# A Yes No Puzzle

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. ]

# The problem of the Hundred Fowls

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

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.