Math Is Fun Forum

I was bored so I searched Anna's old posts for some more mentally challenging puzzles, and I found this. Allow me to present my solution.

So two perfect squares differ by 60. The difference between two perfect squares n² and (n+m)² is  2mn+m². Consider then all possible values of n and m such that 2mn+m²=60. NB: (i) Obviously m has to be even. (ii) We can also assume WLOG than both m and n are positive.

From here on 60−m², and hence n, will always be negative. Hence two are only three such integers, which are

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How about this?

Also

Substituting from [2]:

[1] + [3] ⇒

But we also have

Substituting [5] into [4]:

Finally

[6] & [7] ⇒

Only #1 is correct.

I will give you a hint for #1 and #2 (you already have a hint for  #3). Given a batch of 1000 parts, 700 of them will be from Supplier A and 300 will be from Supplier B. Of the 700 from Supplier A, 14 will be defective, and of the 300 from Supplier B, 15 will be defective. So there will be 29 defective parts altogether. So...

#1. What's the probability of picking one of the 29 defective parts in a batch of 1000?

#2. Of the 29 defective parts, 14 are from Supplier A and 15 are from Supplier B. What is the probability of picking one of the 14 Supplier A parts from the defective bunch?

On the subject of palindromic numbers, note the following:

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These are successive powers of n+1, where n is our base; they're palindromic up to some point depending on n. For base 10 (n=10) the palindromicity breaks down at the fifth power, when carrying-over starts to take over:

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To continue the palindromic sequence, all you need to do is choose a higher base (n>10). For example, consider the powers of 17 in hexadecimal:

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Ta da! Whereas (10+1)⁵ fails to be palindromic in base 10, (16+1)⁵ is palindromic in base 16. Choose n > 20 and (n+1)⁶ will be palindromic in base n too.

I have given you a big enough hint already!

Are you familiar with the fact that a complex number ζ is real if and only if it is equal to its complex conjugate, i.e.

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167 was for if Skippy could make 0 or 1 backward move. However, if Skippy has to make exactly 1 backward move, then 146. I wasn't sure what the wording of the puzzle meant.

What did I leave out? My logic is as follows.

First consider what happens when Skippy didn't make a backward move. To do 7 metres, he can skip 1 metre and then do 6 metres, or skip 2 metres and then do 5 metres. To do 6 metres, he can skip 1 metre and then do 5 metres, or skip 2 metres and then do 4 metres. And so on. So to compute how many ways to do 7 metres without moving backwards, I first compute the number of ways to do 1 and 2 metres 2, then 3 metres, then 4 metres, and so on, up to 7 metres. It turns out to be a case Fibonacci adding: if f(n) is the number of ways to do n metres, then:

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with f(1) = 1 and f(2) = 2.

Now for when Skippy can move backwards. He can do this when he's 1m, 2m, 3m, 4m, 5m or 6m from the starting point. Suppose he skips backwards at m metres from the starting point. He can do the first m metres in f(m) ways; after the backward skip, he has (8−m) metres left to do. Hence, if g(m) is the number of ways to get to the bowl skipping backwards at the m-metre mark, then

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where m = 1, 2, 3, 4, 5 or 6.

Hence the answer to the puzzle is

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This is assuming Skippy can choose not to skip backwards; if he has to make exactly one backward skip, then the answer is

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