1. A=the sum of the digits of the largest 4-digit palindromic number that is divisible by 15.
2. Two sides of a triangle have lengths 4cm and 5cm. The third side has length Xcm, where X is an integer. B=the number of different values that X could take.
3. How many different arrangements are there of the letters in the word GEOCACHE? C=the fourth digit in your answer.
4. If X-1/X = Y-1/Y and X and Y are different, D=X*Y where * means multiply.
5. Becky and Adam are 51 miles apart and cycle towards each other. Starting at noon, Becky cycled at X mph. Starting at 2pm, Adam cycled at Y mph. They meet at 4pm. If they had both started at noon, they would have met at 2.50pm. E=2*Y.
1: 5775 (15*385) so A=24.
2: 2cm to 8cm so B=7.
4: I tried Wolfram: X=1-, Y=1 or vice versa, so C=-1.
5: You can form a pair of simultaneous equations. I got E=21.
In case you're wondering, there are 30 puzzles in this series. 1 to 20 I managed without needing any assistance. However, the questions get harder as you progress. 21 to 24 are with (AllExpert) colleagues of yours, though I am yet to get a reply. I will continue to endeavour to solve what I can for myself.
#1: You are correct. The two other palindromes divisible by 15 are 5115 and 5445.
There are ways to figure this out that include using the "rule of threes" which is that the digits must sum to be divisible by 3, which means if your palindrome is abba, then 2(a+b) is divisible by 3, and so a+b must be divisible by 3. But it's also divisible by 5, so it must end in a 5 (can't end in 0 since then it would only be three digits).
#2: You are correct again. The angle between the two sides given (4 and 5) could be anything, but if that angle is θ, then according to the law of cosines the third side must be:
c = √(a^2 + b^2 - 2ab cos(θ) )
and this expression reaches its larges value at θ=0 and its least value at θ=180. Those triangles are not actually possible (they correspond to laying the two sides of the triangle either overlapping or end-to-end). At those points, c=9 or c=1, so all the integers between (2 through 8) are allowed.
#5: You're right again. The first equation is 4x+2y=51, while the second is (17/6)x+(17/6)y=51.
#4: You're right, but here's an easier way to figure that out:
x - 1/x = y - 1/y
x - 1/x - y + 1/y = 0
(x^2 y - y - xy^2 + x)/xy = 0
(x-y)(1+xy)/xy = 0
This is true if x=y or 1+xy=0. Now the question says x and y are different, so that doesn't work, leaving you with 1+xy=0. It's not hard to see in that case that xy = -1.
Notice! You can't solve for x and y separately! It's lucky that "xy" appeared when we tried to solve the equation, because we could not have solved for x and y and then determined D by multiplying them together.
#3: There is actually a fairly standard way to count these types of rearrangements, called "permutations." You can read more about them in many websites, including:
One way I can explain it really quickly is first let's assume EVERY letter is different in your word. For example, let's say the "word" is really 12345678 = GEOCACHE, so that all the letters are just numbered. I can rearrange 12345678 right?
Well, how many choices can I make for the first letter? 8 choices.
No matter what I choose, I now have 7 choices for my next letter.
I have six choices next. Then five, then four, etc.
The total number of choices is 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40320.
We often denote the product 8! (or in general, n! where n could be 8 or 9 or whatever else).
However, this is NOT quite right. When I do a permutation like:
51278346 = AGEHEOCC
this could be the same as 51278364 (swapping the Cs), or 51872346 (swapping the Es), or 51872364 (swapping both). Every rearrangement of 12345678 really comes in a set of four that are equivalent to some rearrangement of GEOCACHE.
Mathematically, what that means is really this:
Start with 8! (or n! for any number of letters).
For each repeated letter, divide out by k!, where k is the number of times it is repeated.
In your case, that means 8! / (2! 2!) = 40320 / 4 = 10080.
The fourth digit of that number is 8.
Notice that it might sound weird that we are treating the repeated letters different than the non-repeated letters, but that's not entirely true. Technically, we can give all of those guys their own 1! too, but 1! = 1, so we can ignore those factors.