Each letter is a different digit
Judy & Irv
A* WEBTV = DDDDDD
or
(A) (WEBTV) =DDDDDD
Have a good day Judy & Irv
Ernesto
S
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(A) (WEBTV) = DDD,DDD
(7) (95238) = 666,666
Ernesto
cod...@my-dejanews.com
In article <6k9v6r$hhp$1...@newsd-113.bryant.webtv.net>,
jid...@webtv.net (Judy & Irv) wrote:
>
>
> A* WEBTV = DDDDDD
>
> or
>
> (A) (WEBTV) =DDDDDD
>
> Have a good day Judy & Irv
>
7*95238=666666
Unique, I believe.
CO
-----== Posted via Deja News, The Leader in Internet Discussion ==-----
http://www.dejanews.com/ Now offering spam-free web-based newsreading
il...@isgtec.com
In article <6k9v6r$hhp$1...@newsd-113.bryant.webtv.net>,
jid...@webtv.net (Judy & Irv) wrote:
>
>
> A* WEBTV = DDDDDD
>
> or
>
> (A) (WEBTV) =DDDDDD
>
> Have a good day Judy & Irv
>
SPOILER
7 * 95238 = 666666
__/\__
\ /
__/\\ //\__ Ilan Mayer
\ /
/__ __\ Toronto, Canada
/__ __\
Aulis Kellinsalmi
>>
>> A* WEBTV = DDDDDD
>>
>> or
>>
>> (A) (WEBTV) =DDDDDD
>>
>> Have a good day Judy & Irv
>>
>{spoiler}
How about this:
7*95238=666666
AK
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>7*95238=666666
>Unique, I believe.
> CO
>
Imran Ghory
Judy & Irv <jid...@webtv.net> wrote in article
<6k9v6r$hhp$1...@newsd-113.bryant.webtv.net>...
>
> A* WEBTV = DDDDDD
>
> or
>
> (A) (WEBTV) =DDDDDD
How do you work out answers for a question like this ?
Imran
Randy J. Slavey
Imran Ghory wrote in message
<01bd89b8$c0bc5e00$LocalHost@default>...
>
>
>Judy & Irv <jid...@webtv.net> wrote in article
><6k9v6r$hhp$1...@newsd-113.bryant.webtv.net>...
>>
>> A* WEBTV = DDDDDD
>>
>> or
>>
>> (A) (WEBTV) =DDDDDD
Spoiler
Dave Dodson
Imran Ghory wrote:
>
> Judy & Irv <jid...@webtv.net> wrote in article
> <6k9v6r$hhp$1...@newsd-113.bryant.webtv.net>...
> >
> > A* WEBTV = DDDDDD
>
Start by factoring DDDDDD = D * 111111
= D * 3 * 7 * 11 * 13 * 37.
Therefore, A, being a single digit, is either D or 3 or 7.
If A = D, then WEBTV = 111111, which doesn't fit.
If A = 3, then WEBTV = D*222222, which also doesn't fit.
Hence A = 7 (the rest of the possibilities are 2 digits).
So WEBTV = D * 3 * 11 * 13 * 37 = D * 15873.
If D = 1, WEBTV = 15873, which has W = D.
If D = 2, WEBTV = 31746, which has E = D.
If D = 3, WEBTV = 47619, which has T = D.
if D = 4, WEBTV = 63492, which has B = D.
If D = 5, WEBTV = 79365, which has V = D.
If D = 6, WEBTV = 95238, which works.
If D > 6, WEBTV = more than 5 digits.
Hence D = 6 and WEBTV = 95238.
Note that we really didn't need the factors of 111111
that exceeded 1 digit.
Dave
Dave Dodson
I see that I messed up the line beginning "If A = 3, "
That should have said
If A = 3, then WEBTV = D*37037, and:
If D = 1, then WEBTV = 37037 and W = T and E = V.
If D = 2, then WEBTV = 74074 and the same thing.
If D > 2, then WEBTV = more than 5 digits.
Therefore, A is not 3.
Hence A = 7.
Dave
kdg...@dialnet.net
Dave Dodson <dod...@rsn.hp.com> wrote:
> Imran Ghory wrote:
>>
>> Judy & Irv <jid...@webtv.net> wrote in article
>> <6k9v6r$hhp$1...@newsd-113.bryant.webtv.net>...
>> >
>> > A* WEBTV = DDDDDD
>>
>> How do you work out answers for a question like this ?
> Start by factoring DDDDDD = D * 111111
> = D * 3 * 7 * 11 * 13 * 37.
> Therefore, A, being a single digit, is either D or 3 or 7.
Just a quick nitpick. From this so far there's no reason
that A couldn't be 6 or 9. However:
A=6 D=4 WEBTV=74074
A=6 D=2 WEBTV=37037
A=9 D=6 WEBTV=74074
A=9 D=3 WEBTV=37037
> If A = D, then WEBTV = 111111, which doesn't fit.
