There is only one 6digit number enabling this beautiful shifting of the partial products:
mmmmmWINTER
*ICY
mmmmmWINTER
mmmmRWINTE
mmmERWINT
mmmTHERWWWR
What digit is not used in the multiplication above?
Hint : The road being slippery, don't use your own vehicle in the competition:
96075 45979 17$23 07869 546
One can solve most of the crytarithm using almost solely the partial products and product (i.e., without the multiplicands) after noticing that Y must equal 1, and W, I & C <> 0.
The carry in E+carry=T (the E of the third partial product and T of the product) must be 1 or 2, permitting E+1 or E+2 to equal T. T > E, thus E <> 9.
If E = 2, then T = 3 or 4.
If E = 3, then T = 4 or 5.
If E = 4, then T = 5 or 6.
If E = 5, then T = 6 or 7.
If E = 6, then T = 7 or 8.
If E = 7, then T = 8 or 9.
If E = 8, then T = 9.
E+E must be even, therefore W is of the set {2, 4, 6, 8}. E > 1.
If W = 2, then E = 6 with carry 1.
if W = 4, then E = 7 with carry 1 or E = 2 with carry 0.
if W = 6, then E = 8 with carry 1 or E = 3 with carry 0.
if W = 8, then E = 9 with carry 1 or E = 4 with carry 0.
T+T+T+carry = ?W, thus T = (?W  carry)/3. (0 <= ? <= 2).
If W = 2, then T := 1/3, 11/3, or 21/3. T would equal 7 (and E would equal 6).
If W = 4, then T := 3/3, 13/3, 23/3, 4/3, 14/3 or 24/3. T would be 1 or 8.
As T cannot be 1 (as Y = 1), T would be 8 with E+E having carry = 0, but then E would equal 2. which would require T = 3 or 4. Thus W <> 4.
If W = 6, then T := 5/3, 15/3, 25/3, 6/3, 16/3 or 26/3. T would equal 5, with E+E having a carry = 1. Yet, with neither E = 3 or 4 would E+E have a carry, thus W <> 6.
Therefore
W = 2, E = 6 and
T = 7.
N+N+N+carry = ?2. (0 <= ? <= 2). As the carry is 2. N = (?2  2)/3.
Thus, with ?=0 => 0/3, ?=1 => 10/3 and ?=2 => 20/3,
N = 0.
With W = 2 in W+W+W+carry = ?6, the carry resulting from I+I+I+carry is 0, thus I must be less than 4. As I can not be 0, 1 or 2,
I = 3.
With R+R+(0) = 1H. H must be even, as 0, 2, and 6 are already assigned, H is of the set {4, 8}.
If H = 4, then R = 7, but T = 7 thus H <> 4. Thus
H = 8, and therefore
R = 9.
THERWWWR/WINTER = 78692229/230769 = 341. Thus
C = 40 = N
1 = Y
2 = W
3 = I
4 = C
5 = not used in the arithmetic
6 = E
7 = T
8 = H
9 = R
Ciphering the Hint "96075 45979 17$23 07869 546" gives "RENT5 C5RTR YT$WI NTHER 5CE". Substituting "A" for 5 and "O" for $ and parsing, gives "RENT A CAR  TRY TO WIN THE RACE".

Posted by Dej Mar
on 20110825 05:28:36 