Sum Minus Product

First, we note that $x_1 = x_2 = x_3 = x_4 = x_5 = x_6 = x_7 = 1$ solves the equation.

Now we consider at least one integer to be not 1, say $x_7 \neq 1$ :

We can let $x_i = 1+ k_i$ for $i = 1,2,3,4,5,6$ , where $k_i \geq 0$ .

The equation is now $1+k_1+ ...+1+k_6+x_7-(1+k_1)(1+k_2)...(1+k_6)x_7 = 6$

Manipulating this equation, we have

$6 + k_1 + ... + k_6 + x_7 - x_7 - k_1 x_7 - k_2 x_7 - ... - k_6 x_7 - K = 6$ for some non-negative integer K.

$k_1 + ... + k_6 + x_7 - x_7 - k_1 x_7 - k_2 x_7 - ... - k_6 x_7 - K = 0$

$k_1 + ... + k_6 - k_1 x_7 - k_2 x_7 - ... - k_6 x_7 - K = 0$

$k_1 + ... + k_6 - x_7 (k_1 + k_2 + ... + k_6) - K = 0$

$(1 - x_7)(k_1 + k_2 + ... + k_6) - K = 0$

$(1 - x_7)(k_1 + k_2 + ... + k_6) = K \geq 0$

Since $x_7 > 1 \Rightarrow 1 - x_7 < 0$ , so for the equation to hold, $k_1 + k_2 + ... + k_6 = 0$ and thus $k_1 = k_2 = ... = k_6 = 0$ .

For this case, $x_7$ can be any value from $2$ to $8$ since $6 + x_7 - x_7 = 6$ for any $x_7$ .

Any of $x_1, x_2, ..., x_7$ can be the non-"1" integer, so we have $7 \times 7 = 49$ ordered sets.

Adding the case where all of $x_1, x_2, ..., x_7 = 1$ , we have $50$ ordered sets in total.

Note that $(x_1, x_2, \cdots , x_7)= (1, 1, \cdots , 1)$ is a solution to the given equation. We now let any of the numbers not be $1$ . WLOG assume $x_7 \neq 1$ . We make the substitution $x_i= y_i+1$ for $1 \leq i \leq 6$ . Note that the $y_i$ s are non-negative integers. The equation now becomes $\sum \limits_{i=1}^{6} (y_i + 1) + x_7 - \prod \limits_{i=1}^{6}(y_i+1) \times x_7= 6$ Lets look at the expansion of $\prod \limits_{i=1}^{6}(y_i+1) \times x_7$ . It is easy to see that all the terms $y_1x_7$ , $y_2x_7$ , $\cdots$ , $y_7x_7$ must appear when we expand $\prod \limits_{i=1}^{6}(y_i+1) \times x_7$ . We can then write $\prod \limits_{i=1}^{6}(y_i+1) \times x_7 = x_7y_1 + x_7y_2 + \cdots + x_7y_6 + Z$ for some non-negative integer $Z$ . Our equation now becomes $\sum \limits_{i=1}^{6} y_i + 6 - x_7 \times \sum \limits_{i=1}^{6} y_i - Z = 6$ $\implies (1-x_7)\left ( \sum \limits_{i=1}^{6} y_i \right ) = Z$ Note that $Z \geq 0$ . But if $x_7>1$ and $\sum \limits_{i=1}^{6} y_i > 0$ , we have the L.H.S negative, while the R.H.S is still positive, which is a contradiction. So we must have $\sum \limits_{i=1}^{6} y_i= 0$ , i.e $y_1=y_2=\cdots = y_6= 0$ . Note that $x_7$ can be any positive integer from $2$ to $8$ , giving $7$ choices for $x_7$ .

To generalize, we have $7$ choices for the number which is not $1$ , and the other numbers must be $0$ . We can choose the positive number in $7$ ways, giving us $7 \times 7= 49$ choices. Adding it with the trivial solution $(x_1,x_2,\cdots , x_7)= (1,1,\cdots , 1)$ , we get $\boxed{50}$ solutions in all.

