Enigma 1306 - Three all

Enigma 1306 - Three all New Scientist magazine, 11 September 2004. by Adrian Somerfield.

I have in mind three numbers each of three digits (no leading zero) in each of which one digit is 3. Of the following statements about them, three are true and three are false. What is the sum of the three numbers?

(a) The number is a prime. (b) The number is (appropriately) a cube. (c) The middle digit is the average of the other two digits. (d) The third digit differs from the second by 3. (e) The number has as a factor a two-digit prime the difference of whose digits is 3, or whose sum is a cube. (f) The number belongs (appropriately) to the set of triangular numbers 1, 3, 6, 10, 15, 21...

Ciao,

As a guy who won't be able to solve this, I thought I'd say:

I'll be more interested in reading the steps that someone takes to solve this than I will in reading a reply that simply lists the three

This seems slightly 'forced' compared to the usual Enigma number puzzles. Also the wording is a bit odd, in that I find myself having to guess what is meant: the problem being that if we have three numbers A B C, then 'The number is a prime' is not 'a statement about them'.

I will assume that what is meant is that given the three numbers A B C and the six predicates p1 ('is a prime', p2 ('is a cube', etc, exactly three of p1(A), p2(A), ..., p6(A) are true, AND exactly three of p1(, ... p6( are true, AND exactly three of p1(C), ... p6(C) are true.

Furthermore I will assume the obvious grammatical clarification of p5 to 'has as a factor a two-digit prime with digits that differ by 3 or sum to a cube'. This means the valid two digit primes for p5 are 17, 41, 47, 53, 71

A B C are drawn from { 3xy where x, y <> 3 } u { x3y where x <> 0, 3 and y <> 3 } u { xy3 where x<>0, 3, y <> 3 }, a set of size 225.

Swift Excel work demonstrates that there are exactly three numbers that satisfy exactly three of the predicates:

136 meets p4 p5 p6 369 meets p3 p4 p5 630 meets p3 p4 p6

And their sum is 1135.

SPOILER

(a) must be false for all, otherwise (b), (c), (e), and (f) would be all false (b) must be false for all, as the only cube with 3 in it, 343, meets no other conditions (c) Only numbers are 345 357 369 135 234 333 432 531 630 543 753 963 (f) Only numbers are 136 153 231 253 300 325 351 378 435 630 703 903 For each number either (c) or (f) must be true. Examining the numbers listed under (c) and (f) reveals the solution:

136 d,e,f 369 c,d,e 630 c,d,f

Sum is 1135

Please reply to drgmayer at hotmail dot com

Ilan Mayer schrieb:

Eschewing the switch to brute force, I want to continue using logic (assume you're solving the puzzle on the train, without computing machines):

If c) and d) are both true, since one digit is 3, and leading zero is disallowed, the number must be 369, 963 or 630.

369 is not a triangular number, but it is 9*41, which fulfils e). Hence _369_ fulfils c), d) and e), and is part of the solution.

963 is not a triangular number, and it factors to 9*107; 107 is not a 2-digit prime, so that's strike 4. 963 is out.

630 is a triangular number, and it factors to 10*9*7, which means it fulfils c), d) and f) only. _630_ is part of the solution.

Note: the test for triangularity of n boils down to checking floor(sqrt(2n))*ceiling(sqrt(2n)) = 2n.

For the other two numbers, c) and d) can't both be true, so one of them must be true, and e) and f) must jointly be true as well.

A triangular number, multiplied by 2, will produce a rectangle with sides n and n+1; for the number to be in the 100-2000 range, 10 <= n < 45. Also n must be a prime fulfilling e), or a multiple of such a prime.

Primes in that range are 11, 13, 17, 19, 23, 19, 31, 37, 41 and 43. Of these, only 17 and 41 satisfy condition e). n could also be 2*17 = 34.

Generating triangular numbers from these primes, the following products automatically satisfy both e) and f):

16*17/2 = 8*17 = 136 => d)

Basically, I could stop here, because that's the third number, but let's check the rest:

17*18/2 = 17*9 = 153 33*34/2 = 33*17 = 561 34*35/2 = 17*35 = 595 40*41/2 = 20*41 = 820 41*42/2 = 41*21 = 861

Only 136 has a '3' digit and satisfies either c) or d).

The solution is 369+630+136=1135.

Ok, 20/20 hindsight is a nice thing - it wasn't easy to not resort to brute force, which probably got the job done faster!

Cheers

From the wording of the question is it totally clear that the same three statements are true about all three numbers?

cheers

DDEckerslyke schrieb:

It is unclear; but from the facts, there can't be the same three statements true for all numbers, because then there wouldn't be three numbers fulfilling them.

Unless... has anyone checked negative numbers?