Re: Enigma 1457 - Anglo-Italian sums

On Oct 10, 11:48 pm, Ted Schuerzinger <fe...@xxxxxxxxxxxx> wrote:
On Wed, 10 Oct 2007 19:33:58 -0700, Chappy wrote:
Enigma 1457 - Anglo-Italian sums
New Scientist magazine, 25 August 2007.
by Richard England.

Within each of these Anglo-Italian sums,
digits have been consistently replace by
capital letters, different letters used for
different digits. However, the three sums
are entirely distinct - a letter need not
have the same value in one as in either of
the others. No number starts with a zero.

(a) FOUR + FOUR = OTTO
(b) TWO + TRE + FIVE = DIECI
(c) THREE + NOVE = DODICI

What are the numbers represented by OTTO in
(a), DIECI in (b), DODICI in (c)?

Ciao,
Chappy.

Let's start with C first:

THREE
+ NOVE
-------
DODICI

D is obviously 1; O is 0; T is 9. All ten digits are used (CDEIHNORTV)
Since R+0 in the hundreds column doesn't yield R, there's a carry, and I
= R+1 (and I cannot be 2). E cannot be 0, 1, 5, or 6. If E=2, I=4,
R=3; H and N are 5 and 6, at which point we don't get valid values for V
and C.

If E=3, I=6, R=5. There's no valid value for V.
If E=4, I=8, R=7. Again, no valid value for V.
If E=7, I=4, R=3, H and N = 5 and 6. No valid values for V and C.
If E=8, I=6, R=5, H and N are 4 and 7, V=3, and C=2. So, DODICI is
101626.

Now for A:

FOUR
+ FOUR
------
OTTO

We see from U+U = T and O+O = T that the two numbers are 5 apart. Also,
from R+R = O and F+F = O we see that there's no carry from O+O = T, and
F and R are 5 apart. This means that F+5 = R. And, since there's no
carry from O+O = T, O+5 must equal U.

If F=1, R=6, O=2, U=7, T=5. OTTO = 2552. Let's prove this is the only
valid answer:
If F=2, R=7, O=4, T=9. Oops, U would have to be 9, too.
If F=3, R=8, O=6. Oops, T would have to be 3.
If F=4, R=9, O=8. Oops, O is too large

Now, for B:

TWO
+ TRE
+ FIVE
------
DIECI

Obviously, D is 1. I must be 0. (It can't be 2, since the three
numbers being added together can't be more than 11997.) F must be 9 (E
must be at least 2, and 8099 plus two three-digit numbers can't yield
10200). T+T+I only carries a 1.

If O=2, E=4, T=6, and we don't have valid values for CRVW.
If O=8, E=6, T=7; we don't have valid values for CRWV.
If O=6, E=2 or 7. If E=2, T=5, and we need 3478 to map to CRVW. If
E=7, T=8. CRVW map to 2346. This works if 6+2+3 plus a carried 2 is
14.

W, R, V actually are any permutation of 2, 4, 5 and C is 3.


So, C = 4; RVW map to 236 in some order, and DIECI = 10740.

10730


To prove this is the only answer, we need to look at O=4.

O=4, E=3 or 8. If E=3, T=6, and CRVW map to 2578. Doesn't fit.
O=4, E=8, no value for T.

So, we now have unique solutions to all three equations!

In (A), OTTO = 2552
In (B), DIECI = 10740

10730

In (C), DODICI = 101626

--
Ted S.
fedya at bestweb dot net- Hide quoted text -

- Show quoted text -



Please reply to drgmayer at hotmail dot com

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