For all positive integers $x_1, x_2, x_3, x_4, x_5, x_6, x_7$ , $(x_1+1)+x_2+x_3+x_4+x_5+x_6+x_7 - (x_1+1) x_2 x_3 x_4 x_5 x_6 x_7 \leq x_1+x_2+x_3+x_4+x_5+x_6+x_7 - x_1 x_2 x_3 x_4 x_5 x_6 x_7$ . Equality only occurs when $x_2= x_3= x_4,=x_5,=x_6= x_7=1$ . Similarly for other $x_i$ . We also know $x_1=x_2=x_3=x_4=x_5=x_6=x_7=1$ is a solution. So all the solutions are when at most one of $x_i$ is greater than 1. In all other cases, $x_1+x_2+x_3+x_4+x_5+x_6+x_7 - x_1 x_2 x_3 x_4 x_5 x_6 x_7 < 6$ . So the answer is $7\times 7+1=50$ .

It is pretty obvious that if $x_i = 1, \forall i$ , then the equation is satisfied. So, the task now is to find solutions that contain at least one other number that is not equal to 1.

To do this, what I did was to first determine the maximum amount a number can be used as part of the solution. To make this clearer, let us take the case of $2$ . What is the maximum number of $x_i$ 's that can take the value $2$ ?

Obviously, if you want to use $2$ as much as possible, all the other $x_i$ 's not taking the value of $2$ should be equal to $1$ . This is because while the sum $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7$ (hereinafter referred to as the "Sum") grows in a linear fashion, the product $x_1 x_2 x_3 x_4 x_5 x_6 x_7$ (hereinafter referred to as the "Product") grows exponentially. Thus, if you use another value aside from $2$ and $1$ , you wouldn't be able to find how much $2$ 's you can actually use because the product can easily grow farther from the sum. In this case, the intended difference of the sum and the product, which is equal to $6$ , will less likely be achieved.

Hence, solving our dilemma requires solving this equation: $2n + 1 \times (7-n) - 2^n = 6$ , where $n$ is the number of $2$ 's used in the solution. We know that $1 <= n <= 7$ . Simplifying, $n - 2^n +1 = 0$ To find the highest number of $n$ possible, we should try all possible $n$ 's (as there are only seven possible values anyway) and determine the highest possible value of $n$ such that: $n - 2^n +1 >=0$ It is easily verifiable that the $n$ we are looking for is equal to $1$ . Actually, if $n =1$ , the equation above and the equation stated in the problem is satisfied.

Hence, the maximum amount of $2$ 's that can be used is only one. It is also easily verifiable, using the same method, that for $3 <= x_i <=8$ , the intended $n$ is also one. The equation in the problem is also satisfied in such cases.

It is only logical that we also cannot have a solution that has at least two $x_i$ 's whose value is greater than one. This is because the equation is already satisfied if there is exactly one number that is not equal to one. If there are two of such numbers, the difference between the Sum and the Product will decrease, given the fact that the product of two numbers is greater than their sum if both addends/multipliers is greater than 1, except when both addends are equal to $2$ . Such case is not allowed anyway given the proof above.

Hence, the solution required must look like such: there is exactly one $x_i$ that is greater than one. All the other $x_i$ 's are equal to one.

There are $7 \times \frac{7!}{6!1!} = 49$ such solutions. We must also include the first solution mentioned wherein all the $x_i$ 's are equal to one.

Thus, there are $50$ ordered sets of $7$ integers such that $Sum - Product = 6$

Note that we can see 50 solutions easily:

The first 49 are obtained by choosing one of the $x_1$ to $x_7$ and setting it as any integer from 2 to 8.

The last one is where $x_1=x_2=x_3=x_4=x_5=x_6=x_7=1$

We will claim there are no other solutions by proving the following statement:

If any n of $x_1, x_2, x_3, x_4, x_5, x_6, x_7\neq1$ , where $n>1$ $x_1+x_2+x_3+x_4+x_5+x_6+x_7-x_1x_2x_3x_4x_5x_6x_7<6$

We shall induct on n, where n=2 is the base case. When n=2, let the non-one numbers be a and b. (Definition of non-one: A number which is not one.) $x_1+x_2+x_3+x_4+x_5+x_6+x_7-x_1x_2x_3x_4x_5x_6x_7$ $=5+a+b-ab=6-(a-1)(b-1)<6$ . We are done for n=2.

