Algebraic problem solving technique In a "seven-eleven" (7-11) store, a customer selected four items to buy. The check-out clerk says that he multiplied the costs of the items and obtained exactly $7.11$, the very name of the store! The customer tells the clerk that the costs of the items should be added, not multiplied. The clerk then informs the customer that the correct total is also $7.11$. 
 What are the exact costs of the $4$ items?
How to solve this using less calculation? 
 A: Let the amounts (in cents) be $a,b,c$, and $d$.
Then


*

*$\dfrac{a}{100}+\dfrac{b}{100}+\dfrac{c}{100}+\dfrac{d}{100} = 7.11$

*$\dfrac{a}{100}\cdot\dfrac{b}{100}\cdot\dfrac{c}{100}\cdot\dfrac{d}{100} = 7.11$
In other words


*

*$a+b+c+d = 711$

*$abcd = 711000000$
$711000000 = 2^6 \cdot 3^2 \cdot 5^6 \cdot 79$
Also, the product has to be a multiple of $10^6 = 2^6 \cdot 5^6$.
Let's assume that $79 \mid a$ . I would suppose that the units digits of $b, c$ and $d$ are going to end in $0$ and $5$. Two $0s$ and one $5$ would requre that the units digit of $a$ must be $6$.
If this is the case, then $a = 4\cdot 79 = 316$.
Let's suppose that the units digit of $b$ ends in $5$. Since $316$ is almost half of the required sum, I want to keep $b,c$, and $d$ small. So I am going to give each of them two of the required six factors of $5$.
$$a = 316 \quad b = 25b_1 \quad c = 25c_1 \quad d = 25d_1$$
No, that didn't work. So I'll try
$$a = 316 \quad b = 5b_1 \quad c = 25c_1 \quad d = 125$$
So I still need to figure out where two $3s$ and four $2s$ go.
If I give $b$ abd $c$ a $3$each, I get
$$a = 316 \quad b = 15b_2 \quad c = 75c_2 \quad d = 125$$
and I need $b + c = 270$ and I still have to figure out where four factors of $2$ go. I figure I need $\{b_2, c_2\} = \{2, 8\}$. Trying to keep each number small, I try $b_2 = 8$ and $c_2 = 2$. I get
$$a = 316 \quad b = 120 \quad c = 150 \quad d = 125$$
Which add up to $270$!!!!
So the amounts were $\$1.20,\; \$1.25,\; \$150,\; \text{ and }\; \$3.16$ 
