[Guest]

On a remote island all of the natives belong to one of two tribes: the Brights, who are so brilliant at numerical calculations that they always get the correct answer, and the Braves, who bravely rush in to do calculations beyond their ability and never get the right answer. (The Braves are not entirely stupid: they can do simple counting and comparing of numbers, but they always get arithmetic calculations wrong.) Both Brights and Braves pride themselves on their complete honesty. They always tell the truth, or (in the case of Braves) at least what they believe to be the truth; they never purposely tell a lie (unlike the folks on some of those other islands).

One day a group of natives was playing a game of Numberskulls. There were 5 players and a moderator. The moderator, who was a Bright, painted a 3-digit number on each of the players' foreheads, so that each could see all numbers but their own. All 5 of the numbers were different. The moderator would ask them questions in turn about the numbers they could see, and from the answers they would try to deduce what number was on their own forehead. The first to do so was the winner. What follows is a record of the game, with questions omitted and players designated by letters.

(1) A: I see exactly 1 prime number.

(2) B: I see exactly 2 prime numbers.

(3) C: I see exactly 3 perfect squares.

(4) D: I see exactly 3 triangular numbers.

(5) E: I see exactly 3 perfect squares.

(6) A: I see exactly 3 numbers with a digital sum of 10.

(7) B: I see exactly 3 numbers whose square root is more than 25.

(8) C: I see exactly 0 numbers with a digital sum of 10.

(9) D: I see exactly 3 perfect cubes.

(10) E: I see exactly 0 numbers with a digital product of 18.

At this point one of the players announced his number and won. (Of course it was a Bright; for some reason Braves never win these games, a point of much amusement to the Brights!)

What number was on each player's forehead?

[Guest]

was A's number 1?

[superprismatic]

I take it that the numbers have no leading zeros. Am I correct?

[Guest]

I take it that the numbers have no leading zeros. Am I correct?

Yes. None of the numbers can contain any leading zero.

[Guest]

I was unable to find the solution, but I’ll post my work anyway. I suggest I’m missing something or misunderstanding something -- perhaps I'm making an assumption somewhere where I shouldn't.

I would think that since answer (10) gave away the game, that should be a hint that E is a bright. And if E is a Bright, then C is a Bright due to agreement of (3) and (5). And if C is a Bright, A is a Brave due to disagreement of (6) and (8). D’s statements are contradictory (since there are five total numbers, there cannot be both 3 numbers that are triangular numbers and simultaneously 3 numbers that are perect cubes since those numbers do not overlap between 100 and 999), so he apparently is not a bright.

(There are only five perfect three-digit cubes {125, 216, 343, 512, 729}, and the set of triangular numbers you can get from this website http://www.research.att.com/~njas/sequences/A000217.)

Taking only E and C’s information, there is not enough information to determine the solution, so I will assume B is a bright so that we can get the solution.

So I have

A = Brave, B = Bright, C=Bright, D=Brave, E=Bright.

The best hint I could derive was with the key statement (7) which provides six possible numbers for three of the possible three digit numbers of the five.

26 squared is 676

27 squared is 729

28 squared is 784

29 squared is 841

30 squared is 900

31 squared is 961

32 squared is some four digit number.

This agrees with C and E in their statements (3) and (5), so these are the range of the 3 perfect squares out of the set of 6 numbers I provided. Obviously none of these would be prime, so B’s other statement indicates that there are two prime numbers rounding out the other two numbers in the five.

Of the five numbers written on their heads, three are in the set above, and the other two are the large range of three-digit prime numbers. So taking just the statements we know are true:

(1) Brave – thus false

(2) --- there are two prime numbers (there are too many three digit primes to list them out)

(3) --- the three perfect squares (or just squares) are represented by the range given above.

(4) Brave – thus false

(5) --- see (3)

(6) Brave – thus false

(7) --- how I determined the range of squares.

(8) --- actually doesn’t eliminate any of the squares and only a scant few prime numbers (like 109). Doesn’t help as much as you would hope.

(9) Brave – thus false

(10) – even worse a clue than 8, as it hardly eliminates either the squares or the primes.

Maybe I am misunderstanding what a digital product or digital sum is? I assume it anumbrates that one should add the digits (digital sum) or to multiple the digits (digital product) in the number. So 109's digital sum is 10, and digital product is 0. Not sure I get what other information I can gather is. Maybe it has something to do with the fact that A and D were trying to get the right answer and we should be looking at where they made have made their calculation error?

Edited December 5, 2009 by ApologeticJedi

[Guest]

I gather of course all participants know what tribe the others belong to, right? So we don't know if they're brights or braves, but they do, so the winner knowing which number he has in his forehead is not necessarily E but can use the information provided by E (plus he knows whether E is a brave or a bright) to guess the correct answer.

Still working on this, but of course if E's information is a definite clue he must be a bright anyway, cause a brave's data is of course wrong.

[Guest]

Mmm don't think we have information enoughto provide as exact an answer as it's asked...

We assume that since E's answer provides the definite clue, and only clues from Brights are correct, so worth anything, E is a Bright. So there are 3 perfect squares and no number with a digital product of 18, which of all the 3-digit squares eliminates 19 squared, or 361.

C agrees with E in that there are 3 perfect squares, so he's also a Brave. Alas, his other clue doesn't help much with the numbers we have so far (except for 361 which we already took out, there are no other perfect squares for which the digital sum is 10).

Since there are no coincidences in the list of 3-digit perfect cubes or triangular numbers, none of the propositions of D are correct (as there are 5 players, at least one perfect square should also appear in the list of cubes or triangular numbers), so D is a Brave.

