# Wreckage of a ship

The ancestor of the unfamous Boaty McBoatface was wrecked in the Atlantic ocean during the 20th century, the $X$th day of the $Y$th month of the year 1900 + $Z$. It had $U$ propellers but no more than 10, $V$ funels but no more than 10 and there were $W$ crew members.

The product $U*V*W*X*Y*Z$ added to $C$ is equal to 4002331, $C$ being the cube root of the age of the captain of the ship, who was a grandfather.

How many crew members were on the ship?

## 4 Answers

Answer:

101 crew members.

Reasoning:

The age of the captain is 64
Other possible ages would be:
1,8 - you cannot be a grandfather then.
27 - if you were unlucky and your kid was unlucky you might be a grandfather at 27, but I doubt that.
125 and above - Nobody lived that long

This means that $U \times V \times W \times X \times Y \times Z = 4002331 - 4 = 4002327$
$4002327 = 3 * 3 * 7 * 17 * 37 * 101$
There are 6 prime factors in 4002327 and six integer variables.
This means that each letter corresponds to one of the prime factors.
101 fits only for $W$ (crew members) since 101 cannot be a day or a month or the last 2 digits of a year (or 100, because ...yeah...the year 2000 was in the 20th century), or the number of propellers or the number of funnels because they are less than 10.

All values:

Year: $1900 + Z = 1937$
Crew members: $101$
Day of the month: $17^{th}$
Propellers, funnels and month: any combination of $3,3,7$.

• Are you saying it's bad luck to have kids? +1 anyway, good explanation! – TonioElGringo Jul 22 '16 at 8:30
• @TonioElGringo. It's kind of back luck to have kids at 14....just saying. – Marius Jul 22 '16 at 8:50

It could also be

the first of January 1901 in a boat with 1 propeller and 1 funnel and 4002327 crew members (that would be a really big boat tho)

• Ha ha. The question says "$U$ propellers". There is an "s" there that makes it plural. But it could work for day, month and year. – Marius Jul 21 '16 at 13:04
• Lol'd. That works but yes, I was looking for a realistic solution :) – IAmInPLS Jul 21 '16 at 13:32
• Or the ship had 0 propellers and an infinite number of crew members. – Marius Jul 21 '16 at 13:42
• Good point. Even barring the unlikely large crew, there are plausible solutions that use "1". If the plural on propellers and funnels does not imply more than one, then either of these could be 1. And if someone says, e.g., "how many children do you have", few would say that an answer of "1" is some sort of paradox because "children" is plural. The month could certainly be January. – Jay Jul 22 '16 at 4:43

The answer is:

101 Crew Members

Each number is:

C = 4 U = 3 V = 3 W = 101 X = 17 Y = 7 Z = 37 3*3*101*17*7*37+4 = 4002331

My reasoning is:

The age of the captain has to be 64, which is 4 cubed, because 25 is too young to be a grandfather, and 125 is too old to live. The other numbers are just the 6 factors of 4002327, which I found by trial and error on my calculator, though there are probably more efficient ways of doing it!

Edit: Sorry I didn't put my answer into a spoiler, this was my first time on the Puzzling site. I figured it out for the future though!

Edit #2: added my reasoning.

• WHat is your reasoning? – Mithical Jul 21 '16 at 17:08
• And Please use spoiler tags while answering..(>!) – Sid Jul 21 '16 at 17:08

The amount of crew members is any of the following

101 or 303 or 707 or 1717 or 3737 or 5151 or 11211 or 12019 or 26159 or 45481 or 63529 90962 or 190587 or 444703 or 500291 or 1000582

Some logical thinking:

C is either 3 or 4, since C = 2 would make the captain 8 years old (VERY young to be a grandfather, let alone have a licence to captain an ocean-going vessel) and C = 5 would let the captain be 125. Way too old to be alive.

All other variables have a known domain:

X = 1:31 (technically, months can be shorter, but the final result is identical if we limit ourselves to X = 1:28).
Y = 1:12.
Z = 1:100.
U = 2:10, since the question clearly talks about propellors (plural).
V = 2:10

Brute force! If it doesn't work, you don't use enough of it:

Matlab Code time!
clear all;close all;clc;

X = [1:31]'; % Definition of the arrays with all possibilities.
Y = [1:12]';
Z = [1:100]';
U = [2:10]';
V = [2:10]';
C = [3,4];

% Handle C
S = 4002331 - C; % Result is an array of both options for both choices of C

% Handle V
S = repmat(S,[length(V),1]);
V = repmat(V,[1,size(S,2)]);
S = S./V;
S = S(rem(S,1)==0)';

This code replicates the current solutions state for the length of the currently tested variable (in this case V), and does the same for V to have S and V both arrays of equal size. It divides both arrays element-wise. Only divisions which are integer are valid solutions. So we pick out those and use it as our new solutions state. We simply repeat it for all variables and end up with an array of possible solutions for W.

% Handle U
S = repmat(S,[length(U),1]);
U = repmat(U,[1,size(S,2)]);
S = S./U;
S = S(rem(S,1)==0)';

% Handle X
S = repmat(S,[length(X),1]);
X = repmat(X,[1,size(S,2)]);
S = S./X;
S = S(rem(S,1)==0)';

% Handle Y
S = repmat(S,[length(Y),1]);
Y = repmat(Y,[1,size(S,2)]);
S = S./Y;
S = S(rem(S,1)==0)';

% Handle Z
S = repmat(S,[length(Z),1]);
Z = repmat(Z,[1,size(S,2)]);
S = S./Z;
S = S(rem(S,1)==0)';

S = unique(S)

And it turns out that

If the domain of X is limited to 1:28, the resulting solution state S is exactly the same. So we can safely ignore all the hassle associated with dates like the 31st of February, since every month will always have at least 28 days.