THY
× CAR
────────
THY
GAZ
RRZC
────────
RTZTRY
On a 14067 60932 the 2841 (=k069) asked: "…Me, to pay for the fuel? What is this 51487 23069?"
Fellow floobers are requested to make sense of the above.
THY
× CAR
--------
THY
GAZ
RRZC
--------
RTZTRY
First, since THY×R=THY, R=1.
THY
× CA1
--------
THY
GAZ
11ZC
--------
1TZT1Y
In the penultimate column of addition, T cannot equal 1, so T must represent 2 and there is a carry digit from the next column.
2HY
× CA1
--------
2HY
GAZ
11ZC
--------
12Z21Y
Since there is a carry digit, G does not equal 0; rather, G is 9 and there is another carry digit from the next column:
2HY
× CA1
--------
2HY
9AZ
11ZC
--------
12Z21Y
From the second product, 2HY*A=9AZ, which is only possible if A=3 or A=4.
From the third product, 2HY*C=11ZC, which is only possible if C is 3,4, or 5.
From the second column of addition, neither H nor Z may represent 1, so H+Z=11, and there is a carry digit into the sum 1+2+A+C=10+2, which means A+C=9.
From these three facts, the only possible values of A and C are A=4 and C=5.
2HY
× 541
--------
2HY
94Z
11Z5
--------
12Z21Y
From the third product, 2HY*5=11Z5, which is only possible if H=2 or H=3.
From the same product, Y must be odd (because the resultant product ends in 5).
Since 2 is already represented by T, H must equal 3. As the only other odd digit is 7, we have that Y=7, and by doing the mutliplication, Z=8.
237
× 541
--------
237
948
1185
--------
128217
These are all the values in the equation, so we can plug those into the sentence:
On a RA06Y 60GHT the TZAR (=k06G) asked: "…Me, to pay for the fuel? What is this CRAZY TH06G?"
The only substitutions that make sense here are 0=I and 6=N, giving:
On a rainy night the tzar (=king) asked: "…Me, to pay for the fuel? What is this crazy thing?"
0=I
1=R
2=T
3=H
4=A
5=C
6=N
7=Y
8=Z
9=G
|
Posted by DJ
on 2011-05-13 21:49:21 |