- 09 Jul '10 18:29 / 1 editI found this interesting problem in New Scientist : http://www.newscientist.com/article/mg20727670.800-enigma-number-1602.html

Solve it and be £15 richer.

Note: I don't have the solution.

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Miscount

I have assigned a whole number to each letter of the alphabet. These numbers are not necessarily different and they include negatives and zero. With these numbers I found that

O+N+E = 1

T+W+O = 2

T+H+R+E+E = 3

and so on all the way up to

T+W+E+N+T+Y+N+I+N+E = 29.

Unfortunately, I now find that I made one slip in my additions. Just one of my 29 equations was wrong.

Which equation is wrong?

What should the sum of the letters equal in that case?

WIN £15 will be awarded to the sender of the first correct answer opened on Wednesday 4 August. The Editor's decision is final. Please send entries to Enigma 1602, New Scientist, Lacon House, 84 Theobald's Road, London WC1X 8NS, or to enigma@newscientist.com (please include your postal address). - 09 Jul '10 21:43

What if twenty is wrong?*Originally posted by FabianFnas***I found this interesting problem in New Scientist : http://www.newscientist.com/article/mg20727670.800-enigma-number-1602.html**

Solve it and be £15 richer.

Note: I don't have the solution.

- - - - - - - - - - - - - - - - - - - -

Miscount

I have assigned a whole number to each letter of the alphabet. These numbers are not necessarily different and ...[text shortened]... 's Road, London WC1X 8NS, or to enigma@newscientist.com (please include your postal address).

If twenty is wrong, then:

- all other are correct

- t+w+e+n+t+y is not 20

o+n+e = 1 and t+w+e+n+t+y+o+n+e = 21

the difference is t+w+e+n+t+y = 20

So, twenty must be correct, and so are 1 - 9 and 21 - 29 (if two is wrong, twentytwo must also be wrong).

For the same reason, 14, 16, 17, and 19 are correct.

Only 10, 11, 12, 13, 15, and 18 can be wrong.

What if 13 is wrong? Then:

- all other are correct

- t+h+i+r+t+e+e+n is not 13

s+e+v+e+n+t+e+e+n = 17

s+e+v+e+n = 7

This makes, that t+e+e+n = 10 = t+e+n

t + 2e + n = t+e+n

2e = e

e = 0

e+i+g+h+t+e+e+n = 18

e+i+g+h+t = 8

This makes, that e+e+n = 10

e = 0 --> 0+0+n = 10 --> n = 10

e+e+n = 10 = t+e+e+n

t = 0

n+i+n+e = 9

10 + i + 10 + 0 = 9

i = -11

t+h+r+e+e = 3

0 + h + r + 0 + 0 = 3

h + r = 3

t+h+i+r+t+e+e+n is not 13

0+ (h+r) -11 + 0 + 0 + 0 + 10 is not 13

3 - 11 + 10 = 2 is not 13

My presumption that 13 is wrong, is true. - 09 Jul '10 21:55

If you're correct in this, then mail the solution and the £15 is yours!*Originally posted by Thomaster***What if twenty is wrong?**

If twenty is wrong, then:

- all other are correct

- t+w+e+n+t+y is not 20

o+n+e = 1 and t+w+e+n+t+y+o+n+e = 21

the difference is t+w+e+n+t+y = 20

So, twenty must be correct, and so are 1 - 9 and 21 - 29 (if two is wrong, twentytwo must also be wrong).

For the same reason, 14, 16, 17, and 19 are correct.

Only 10, 11, 12, 13 ...[text shortened]... + 0 + 0 + 0 + 10 is not 13

3 - 11 + 10 = 2 is not 13

My presumption that 13 is wrong, is true. - 10 Jul '10 19:22 / 5 edits

nevermind...thought you were *assuming* 13 false! It *is* false as you have indeed shown:]*Originally posted by Thomaster***What if twenty is wrong?**

If twenty is wrong, then:

- all other are correct

- t+w+e+n+t+y is not 20

o+n+e = 1 and t+w+e+n+t+y+o+n+e = 21

the difference is t+w+e+n+t+y = 20

So, twenty must be correct, and so are 1 - 9 and 21 - 29 (if two is wrong, twentytwo must also be wrong).

For the same reason, 14, 16, 17, and 19 are correct.

Only 10, 11, 12, 13 ...[text shortened]... + 0 + 0 + 0 + 10 is not 13

3 - 11 + 10 = 2 is not 13

My presumption that 13 is wrong, is true.