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The Love Bug

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What the... [May. 31st, 2002|08:01 pm]
The Love Bug
[Current Mood |confusedconfused]

  T H I S
+     I S
+ V E R Y
  -------
  E A S Y


We are told that:
  • T = 3
  • Y = 6

We can deduce that:
  • S = 5 or 0
  • E >= 4
  • V <=5

We are also told that no letter can have the same value as another letter.

Therefore, the question is what values do each of the letters above have, so that the sum at the top is correct.

  A = ?
  E = ?
  H = ?
  I = ?
  R = ?
  S = ?
  T = ?
  V = ?
  Y = ?


By the way, this is supposed to be a question that my sister (aged 13) got for her homework!!! :-$
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Comments:
From: bratstreet
2002-05-31 12:18 pm (UTC)

Good lord..

I think I remember getting problems like that at that age.

This is why it took me 3 years to pass the math course I needed for graduation..

and thus why it took 3 years to graduate.

lol.

I have not the mind of a mathematician.
(Reply) (Thread)
[User Picture]From: jena
2002-05-31 12:54 pm (UTC)

damn my Mensa brain.

A = 7
E = 8
H = 9
I = 1
R = 2
S = 5
T = 3
V = 4
Y = 6

I can never turn down a challenge, and even though I tried to ignore it and get some work done, I just couldn't.
(Reply) (Thread)
[User Picture]From: thelovebug
2002-05-31 12:59 pm (UTC)

Re: damn my Mensa brain.

Was that purely mathematical, or was there some lateral thinking in that?
(Reply) (Parent) (Thread)
[User Picture]From: jena
2002-05-31 01:20 pm (UTC)

Re: damn my Mensa brain.

Purely mathematical. S had to be 5, because you have nine letters, therefore the puzzle will use the nine digits and no zero. V had to be 5 or less, but 5 is used by S, so V={1,2,4}. The tens column (I+I+R=5 or 15) already has 1 added into it from the ones column equaling 16, so it becomes I+I+R=4 or 14. It is impossible to solve it for 14, so it must be 4, thus I=1 and R=2. You are left with the letters H, E, V, and A, and the numbers 4, 7, 8, and 9. The only possibility for V is 4, because of the options, it is the only number that results in a single digit for the sum of T and V. The size of the remaining numbers implies that there will be 1 carried from the hundreds column to the thousands column, thus 1+3+4=8. All that is left is A, H, 7 and 9. If H were 7, A would have to be 6. However, if H were 9, A would be 7.

Q.E.D.

THIS + IS + VERY = EASY ==>
3915 + 15 + 4826 = 8756
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