR -  3

SLABS -  1

STEAM -  3

SHEAF -  2

CHARM -  2 +5 pg

SPERM -  5 +15 pg

+10 ARAVER

maurice -  227 + 9 = 236

plainglazed -   162 + 20 = 182

phaze -  169
Logophobic - 143
araver -  64 + 10 = 74
nana - 10

_ _ _ _ _

QUOTE

TOPIC

_ _ _ _ _

QUOTE - 0
TOPIC - 0

BRUTE

FORCE

_ _ _ _ _

QUOTE - 0
TOPIC - 0
BRUTE - 0
FORCE - 0

SCANT

PEARS

_ _ _ _ _

QUOTE - 0
TOPIC - 0
BRUTE - 0
FORCE - 0
SCANT - 1
PEARS - 2

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SCARS if 3 then ---RS since scant =1 and two changes made

_ _ _ _ _

QUOTE - 0
TOPIC - 0
BRUTE - 0
FORCE - 0
SCANT - 1
PEARS - 2
SCARS - 1

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BRACE if 1 then --A-- since BRUTE=0 and FORCE=0.

Spoiler

Assume ---NT=0 then ---RS=0. From PEARS=2 and SCANT=1 we get PEA--=2 and SCA--=1. If S----=1 then --A--=0 since SCA--=1, but this implies PE---=2 from PEA--=2, which contradicts S----=1. So S----=0. Assume -C---=1 then --A--=0 since SCA--=1, but this implies PE---=2 from PEA--=2, which contradicts -C---=1. Therefore -C---=0, so only --A--=1 is possible from SCA--=1.
If ---NT is not 0 then one of ---N- and ----T and also one of ---R- and ----S and also one of P---- and -E---.

So, Occam made me bet on --A--, hence the guess.

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_ _ A _ _

QUOTE - 0
TOPIC - 0
BRUTE - 0
FORCE - 0
SCANT - 1
PEARS - 2
SCARS - 1
BRACE - 1 +5 araver

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PEARS = 2 &  SCARS = 1,  the 1 being A and not RS means P 1st or E 2nd

FEARS

if 3 then FE are 1st 2nd from above  (not RS)

If 2 THEN E is 2nd from above and now we know P is not first since PEARS = FEARS

if 1 then P is 1st since only change from PEARS = 2

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