Wikipedia:Reference desk/Archives/Mathematics/2007 October 7

= October 7 =

French Franc 1965
Shall appreciate if anyone can tell me today's value in Euros of 15,000 1965 French Francs. Info needed to illustrate history of a small village.86.197.151.8 08:52, 7 October 2007 (UTC)petitmichel


 * I have moved this question to the Humanities section, here. --Lambiam 12:57, 7 October 2007 (UTC)

Thanks. Just for interest the equivalent sum in 2006 (old francs) would have been approx. 115,500 francs! Appreciation of 100,000! Whow! (Yes, this is a crude figure, but sufficient for my needs.86.200.4.62 14:33, 7 October 2007 (UTC)petitmichel

Proof question
Hello. I just did a question that I would like someone to check for me. It would be much appreciated.

The positive integers can be split into five distinct arithmetic progressions, as shown:

A: 1, 6, 11, 16 B: 2, 7, 12, 17 C: 3, 8, 13, 18 D: 4, 9, 14, 19 E: 5, 10, 15, 20

a) Write down an expression for the value of the general term in each progression. b) Hence prove that the sum of any term in B and any term in C is a term in E. c) Prove that the square of every term in B is a term in D. d) State and prove a similar claim about the square of every term in C.

a)

$$ A=5n-4 $$

$$ B=5n-3 $$

$$ C=5n-2 $$

$$ D=5n-1 $$

$$ E=5n $$

b)

\begin{align}

B + C &= (5m-3) + (5m-2) \\ &= (5m+5m)+(-2-3)\\ &= 10m-5\\ &=5(2m-1)\\ \end{align} $$

Since this number is multiple of 5, it is a term in E.

c)

\begin{align} B^2 &= (5n-3)^2\\ &= 25n^2-30n+9\\ \end{align} $$



\begin{align}

25n^2 &\equiv 0 \text{ (mod } 5)\\ -30n &\equiv 0 \text{ (mod } 5)\\ 9 &\equiv 4 \text{ (mod } 5)\\ \implies 25n^2-30n+9 &\equiv 4 \text{ (mod } 5)\\

\end{align} $$



\begin{align} D&=5n-1\\ 5n &\equiv 0 \text{ (mod } 5)\\ -1 &\equiv 4 \text{ (mod } 5)\\ \implies 5n-1&\equiv 4 \text{ (mod } 5)\\ \end{align} $$

Therefore since both are congruent to 4 (mod 5) it is true.

d)

\begin{align} C^2 &= (5n-2)^2\\ &= 25n^2-20n+4\\ \end{align} $$



\begin{align} 25n^2 &\equiv 0 \text{ (mod } 5)\\ -20n &\equiv 0 \text{ (mod } 5)\\ 4 &\equiv 4 \text{ (mod } 5)\\ \implies 25n^2-20n+4&\equiv 4 \text{ (mod } 5)\\ \end{align} $$

So, as proved above, the square of any term in C is also a term in D. —Preceding unsigned comment added by 172.200.78.35 (talk) 15:08, 7 October 2007 (UTC)


 * a) and b) are good. b) would be even better if you remark that $$2m-1$$ must be a positive integer (for example, $$5(-1) = -5$$ is not in E, and you need to mention that this doesn't happen here).
 * In c) and d), you have the right idea, but the way you have written it doesn't really prove the needed result. You say "every square of an element of B is congruent to 4 modulo 5, and every element of D is congruent to 4 modulo 5, hence every square of an element of B is is an element of D", which is a fallacy (for example, every man breathes, and every woman breathes, but not every man is a woman). What you should have shown is the converse, that every number congruent to 4 modulo 5 is an element of D. However, I think this just complicates matters; it is easier to just say that $$(5n-3)^2 = 5(5n^2-6n+2) - 1$$ and show that $$5n^2-6n+2$$ is a positive integer, and hence this is in D. The same goes for part d). -- Meni Rosenfeld (talk) 15:25, 7 October 2007 (UTC)

Thank you Meni, I'll take that on board. 172.200.78.35 15:33, 7 October 2007 (UTC)

Just an after thought, would my problem with b) and your solution to c) and d) have been solved if I said n>o? 172.200.78.35 15:38, 7 October 2007 (UTC)
 * First, let me correct myself; there is a relatively serious error in your solution to b). You have an element of B and an element of C; You know that the former is some multiple of 5 minus 3, and and latter is some multiple of 5 minus 2. But it doesn't have to be the same multiple. So your numbers are not $$5m-3$$ and $$5m-2$$ but rather $$5n-3$$ and $$5m-2$$, where n and m are positive integers which may or may not be the same.
 * Now, stating that n (and m) is positive is just the start. From a) we learn that:
 * If n is a positive integer, then $$5n-4$$ is in A, and so on;
 * If some number is in A, then there is some positive integer $$n$$ such that the number is equal to $$5n-4$$, and so on.
 * So a correct solution to b) would be: Let x be in B and y be in C. Then there are some positive integers n and m such that $$x=5n-3$$ and $$y=5m-2$$. So $$x+y = 5(m+n-1)$$, and $$m+n-1$$ is a positive integer. Therefore, $$x+y$$ is in E. -- Meni Rosenfeld (talk) 16:15, 7 October 2007 (UTC)