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 = ?
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.
Now to answer the questions:
(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 :)
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