> If A = 3, then WEBTV = D*222222, which also doesn't fit.
A=3 D=2 WEBTV=74074
A=3 D=1 WEBTV=37037
While these don't work, I don't see how you can rule them out
without checking.
Kev
Torkel Franzen
"Imran Ghory" <imr...@btinternet.com> writes:
> > A* WEBTV = DDDDDD
> How do you work out answers for a question like this ?
There is only one good way of solving stupid problems like this: let a
computer do it. Thus:
solve(A,W,E,B,T,V,D):-
L=[1,2,3,4,5,6,7,8,9],
membrest(A,L,L1),
membrest(D,L1,L2),
membrest(W,L2,L3),
membrest(E,[0|L3],L4),
membrest(B,L4,L5),
membrest(T,L5,L6),
membrest(V,L6,_),
X1 is A*(V+10*T+100*B+1000*E+10000*W),
X1 is D+10*D+100*D+1000*D+10000*D+100000*D.
membrest(X,[X|L],L).
membrest(X,[Y|L],[Y|M]):-
membrest(X,L,M).
This yields
| ?- solve(A,W,E,B,T,V,D).
A = 7,
B = 2,
D = 6,
E = 5,
T = 3,
V = 8,
W = 9 ? ;
no
That is, the only solution is 7*95238=666666.
Matthew Daly
I'll never forget the time that Torkel Franzen <tor...@sm.luth.se> said:
>"Imran Ghory" <imr...@btinternet.com> writes:
>
> > > A* WEBTV = DDDDDD
> > How do you work out answers for a question like this ?
>
> There is only one good way of solving stupid problems like this: let a
>computer do it.
Yuck, absolutely not. Especially in this case, a pocket calculator could
solve the problem faster than you can type in and run your program.
DDDDDD = 111111*D, and 111111 is seen to be divisible by 3 and 7. So our
first instinct is to divide 111111 and 222222 by 3 and 111111, 222222,
333333, 444444, 555555, and 666666 by 7 on the calculator to see if we can
find a quotient that contains 5 digits different from each other, A, and
D. We find in under a minute that 7 * 95238 = 666666.
I would almost wager that you can solve it on paper faster than you can
write a program.
-Matthew
Torkel Franzen
>DDDDDD = 111111*D, and 111111 is seen to be divisible by 3 and 7. So our
>first instinct is to divide 111111 and 222222 by 3 and 111111, 222222,
>333333, 444444, 555555, and 666666 by 7 on the calculator to see if we can
>find a quotient that contains 5 digits different from each other, A, and
>D.
That's work. If I don't have a computer at hand, my first instinct
is to ignore problems of this kind.
>I would almost wager that you can solve it on paper faster than you can
>write a program.
No doubt there are people who could solve this particular problem
faster than I could write a program to solve it. It's a simple matter,
however, to produce similar problems for which this is no longer true.
The point of my comment was that this type of problem is essentially
trivial. Solving it doesn't tell us anything we didn't know, but is
merely a lot of boring work. It's the kind of thing that should be
left to computers.
cod...@my-dejanews.com
In article <TORKEL.98M...@anhur.sics.se>,
tor...@sics.se (Torkel Franzen) wrote:
>
> In article <35701c6b....@news.frontiernet.net> mwd...@pobox.com
(Matthew Daly) writes:
>
> >DDDDDD = 111111*D, and 111111 is seen to be divisible by 3 and 7. So
our
> >first instinct is to divide 111111 and 222222 by 3 and 111111, 222222,
> >333333, 444444, 555555, and 666666 by 7 on the calculator to see if we
can
> >find a quotient that contains 5 digits different from each other, A, and
> >D.
>
> That's work. If I don't have a computer at hand, my first instinct
> is to ignore problems of this kind.
Thank you. Please continue your benevolence.
>
> >I would almost wager that you can solve it on paper faster than you can
> >write a program.
>
> No doubt there are people who could solve this particular problem
> faster than I could write a program to solve it. It's a simple matter,
> however, to produce similar problems for which this is no longer true.
> The point of my comment was that this type of problem is essentially
> trivial. Solving it doesn't tell us anything we didn't know, but is
> merely a lot of boring work. It's the kind of thing that should be
> left to computers.
>
Some of us enjoy the "boring work" and can't imagine anything more boring
than plugging the numbers into a computer to solve.
As for triviality, nothing in the NG is likely to produce a cancer cure or
end world hunger. That's why it's REC.puzzles.
Torkel Franzen
cod...@my-dejanews.com writes:
>Some of us enjoy the "boring work" and can't imagine anything more boring
>than plugging the numbers into a computer to solve.
To each his own. To my mind, these "each letter is a different
digit" puzzles, along with a slew of others that can be trivially
reduced to trying out combinations, are just a source of programming
exercises.