Next, we know from Bright C that there are no numbers with a digital sum of 10, so A is not correct when he sees 3 of them... So he's a Brave, and his other statement, there is one prime number, is also false.

There's not enough information to determine which tribe does B belong to. He could be a Bright because in the list of 3-digit perfect squares there are indeed 3 numbers (more) with square roots higher than 25. If he's a Bright, his other statement holds, and the other 2 numbers that are not perfect squares of numbers at or above 25, are prime numbers. But. There are a lot of perfect squares with square roots below 25, so he could also be a Brave and we wouldn't have any more information on the other two numbers. Personally, I'll cheat and assume that he's a Bright, since he's answers then help us a narrow a little more our options.

So basically we do know that 3 of the numbers are in this list: 625, 676, 729, 784, 841, 900 and 961, which are the 3-digit perfect squares that result from operating with numbers at or above 25. The other 2 numbers are primes, so they're in this list (no menial task, to narrow down this one, though I've taken out the ones with a digital sum of 10, of course, and those with a digital product of 18): 101, 103, 107, 113, 131, 137, 149, 151, 157, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 239, 241, 251, 257, 263, 269, 277, 281, 283, 293, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 617, 619, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997.

So we'd have to guess 3 out of 6 perfect squares and 2 out of 131...

So yep, it seems I fully agree with apologeticjedi :-)

Edited December 5, 2009 by RoteFreiherr

[superprismatic]

Great problem! Hard, but fun to work. Thanks, KS!

A 631 Bright
B 136 Brave
C 233 Brave
D 361 Brave
E 343 Brave

Consider A. If he is a Bright, then he correctly
computes that one of the numbers he sees is prime.
However, If he is a Brave, he must be presented with
3 primes (which he calculates to be composite) and
1 composit (which he calculates to be prime).

Now consider B. If he is a Bright, he correctly
computes that two of the numbers he sees are prime.
If, however, he is a Brave, then he must be presented
with 2 composits (which he calculates to be prime)
and 2 primes (which he calculates to be composit).
So, whether he's a Bright or a Brave, we may conclude
that he sees 2 primes and 2 composits.

Arguing in this way, we can make a table of what
each person sees in both the case he is a Bright
and the case he is a Brave:

Bright Brave
1 A 1 Prime 3 Primes
2 B 2 Primes 2 Primes
3 C 3 Squares 1 Square
4 D 3 Triangular 1 Triangular
5 E 3 Squares 1 Square
6 A 3 Dig Sum of 10 1 Dig Sum of 10
7 B 3 numbers > 625 1 number > 625
8 C 0 Dig Sum of 10 4 Dig Sum of 10
9 D 3 Cubes 1 Cube
10 E 0 Dig Prod of 18 4 Dig Prod of 18

Line 1 tells us that there are 2 primes no matter
what B is.
Ergo, at least 2 of the numbers are prime

Lines 4 and 9 tell us that D is a Brave because
Triangular numbers and Cubic numbers have no overlap
here and there must be at least one overlap if he were
actually presented with 3 of each.
Ergo (line 4), at least 1 number is triangular
Ergo (line 9), at least 1 number is a cube

Now, there cannot be 3 squares because the Ergos
above account for 4 of the 5 numbers and squares
and triangles don't overlap here, although squares
and cubes overlap at 729. In any case, no person
could see more than 2 squares.
Ergo (line 3&5), at least 1 number is a square
Ergo (line 8), at least 4 numbers have Dig Sum of 10
Ergo (line 10), at least 4 numbers have Dig Prod of 18

So, A must see at least 3 numbers with a Dig Sum of 10.
Ergo (line 6), A is a Bright

Now because A sees 3 of the 4 numbers with Dig Sum of
10, He must have a number with a Dig Sum of 10 on his
head. Also, since A sees 1 prime and B sees 2, A
must have a prime on his head.

We know by the Ergo for line 10 that A,B,C, and D all
have numbers whose Dig Prod is 18. Of squares, cubes,
and triangular numbers in the range of this problem's
numbers only 2 have a Dig Prod of 18 they are
361 (a square) and 136 (a triangular number). So,
both primes must have a digital product of 18. All
of these primes are 163, 233, 613, and 631. So, from
line 7, we have that it is impossible to get 3 numbers
greater than 625.
Ergo (line 7), B is a Brave

So, the only Bright is A!

To summarize, the 5 numbers are 2 primes, 1 triangular
number, 1 square, and 1 cube. The cube may also be a
square, but it would have to be 729. Here are the
properties of each person's number:
A prime, Dig Sum of 10, Dig Prod of 18
B not prime, Dig Sum of 10, Dig Prod of 18
C Dig Prod of 18, Dig Sum not 10
D not triangular, Dig Sum of 10, Dig Prod of 18
E Dig Sum 10
So,
A possibilities are 163, 613, and 631
B possibilities are 136, 361
C 233 (Dig Sum 18 => not a cube,
No triangular or square has required Dig Sum
and Dig Prod, so must be a prime & only one
w/required Dig Sum & Dig Prod)
D 361 only square with appropriate properties
(reason as in C above) => B is 136
E 343 the only cube with Dig Sum of 10
So, finally, since line 7 has B seeing one number
greater than 625, A must be 631 and we have
A 631 Bright
B 136 Brave
C 233 Brave
D 361 Brave
E 343 Brave
A wins and used lines 2,7,8, and 10 to deduce his
number of 631.