# Math Help - Cryptic Messages

1. ## Cryptic Messages

Hiya, I'm having troulbe figureing these ones out. They're for extra credit...which I desperately need...

LKG WUAL LFVYOLAVZMGE ZPVG ZPL EQI BPU ZPMGSY PL SGUBY MZ VRR MY ZPL UGL BPU ALVRRI NULY
- VR CLAGY ZLMG

Below, each letter stands for a specific number. Once a number is used it cannot be used again. Different numbers must be assigned to each letter. There are no zeros in the most left column. Find the value of each letter. (1-9 are used.)

D O S
D O S
+ T R E S
------------------
S I E T E

Thanks so much!

2. Originally Posted by Chinchilla Babii
LKG WUAL LFVYOLAVZMGE ZPVG ZPL EQI BPU ZPMGSY PL SGUBY MZ VRR MY ZPL UGL BPU ALVRRI NULY
- VR CLAGY ZLMG

This is a mono-alphabetic substitution cypher, to get you started ZPL probably is THE, VRR=ALL, LFVYOLAVZMGE=EXASPERATING.

(and are the first three characters correct?)

RonL

3. ## typo

the first word is supposed to be LKLG...

4. Hello, Chinchilla Babii!

If you've never done crytograms, the first one is probably impossible,
. . especially with the typos.

LKLG WUAL LFVYOLAVZMGE ZPVG ZPL EQI BPU ZPMGSY
PL SGUBY MZ VRR MY ZPL UGL BPU ALVRRI NULY
- VR CLAGYZLMG
Drag your cursor between the asterisks.
*
Even more exasperating than the guy who thinks
he knows it all is the one who really does.
- Al Bernstein

*

Below, each letter stands for a specific number.
Different numbers must be assigned to each letter.
There are no zeros in the most left column.
Find the value of each letter. (0-9 are used.)

$\begin{array}{ccccc}
^1 & ^2 & ^3 & ^4 & ^5 \\
& & D & O & S \\
& & D & O & S \\
+ & T & R & E & S \\ \hline
S & I & E & T & E \end{array}$

In column-1, $\boxed{S = 1}$
Hence: . $\boxed{T= 9}$ with a "carry" from column-3: . $\boxed{I = 0}$

$\begin{array}{ccccc}
^1 & ^2 & ^3 & ^4 & ^5 \\
& & D & O & 1 \\
& & D & O & 1 \\
+ & 9 & R & E & 1 \\ \hline
1 & 0 & E & 9 & E\end{array}$

Obviously, $\boxed{E = 3}$ and we have:

$\begin{array}{ccccc}
^1 & ^2 & ^3 & ^4 & ^5 \\
& & D & O & 1 \\
& & D & O & 1 \\
+ & 9 & R & 3 & 1 \\ \hline
1 & 0 & 3 & 9 & 3\end{array}$

In column-4, we have:. $2O + 3 \to 9$
. . This is true if: . $O = 3\text{ or }O = 8$
Since $E = 3$, then $\boxed{O = 8}$

$\begin{array}{ccccc}
^1 & ^2 & ^3 & ^4 & ^5 \\
& & D & 8 & 1 \\
& & D & 8 & 1 \\
+ & 9 & R & 3 & 1 \\ \hline
1 & 0 & 3 & 9 & 3\end{array}$

In column-3: . $2D + R + 1 \:=\:13\quad\Rightarrow\quad 2D + R \;=\;12$
. . The only solution is: . $\boxed{D = 5}\;\boxed{R = 2}$

Solution: . $\begin{array}{ccccc}
& & 5 & 8 & 1 \\
& & 5 & 8 & 1 \\
+ & 9 & 2 & 3 & 1 \\ \hline
1 & 0 & 3 & 9 & 3\end{array}$

Some obvious advice . . .
Don't even try to convince your teacher that you solved these yourself.

He/she will probably give an additional problem to solve,
. . just to observe your technique.
And you'll prove to everyone that you are a Cheat, a Liar, and a Complete Phony.

5. Hello Soroban,

I'd like to know how you can state from the beginning that S=1 ?

And the "Hence T=9" ?

6. Hello, Moo!

I'd like to know how you can state that $S=1 \text{ and }T=9$ ?

We have:
. . . $\begin{array}{ccccc} &&D&O&S\\&&D&O&S \\ &T&R&E&S \\ \hline S&I&E&T&E \end{array}$

We have some three-digit numbers and one four-digit number.
. . And their sum is a five-digit number.

If the upper digits were all 9's, we'd have:. $999 + 999 + 9999 \:=\:11997$ ... maximum sum.
. . Hence: . $SIETE \:<\:{\color{red}1}1997$

Then the sum begins with 1: . $S = 1$

Its second digit is $\leq 1\!:\;I = 0$

So far, we have:

. . $\begin{array}{ccccc} &&D&O&1\\&&D&O&1 \\ &T&R&E&1 \\ \hline 1&0&E&T&E \end{array}$

And we see immediately that $E = 3$

. . $\begin{array}{ccccc}^1 & ^2 & ^3 & ^4 & ^5 \\ &&D&O&1\\&&D&O&1 \\ &T&R&3&1 \\ \hline 1&0&3&T&3 \end{array}$

How can the sum begin with $"10"$ ?
There are two ways;
. . $T = 8$, and 2 is "carried" from column-3
. . $T = 9$, and 1 is "carried" from column-3.

Suppose $T = 8$, then we have:

. . $\begin{array}{ccccc}^1&^2&^3&^4&^5 \\ &&D&O&1\\&&D&O&1 \\ &8&R&3&1 \\ \hline 1&0&3&8&3 \end{array}$

In column-4, we see that: . $2O + 3\text{ ends in }8 \quad\Rightarrow\quad 2O\text{ ends in }5$
. . Since $2O$ is an even number, this is impossible.

Therefore: . $T = 9$ . . . etc.