1. ## algebra problem

Here is a quick mathematics problem that requires no PhD level calculations; just some high school algebra and a little ingenuity.

Take any prime number greater than 2. .... Square it. .... Subtract 1 from the result.
Prove that the final result is evenly divisible by 8.

So if P is our prime, prove that ( P^2 - 1 ) is divisible by 8.

2. Hint: P being prime is a red herring. This result is true for any odd number.

3. Ok, here's my question....how is this a game????

4. Math's my name, math's my game.

5. As MM said, any prime number greater than 2 is an odd number by definition.

So, (P^2 – 1) can be rewritten as ((2N+1)^2 – 1), where N is a positive integer.

(2N+1)^2 – 1 = 4N^2 + 4N = 4N(N+1)

Since N is a positive integer, either N or N+1 should be an even number, thus N(N+1) can be rewritten as 2M, where M is another positive integer.

So, 4N(N+1) = 4(2M) = voila! 8M.

After MM’s hint, it’s become so easy I feel guilty.

6. Originally Posted by Mathman
Math's my name, math's my game.

7. Originally Posted by lowtherlore
As MM said, any prime number greater than 2 is an odd number by definition.

So, (P^2 – 1) can be rewritten as ((2N+1)^2 – 1), where N is a positive integer.

(2N+1)^2 – 1 = 4N^2 + 4N = 4N(N+1)

Since N is a positive integer, either N or N+1 should be an even number, thus N(N+1) can be rewritten as 2M, where M is another positive integer.

So, 4N(N+1) = 4(2M) = voila! 8M.

After MM’s hint, it’s become so easy I feel guilty.
Very good.

8. Here is another quick mathematics problem that requires no PhD level calculations: Prove that the accounting imbalance due to a transposition error (an error caused by switching the position of two adjacent digits) is always divisible by 9 (e.g., \$5,400 accidentally recorded as \$4,500 will result a \$900 imbalance, which is divisible by 9). Or you may choose to answer this one: How can I prove to my cat that 1 + 1 = 2?

My suggestion: Whoever answers the above question(s) should give another math-related question, so it becomes a game like other threads that can go on and on.

9. Originally Posted by skatinginbc
Here is another quick mathematics problem that requires no PhD level calculations: Prove that the accounting imbalance due to a transposition error (an error caused by switching the position of two adjacent digits) is always divisible by 9 (e.g., \$5,400 accidentally recorded as \$4,500 will result a \$900 imbalance, which is divisible by 9). Or you may choose to answer this one: How can I prove to my cat that 1 + 1 = 2?

My suggestion: Whoever answers the above question(s) should give another math-related question, so it becomes a game like other threads that can go on and on.
Solution to the first problem:

X is the accounting number:
X = a*10^n + b*10^(n-1) + c (a and b are the adjacent digits)

Y is the number with a transposition error with a and b:
Y = b*10^n + a*10^(n-1) + c

The difference between X and Y is:

X – Y
= (a – b)*10^n + (b – a)*10^(n – 1)
= (a – b)*10*10^(n – 1) – (a – b)*10^(n – 1)
= (10*(a – b) – (a – b))*10^(n – 1)
= Voila! 9*(a – b)*10^(n – 1)

I have no clue on how to educate your cat.

OK, this is not algebra or a typical numerical series, but here’s another quiz:

1
1 1
1 2
1 1 2 1
1 2 2 1 1 1
1 1 2 2 1 3
1 2 2 2 1 1 3 1
( ? )

What’s to come in the eighth row?

10. Originally Posted by lowtherlore
OK, this is not algebra or a typical numerical series, but here’s another quiz:

1
1 1
1 2
1 1 2 1
1 2 2 1 1 1
1 1 2 2 1 3
1 2 2 2 1 1 3 1
( ? )

What’s to come in the eighth row?
My answer: 1 1 2 3 1 3 3 1 1 1
Why: I just picked an answer that seems to fit a pattern established by the previous series of numbers. I'm not sure if there is only one pattern and therefore not certain if there is only one answer. Let me know if my answer is not the "right" one.

Ok, here comes the next challenge (challenging for me to think of an interesting math question): Prove that the accounting imbalance caused by an error of reversal (e.g., \$1,000 credit was recorded as \$1,000 debit) is always divisible by 2.

11. You came so close I can’t tell whether you got it wrong or you just made a mistake in writing your answer.

To give a hint, and to make sure there’s only one pattern in the series, I’ll give the eight row.

1
1 1
1 2
1 1 2 1
1 2 2 1 1 1
1 1 2 2 1 3
1 2 2 2 1 1 3 1
1 1 2 3 1 2 3 1 1 1
( ? )

Now, what’s in the ninth row?

The numerical series appeared in Clifford Stoll’s The Cuckoo's Egg: Tracking a Spy Through the Maze of Computer Espionage and later in Bernard Werber’s Les Fourmis (The Ants).

Another hint: In its original form, each row of the series reads from right to left.

12. 1 2 2 1 3 1 1 1 2 1 3 1 1 3

Now, prove that the accounting imbalance caused by an error of reversal (e.g., \$1,000 credit was recorded as \$1,000 debit) is always divisible by 2. I know it's too easy, but it's just a game that everyone can play.

13. Originally Posted by skatinginbc
1 2 2 1 3 1 1 1 2 1 3 1 1 3

If \$ is the accounting entry (in whole dollar terms), then the absolute value of the discrepancy in the ledger balance caused by a wrong credit or debit entry is 2\$, thus divisible by two. Too simplified?

Another puzzle:

There are two integers, X and Y, between 2 and 20, inclusive, and two people, P and S.

P is told the value of the product X*Y.
S is told the value of the sum X+Y.
Neither of P, S is told the individual values of X, Y.

P and S have the following conversation:

S: I cannot determine X, Y.
P: I cannot determine X, Y.
P: In that case, I now know X, Y.
S: In that case, I too now know X, Y.

What are X, Y?

14. 4 and 13 look good to me.

Now here comes my puzzle challenge:
X denotes the total Olympic medals that Skater A and Skater B had won and Y the total World medals that the two skaters had won.
X * Y = 0
X + Y = 10
Who are the skaters?

15. Originally Posted by skatinginbc
4 and 13 look good to me.
Correct.

It would be perfect if you share the solution, as it has I think beautiful two-step logic in it.

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