# LOVES+LIVE=THERE. How many “loves” are “there”?

This is a problem from Mathematical Circles ( Chapter 0, Problem#17 ). It goes like this:-

The answer is that there are 95343 "loves" in "there".

Now, this is something that I am unable to comprehend. The last statement of the question states that the answer is the maximum possible value of word THERE.

I have used the following approach to calculate the maximum possible value of THERE:-

$LOVES+LIVE=THERE$. Expanding it in decimal form, I have $(L\times10^4 + O\times10^3 + V\times10^2 + E\times10^1 + S\times10^0) + (L\times10^3 + I\times10^2 + V\times10^1 + E\times10^0) = (T\times10^4 + H\times10^3 + E\times10^2 + R\times10^1 + E\times10^0)$

$\Rightarrow L\times10^4 + (O+L)\times10^3 + (V+I)\times10^2 + (E+V)\times10^1 + (S+E)\times10^0 = (T\times10^4 + H\times10^3 + E\times10^2 + R\times10^1 + E\times10^0)$

$\Rightarrow L=T \text{ }$ which is impossbile because it is already said that two letters can never represent the same number. Hence, assume that $L \neq T$. This clearly means that there is some kind of carrying operation from the preceding additions. Also,I equated the like powers and obtained the upper that every letter could have. This brought me to the following conclusions ( by equating like powers ):-

Since, $S=S+E$, we have $S=0$ and $E\le9$. Hence the maximum possible value of $E$ is $9$.

Now, $E+V=R$. This means that $R\le8$ and $V\le9$

Now, for $V+I=E$. Since, $E$ can atmost $9$ and $V$ can be atmost $9$, I assumed that $I\le9$ because only by this assumption I can justify the carry over.

Now, considering $T$, it can have only $9$ values - $1,2,3\ldots9$. Hence, $T\le9$. To make this possible $L\le8$ and $0{\le}O\le9$ and hence $H\le8$.

Hence, from all these deductions, I can say that $T\le9$, $H\le8$, $E\le9$ and $R\le8$. Hence, the maximum possible value of THERE is $98989$. This is even working out if I consider $LOVES = 89990$ and $LIKE=8999$, I get $98989$ as the answer.

My question is :- "WHAT AM I NOT ABLE TO UNDERSTAND ABOUT THE QUESTION?" and "WHAT AM I DOING WRONG?"

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You can't take LOVES=89990, all the digits must be different. –  gev Mar 24 '13 at 10:37
"(equal digits by same letters, and different digits by different letters)" You have T, E, O, V, I, K all representing 9 and H, R, L all representing 8, but the problem requires different letters to represent different digits. –  Ivan Loh Mar 24 '13 at 10:40
and hence I can also say that $S\neq0$. I am so sorry for this mistake. –  kusur Mar 24 '13 at 10:41
Also, you started maximization of value from lower decimal places, you should start from higher decimal places, E$\neq$9 –  user45099 Mar 24 '13 at 10:56

$LOVES+LIVE=THERE$. Expanding it in decimal form, I have $(L\times10^4 + O\times10^3 + V\times10^2 + E\times10^1 + S\times10^0) + (L\times10^3 + I\times10^2 + V\times10^1 + E\times10^0) = (T\times10^4 + H\times10^3 + E\times10^2 + R\times10^1 + E\times10^0)$

$\Rightarrow L\times10^4 + (O+L)\times10^3 + (V+I)\times10^2 + (E+V)\times10^1 + (S+E)\times10^0 = (T\times10^4 + H\times10^3 + E\times10^2 + R\times10^1 + E\times10^0)$

$\Rightarrow L=T \text{ }$ which is impossbile because it is already said that two letters can never represent the same number. Hence, assume that $L \neq T$. This clearly means that there is some kind of carrying operation from the preceding additions. Also,I equated the like powers and obtained the upper that every letter could have. This brought me to the following conclusions ( by equating like powers ):-

Since, $L\neq T$, we have to have a carry operation and hence the maximum possible of $T=9$ and for this we have to use the maximum possible value of $L=8$.

Now, $O+L=H$. For this, we have to calculate the maximum possible values of $O$ and $H$. This means that $O=7$ and $H=5$ because $1$ will be carried over to the higher place and hence used for this calculation of $T$.

Now, the trick is to solve for $V+I=E$.Here, we have 5 choices for $V$ and $I$ each - $1,2,3,4,6$. Here we need to perform some trial and error to eliminate some of the choices. Here, most of cases are not suitable because they lead to redundancy in the digits. These possible combinations of $(V,I)$ are $(6,4); (4,6); (6,3); (3,6); (6,2); (2,6); (6,1); (1,6); (4,1); (1,4); (4,2); (2,4); (4,3); (3,4); (3,1); (1,3); (3,2); (2,3)$. The remaining pairs are $(1,2)$ and $(2,1)$. Out of these the pair $(2,1)$ is useless because this makes $E=3$ which in turn makes $R=5$, again a redundancy. Hence, the most viable option is the pair $(1,2)$ for which $E=3$ and hence $R=4$. Also, since $S+E=E$, we can say that $S=0$ and hence the $E$ at unit's place is also $3$. Hence, our number is 95343. Hence the maximum value of THERE is 95343. Hence there are 95343 LOVES in THERE.

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