Please can you help

[granny grump]

Question 1 - Far Far Away
Alpha Centauri, the closest star to the sun, is 4.3 light years away. A starship is traveling at 1516 times the speed of light. How long will it take for it to get from the sun to Alpha Centauri

Question 2 - Ear We Go
Yuri and his space dog Laika travel to a moon base. Before they have arrived, there were 52 ears at the base; after they arrived, there were 43 pairs of legs at the base. How many cosmonauts and how many space dogs are there now ate the base?

Question 3 - Prime Directive
Just before Jonthan and Jim step out for a space walk, Jean-Luc decides to give them a challenge:
'I'm thinking of a prime number less than 10,000. The product of it's digits is zero, and it's first digit is equal to the sum of the other digits.'
Jonathan replies that this is insufficient information to work out the number.
Jean-Luc adds:
'reversing the digits gives you a number which is divisible by exactly
three prime factors, two of which are 19 and 29.'
Jim states that is now enough information.
What number is Jean-Luc thinking of?

[factor-fiction]

Number of years = 4.3/1516
number of days = 365.25 x 4.3/1516 = 1.036

[factor-fiction]

There's now 15 dogs and 13 cosmonauts at the base

[granny grump]

Thank you factor30 - I am trying to help my grandson - can you explain how you got the answer - I don't want to give him the answer but explain how to work it out - I'm useless - he thinks his nan knows everrything but she doesn't!

[Fibonacci]

q3 2011
Since product of digits is zero one of the digits (number < 10000) must be 0
the number reversed is a mutiple of 551 (19x29)
551 x 2= 1102 2011 - is prime and 2 = 0+1+1

[Fibonacci]

q2 How many cosmonauts and how many space dogs are there now ATE the base? - serious space problem too much time in space!!

originally
let number of cosmonauts = c
let number of dogs = d

before arrival
2c + 2d = 52
therefore c + d = 26 (1)

after arrival
2(c + 1) + 4(d+1) = 86 (43 x 2 legs)
2c + 2 + 4d + 4 = 86
2c + 4d = 80
Therefore c + 2d = 40 (2)

Comparing (1) & (2) gives d = 14 and c =12
Now there is one more of each
So cosmonauts = 13 and dogs = 15

[granny grump]

Thank you fibonacci

[factor-fiction]

For 2 I used a simultaneous equations.
But I started with the fact that there were 52 pairs of ears so the number of cosmonauts plus the number of dogs was 26.


After they land there are C men and D dogs
We also know each man has 2 legs and each dog has 4 legs
We know that afterwards there are 43 pairs of legs- that equals 86 legs.
So: the number of legs is 2C +4D =86
We already know C+D = 28 (the original 26 plus the new man and dog)

So by doubling up we know 2C +2D = 56
So using the simultaneous equations we can see the difference of 2D = 86-56=30

So D =15
So C= 28-15=13

[factor-fiction]

Drat- took me 10 minutes to type it!

[granny grump]

Thank you for all your trouble factor30