A Yes No Puzzle

Posted on Friday, October 7th, 2011 by Gaurav Tiwari

This is not just math, but a very good test for linguistic reasoning. If you are serious about this test and think that you’ve a sharp [at least average] brain then read the statement (only) below –summarize it –find the conclusion and then answer that whether summary of the statement is Yes or No.
[And if you're not serious about the test ...then read the whole post to know what the stupid author was trying to tell you.

]
STATEMENT: If the question you answered before you answered the question you answered after you answered the question you answered before you answered this one, was harder than the question you answered after you answered the question you answered before you answered this one, was the question you answered before you answered this one harder than this one? YES or NO?

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The problem of the Hundred Fowls

Posted on Saturday, October 1st, 2011 by Gaurav Tiwari

This is a popular Chinese problem, on Linear Diophantine equations, which in wording seems as a puzzle or riddle. However, when used algebraic notations, it looks obvious. The problems states :

If a cock is worth 5 coins, a hen 3 coins, and three chickens together 1 coin, how many cocks, hens and chickens, totaling 100 in number, can be bought for 100 coins?

This puzzle in terms of algebraic equations can be written as $5x+3y+\frac{1}{3}z=100$ and $x+y+z=100$
where $x, y, z$ being the number of cocks, hens and chicks respectively.
We find that there are two equations with three unknown quantities. So eliminating one of the unknowns, by putting $z=100-x-y$ from second equation into first one such that $5x+3y+\frac{1}{3} (100-x-y)=100$
or, $15x+9y+100-x-y=300$
or, $14x+8y=200$
or, $7x+4y=100$ .
Which is a linear Diophantine equation (with only two unknown quantities).
The equation $7x+4y=100$ has the general solution [links to WolframAlpha] $x=4 t$ and $y=25-7t$ , so that $z=75+3t$ where $t$ is an arbitrary integer.
Now, since $x, y, z$ are the number of creatures, hence $x, y, z >0$ and thus $4t >0$ , $25-7t >0$ and $75+3t >0$ which imply that $0 < t < 3\frac{4}{7}$ . And because t must have integer values, we have $t=1,2,3$ . Which gives the following three solutions:

Values of $t$	No. Of cocks ( $x=4 t$ )	No. Of hens ( $y=25-7t$ )	No. Of chicks ( $z=75+3t$ )
1	4	18	78
2	8	11	81
3	12	4	84

So there are the three ways to chose the number of cocks, hens and chicken totaling 100 to buy for 100 coins.

Problem Sources:
Elementary Number Theory
David M. Burton, 2006
McGrawHill Publications

Wikipedia article on Diophantine Equations

Image Credit

Six Puzzles

Posted on Friday, February 25th, 2011 by Gaurav Tiwari

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. $8648586485$ $=11 \times 9091 \times 86485$