The alphametic:
ED
_____
PEGG)MATHP
UZ..
----
Z...
.E..
----
.1..
Since MATH - UZ.. = Z.., M = U+1 and U is at most 8. Also, Z+Z = A+10
or A+9; in any case, Z is 5 or more.
Since E x PEGG = UZ.., E is at least 2 and at most 5. (If E=6, E x
PEGG is at least 9600, but UZ.. is at most 8999.)
Suppose E=5. Then P=1, U=7, M=8. Z is 6 or 9, but when Z=9, there is
no suitable number for A, so Z=6 and A is 2 or 3. Since D x 15GG =
.5.., D must be 3 (1 and 5 are taken, 2 and 4 cannot produce the 5 in
the product, and other digits are too large) so A=2. But there is no
suitable value for G. G=0 does not allow 5 x 15GG = 76.., while G=4 or
larger does not allow D x 15GG = .5.., and 1, 2, and 3 are taken.
Suppose E=4. Again P=1 (P=2 makes U=9, too large), U=5, M=6. Z is 7,
8, or 9. Since D x 14GG = .4.., D must be 3 (D=2 gives a product of
28.. or 29.., 7 is too large, and 1, 4, 5, 6 are used). But since Z is
at least 7, 3 is the wrong number for the quotient here; 14GG x 4 is
less than 7000 regardless of G.
Suppose E=3. If P=1, the product E x PEGG = UZ.. is 39.., 40.., or
41.. but 3 is already used and cannot be U, and Z must be at least 5.
So P=2, and U=6, Z=9, M=7, A=8. For D x 23.. = .3.. D must be 1 or 4.
But the subtraction 9... - .3.. = .1.. fails; D=1 is far too small a
value for this term in the quotient, while D=4 forces 9... - 93..
which cannot have a four-digit difference.
So E=2. P=1 fails because U is forced to be 2. P=3 works, giving U=6,
M=7. P=4 also works, giving U=8, M=9. In both cases Z is forced to be
5, A is 0 or 1, and only D=1 allows D x PEGG = .E.. so A=0. G is one
of the unused digits larger than 5. There are now four cases for PEGG:
3288, 3299, 4266, and 4277. The corresponding values of UZ.. are 6576,
6598, 8532, and 8554.
Since D=1, the .E.. is PEGG and E=2. Consider 5... - P2GG = .1..; the
first digit after the 5 must be a 3 or 4. Now consider M0TH - U5.. =
5?. where ? is a 3 or 4 and U5.. is one of the numbers from the end of
the previous paragraph. There is no borrowing from the hundreds so T =
? + the tens digit of U5?? possibly plus 1. For U5.. = 6576 or 6598 T
is larger than 9 and this fails. For U5.. = 8554 T is at least 8 but
in that case 8 are 9 are both used as U and M. Only U5.. = 8532 works.
In this case the available numbers for T and H are 7 and 3 so T=7,
H=3, and the rest of the math can be worked out.
21
_____
4266)90734
8532
----
5414
4266
----
1148
---------------------------------------------------------
Dear Ed,
Here's an answer for the new alphametic on your site. I must confess,
though, that I didn't find the solving to be gratifiying. I used brute
force to solve it, and 5 was the smallest sensible value for Z.
21
4266)90734
8532
5414
4266
1148
Sincerely,
Bryce Herdt
--------------------------------------------------------
Hi Ed,
Assuming the usual rules for alphametics (no leading zeros, each letter is a
different number), I found two solutions (both with D=1 but no other assignments
in common):
0123456789
GDPETHUMAZ
ADEHPZGTUM
I found these by starting with the observation that U<9 so PE*E < 90 which
means E<6 (and obviously E>1). D*P5GG=?5?? leads to D=1 or 6, both of which are
easily eliminated, so 2 <= E <= 4. I then just considered all the remaining
possible combinations of P and E and what those implied for the other numbers (Z,
A, U, M, G), eliminating cases that would cause two different letters to have
the same numeric value, until only the two solutions above remained.
Here's another fun one for you:
ED * MATH = PUZZLE
Cheers,
Michael Brundage
Software Design Engineer
Xbox System Software
Microsoft
--------------------------------
Jeff Smith
-------------------------------=
Darrel C Jones
> A lovely one it is, indeed. I can't easily type a detailed solution,
> but the answer took about an hour of effort by hand.
>
> The key:
>
> A=0, D=1, E=2, H=3, P=4, Z=5, G=6, T=7, U=8, M=9
>
> The division is 90734 / 4266, with a quotient of 21 remainder 1148.
>
> At first, I thought the "1" given in the remainder was superfluous,
> as all other options fitting the given letters
> had three-digit remainders. Until I realized there was a second
> solution with a four-digit remainder,
> namely 90943 / 3288 = 21 remainder 1895. The "1", of course, rules
> this one out.
Dear Ed,
The solution to the cryptomath puzzle of 12/23/04 is
2 1 E D
4 2 6 6 | 9 0 7 3 4 P E G G | M A T H P
8 5 3 2 U Z - -
5 4 1 4 Z - - P
4 2 6 6 - E - -
1 1 4 8 - 1 - -
I'd like to claim I used some clever process to figure out the answer,
but I cheated and wrote a short program to brute force it with very little finesse.
Warren Phillips