This is an alphametic - Each of the alphabets in the below represents a unique digit in base 10 such that the following representation holds true.

            T   E   N
+   T   E   N
+  F   O   R   T   Y
------------------------
S    I   X   T   Y

(i)   Value of F+Y = ?
(ii)  The value of TOY = ?
(iii) Value of E+O+T = ?
(iv)  Value of R*T = ?
(v)   Value of S+I+X+T+Y = ?

• N is 5, rest i am sleepy... – ABcDexter Jul 5 '16 at 19:40
• @ABcDexter I think you're just sleepy. – LeppyR64 Jul 5 '16 at 19:41
• @LeppyR64 haha yup! – ABcDexter Jul 6 '16 at 7:14

Solution:

    850 +
850 +
29786 =
31486
$T = 8, E = 5, N = 0, F = 2, O = 9, R = 7, Y = 6, S = 3. I = 1, X = 4$

Reasoning:

$N + N + Y$ has the last digit $Y$.
This means $N$ is one of $0,5$.
but $E+E+T +$(possibly a carriage of 1) has the last digit $T$.
The 2 above make me thing that $N = 0$ and $E = 5$

Now we have

       T50
+T50
+  FORTY
-----
SIXTY


Next:

$O$ has to be a big digit since the value of the 10 thousands changes ($F$ and $S$).
Since the max carriage from $T+T+R$ can be $2$, $O$ can be one of $8,9$.
but if $O$ is $8$ it means $I$ is $0$ and that cannot be because $N = 0$.
So $O=9$ and the carriage from the hundreds must be 2 otherwise $I=0$ that cannot be.
In this case $I = 1$.

Now we have:

       T50
+T50
+  F9RTY
-----
S1XTY


Continuing:

If the carriage return from hundreds must be 2 and we can see that the carriage from the tens is 1 it means that:
$T+T+R + 1 >= 20$.
This means that $[T,R]$ can be one of $[6, 8], [6,7], [7,6], [7, 8], [8, 4], [8,6], [8,7]$.

Taking one by one:

$T = 6, R = 8$ can't be because it will result in $X = 1$ but $I = 1$.
$T = 6, R = 7$ can't be because it will result in $X = 0$ but $N = 0$.
$T = 7, R = 6$ can't be because it will result in $X = 1$ but $I = 1$.
$T = 7, R = 8$ results in

       750
+750
+  F987Y
-----
S137Y


Since $S = F+1$, we are left with digits $2, 4, 6$ we don't have 2 consecutive digits left.
$T = 8, R = 4$ results in $X = 1$ that is not possible because $I = 1$.
$T = 8, R = 6$ results in same thing as for $T = 7, R = 8$.
$T = 8, R = 7$ we get.
       850
+850
+  F978Y
-----
S148Y


We are now left with digits 2,3 and 6 to fill in.
Since $S = F+1$ it means $S = 3, F = 2$ and $Y = 6$ and we reach the solution listed above.

(i) Value of F+Y = ?

8

(ii) The value of TOY = ?

896

(iii) Value of E+O+T = ?

22

(iv) Value of R*T = ?

56

(v) Value of S+I+X+T+Y = ?

22

850
+
850
+
29786
=
31486

So

(i) F + Y = 8
(ii) TOY = 432
(iii) E + O + T = 22
(iv) RT = 56
(v) S + I + X + T + Y = 22

Someone correct me if my mental maths is wrong :)

• I love these puzzles so much – Beastly Gerbil Jul 5 '16 at 20:02
• I believe the last sum should be 22, not 32. – gtwebb Jul 5 '16 at 20:03
• @gtwebb, thanks for some reason i did F + O + R + T + Y instead of sixty. Oops – Beastly Gerbil Jul 5 '16 at 20:04
• @RahulSinha, is this correct? – Beastly Gerbil Jul 5 '16 at 20:15
• I don't mean to be picky, but I'm going to be (actually I meant it). For alphametic puzzles shouldn't there be a reasoning for the value of the symbols? – Marius Jul 5 '16 at 20:44

First, $N+N+Y$ yields $Y$.

So $N+N \in \{0,10\}$ which means that $N \in \{0,5\}$.

If $c_1$ is the carry over from the ones column into the tens, then $c+E+E+T \implies T$.

This means that $c_1+E+E \in \{0,10\}$. This is impossible unless $c_1=0$. Therefore, $N=0$ to get no carry over, and $E=5$ since 0 is taken. The carry over into the hundreds column is then 1.

We also know that

$1+T+T+R$ can carry over at most 2 if $T$ and $R$ are maxed. Lets call this carry over $c_2$. Thus, $O+c_2$ carries over, and it can be at most 1. Thus, $O\in \{8,9\}$, $I\in \{0,1\}$, and $S=F+1$.

But since

0 is already taken, we know $I=1$. Thus, $O=9$ and $c_2=2$. The lowest value for $X$ is then 2. Thus, $1+T+T+R \ge 22$. If $T=8$, then $R \in \{6,7\}$. If $T=7$, then $R=8$. $T=6$ does would require $R=9$ which is already taken.

We know that $S$ and $F$ must be

consecutive, and 7 is already taken. thus, $S \in \{3,4\}$ and $F \in \{2,3\}$. In both cases, 3 is taken, so $T=8, R=6$ cannot work because that would make $X=3$. Similarly, $T=7,R=8$ makes $X=3$, so that is also not an option.

Therefore,

$T=8$ and $R=7$ which makes $X=4$. Thus, $S=3$ and $F=2$. The only remaining value for $Y$ is 6.

Solution is:

0=N, 1=I, 2=F, 3=S, 4=X, 5=E, 6=Y, 7=R, 8=T, 9=O

850
+ 850
+ 39786
-------
41486