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Six Puzzles

Friday, February 25th, 2011 08:25 / 2 Comments

Assume that the English letters are digits (from 0 to 9 ) and they satisfy the given relations, then you have to solve each equation for these letters.

For Example:
$ABCDE \times ABCDE = FDBABCDE$ can have a solution:
$09376 \times 09376 = 87909376$

Similarly, Try these:

1. $ABCDEEABCD$ $\times$ $FEC$ = $AAAAAAAAAAAA$

2. $(A+B+C+D+E)$ $\times$ $(A+B+C+D+E)$ $\times$ $(A+B+C+D+E)$ = $ABCDE$

3. $6 \times ABCDEF =DEFABC$

4. $ABCDEABCDEABCDE$ = $C \times CCCCCGGGGH \times ABCDE$

5. $ABCDEABCDE$ $= FF \times GHGF \times 86485$

Solution:

1. $8547008547 \times 104$ $=888888888888$
2. $(1+9+6+8+3) \times (1+9+6+8+3) \times (1+9+6+8+3)$ $=19683$
3. $6 \times 14857$ $857142$
4. $283512835128351$ $=3 \times 3333366667 \times 28351$
5. $8648586485$ $=11 \times 9091 \times 86485$

Puzzle Idea: Mr. Sawinder Singh, Gurdaspur, Punjab (INDIA)
Note:
There may be many other solutions for these puzzles too.

Dirichlet’s Theorem and Liouville’s Extension of Dirichlet’s Theorem

Thursday, February 24th, 2011 17:30 / Leave a Comment

Topic

Beta & Gamma functions

Statement

$\int \, \int \, \int_{V} \, x^{l-1} y^{m-1} z^{n-1} dx \, dy \,dz = \frac { \Gamma {(l)} \Gamma {(m)} \Gamma {(n)} }{ \Gamma{(l+m+n+1)} }$
where V is the region given by $x \ge 0, y \ge 0, z \ge 0, \, x+y+z \le 1$ .