16
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Obviously, FOUR + FIVE = NINE, but what if each letter is assigned a digit (0, 1, 2, 3, 4, 5, 6, 7, 8, 9), and two different letters can't be assigned the same digit?

Fill in numbers for the different letters to get the smallest value for NINE in this "equation":

  F O U R
+ F I V E
= N I N E
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7
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I think there are multiple answers:

1980+1254=3234
1970+1264=3234
1960+1274=3234
1950+1284=3234

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  • 5
    $\begingroup$ Care to show your logic to arrive at the numbers you did? $\endgroup$ – Jeff Zeitlin Dec 27 '19 at 20:45
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    $\begingroup$ I used the same logic as Grant's answer. $\endgroup$ – Duck Dec 28 '19 at 1:15
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    $\begingroup$ @Duck there are 72 answers $\endgroup$ – DrD Dec 28 '19 at 12:45
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    $\begingroup$ @DEEM There are only 4 answers given the "smallest value of NINE" requirement, and they are the ones Duck provided. $\endgroup$ – Rain Dec 29 '19 at 19:24
17
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I also came to the answer:

3234

Reasoning:

Starting from the least significant digit, as we are doing basic addition, we have $R + E = E$. That gives us our first known digit, $R = 0$. And we can fill in $E$ later as it won't effect anything else.

The next thing to look at is the similar equation $O + I = I$. We have already used $0$, so the only way for this equation to become true is if it is actually $O + I + 1 = I$, with a carry from the previous column. We simplify this to be $O + 1 = 0$, or $O + 1 = 10$, and we now know that $O = 9$.

From here we have two equations which equate to $N$. $1 + F + F = N$ and $U + V = N - 10$ (Due to it needing to carry a $1$). We can simply start filling in numbers to find a combination that holds true. If we set $F = 1$, then $N$ must be $3$, which then means $U + V = 13$, which can be accomplished with either $7 + 6$ or $8 + 5$. This leaves $2$ and $4$ free to be assigned to $E$ and $I$. We make $E = 2$ as it is the more significant digit, therefore gives the smaller total.

That brings us to the outcome that Duck has already provided:

1980+1254=3234
1970+1264=3234
1960+1274=3234
1950+1284=3234

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  • $\begingroup$ I seem to have messed up my spoiler tags, and Im not sure how...but they are the same answers that other users have found $\endgroup$ – Grant Dec 27 '19 at 22:04
  • $\begingroup$ Those spoiler tags are a bit of a menace. I noticed yours pointed the wrong way and tried to fix them, but it didn’t work so I reverted the changes. Since the results are already known, consider just leaving them out in the open. You can use 2 spaces at the end of a line to force a line break, or just leave an empty line in between, or use the “-“ character at the start of each line to produce a bullet list. $\endgroup$ – Lawrence Dec 27 '19 at 22:19
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    $\begingroup$ Welcome to Puzzling! Our answers are similar, but although I pipped you at the post (unknowingly!), I think yours was expressed more elegantly. +1 from me. $\endgroup$ – Lawrence Dec 27 '19 at 22:23
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    $\begingroup$ Nice explanation. I think the last sentence should say: "make I=2", not E. $\endgroup$ – CG. Dec 27 '19 at 23:09
3
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Logic:

The smallest 4-digit number starts with digit 1. From the left-most column (col 1), $N \geq 2$, so let’s try $F=1$.

Then $N = 2$ or $3$.

Now, in col 4, since $R+E=E$ (or $10+E$), $R$ must be $0$ (or $10$, but that’s 2 digits).

There is no carry to col 2. So $U+V=2$ doesn’t work ($0+2=2$, but $0$ has been taken). $U+V=12$ could work, but that would result in a carry to col 2. Since $O+I+1>I$, it must also have a carry to col 1, so $N>2$. No solution with $N=2$.

Try $N=3$. We still have $R=0$ with no carry from col 4 to col 3. Col 3 is now $U+V=3$ or $13$. It must be $13$ since $0$ and $1$ have been taken.

There being a carry from col 3 to col 2, $O=9$ so that $O+I+1=I+10$.

The smallest value remaining for $I$ is $2$.

So $U+V=13$ requires $5+8$ or $6+7$ or $7+6$ or $8+5$. Since $FOUR=19xx$ and $FIVE=12xx$, minimising them together (smallest product) requires $U=8, V=5$. Or if we want the smallest $max(FOUR,FIVE,NINE)$ we would have $U=5, V=8$.

Either way, we have a free choice for $E$, the smallest value available being $E=4$.

Putting it all together, we have, depending on how we define ‘smallest’:

1980+1254=3234,
1950+1284=3234.

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2
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If each digit represented like below.

E = 5; F = 1; I = 2; N = 3; O = 9; R = 0; U = 6; V = 7

and

E = 4; F = 1; I = 2; N = 3; O = 9; R = 0; U = 6; V = 7

Answers:

1 9 6 0 +
1 2 7 5
-------
3 2 3 5

and

1 9 6 0 +
1 2 7 4
-------
3 2 3 4

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  • $\begingroup$ @CG. I got a smaller solution. $\endgroup$ – Duck Dec 27 '19 at 20:24
  • $\begingroup$ check it now, Probably I am more closer to smaller value. $\endgroup$ – CR241 Dec 27 '19 at 20:27
  • $\begingroup$ @JeffZeitlin updated. I only got 3235 $\endgroup$ – CR241 Dec 27 '19 at 20:31
  • $\begingroup$ Consider (a) showing your work, and (b) putting the whole thing behind spoiler tags >! $\endgroup$ – Jeff Zeitlin Dec 27 '19 at 20:32
  • $\begingroup$ I have a smaller solution. It is very close to yours. $\endgroup$ – Duck Dec 27 '19 at 20:35
1
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This puzzle has 72 solutions

E=4 F=1 I=2 N=3 O=9 R=0 U=8 V=5

E=6 F=1 I=2 N=3 O=9 R=0 U=8 V=5

E=7 F=1 I=2 N=3 O=9 R=0 U=8 V=5

E=4 F=1 I=2 N=3 O=9 R=0 U=7 V=6

E=5 F=1 I=2 N=3 O=9 R=0 U=7 V=6

E=8 F=1 I=2 N=3 O=9 R=0 U=7 V=6

E=4 F=1 I=2 N=3 O=9 R=0 U=6 V=7

E=5 F=1 I=2 N=3 O=9 R=0 U=6 V=7

E=8 F=1 I=2 N=3 O=9 R=0 U=6 V=7

E=4 F=1 I=2 N=3 O=9 R=0 U=5 V=8

AND ON AND ON

But the smallest solution is 3234 as @Duck has said

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