Suppose when n=k, the statement is true. Then let the sum of the k non-one numbers be a, product of the k non-one numbers be b. We let the (k+1)th non-one number be c, Note that $(7-k)+a-b<6$ . This implies that $5+a<6+b$ , and subsequently, $a \leq b$ . $[7-(k+1)]+c+a-bc=(7-k)-(bc-c-a+1)$ $\leq 5-(ac-c-a+1)=5-(a-1)(c-1)<6$

Hence for k+1, the statement is true.

Thus we proven the statement.

Thus, we can see that since at most 1 number among $x_1, x_2, x_3, x_4, x_5, x_6, x_7 \neq 1$ . There are 50 such cases and it is easy to test that all of them work.

We claim that at most a single $x_i \geq 2$ for $1 \leq i \leq 8$ .

Proof: For the sake of contradiction,assume $x_1 \geq x_2 \geq 2$ ,while $x_i=1$ for $3 \leq i \leq 7$ . Then,

$x_1 + x_2 + x_3 + ... + x_7 - x_1x_2x_3...x_7 = (x_3 + ... + x_7) + ((x_1+x_2) - (x_1x_2)x_3...x_7) \leq (x_3 + ... + x_7) + (4 - 4 (x_3...x_7)) \leq (x_3 + ... + x_7) + 0 \leq 5 \neq 6$

Here we used the fact that if $a,b \geq 2 \Rightarrow ab \geq (a+b)$ . Since this is true when $x_3=x_4=...=x_7=1$ , & incrementing any $x_j$ increases the product substantially,we conclude atmost a single $x_i \geq 1$ .

Each required ordered tuple is some permutation of $\{a,1,1,1,1,1,1\}$ where $1 \leq a \leq 8$ yielding $1 + 7 \times 7 = 50$ cases.

for this moment just think that x1=x2=x3=x4=x5=x6=1 then the left side of the equation stands like this
(1+1+1+1+1+1+x7) - x7 = 6+x7 -x7 = 6 = right side

which means that when x1=x2=x3=x4=x5=x6=1 . the equation is true for any value of x7 .

as it is given that 1≤x i ≤8 . so the possible values of x7 with in the given range is 1,2,3,4,5,6,7,8. when x7=1 the amount of ordered tuples = 1 when x7=2 ordered tuples = 7 when x7=3 ordered tuples = 7 when x7=4 ordered tuples = 7 when x7=5 ordered tuples = 7 when x7=6 ordered tuples = 7 when x7=7 ordered tuples = 7 when x7=8 ordered tuples = 7

we can find all these ordered tuples from simple (permutation -combination)

so now the possible answers are 1+(7*7) = 50

but now the question is that we only have solved the equation for the only case when x1=x2=x3=x4=x5=x6=1 .what about the other cases ?

then assume that this time only x1=x2=x3=x4=x5=1 so both x6 and x7 are variables .and x6,x7 are not equal to 1. as we have done that case in our original solve at the beginning .

thus the equation stands like this

5+x6+x7- (x6 x7) =6 or x6+x7- (x6 x7) = 1 that means (x6+x7) > (x6 x7) but for any positive integer a,b where a,b are not equal to 1 always (a+b) ≤ (ab) . so there is no possible intiger value of x6,x7 when they are both greater than 1 .

same way for any any positive integer a,b,c,d,e,f,g not equal to1. the law is true such a+b+c < abc a+b+c+d < abcd a+b+c+d+e < abcd a+b+c+d+e+f < abcdef a+b+c+d+e+f+g < abcdefg but then that means there is no possible value
of x1,x2,x3,x4,x5,x6,x7 . there is possible values of x1,x2,x3,x4,x5,x6,x7 when atleast six of them have the value equal to 1.so we did the original solution assuming that x1=x2=x3=x4=x5=x6=1 .

We first notice that the 7-tuple $(1,1,1,1,1,1,1)$ is a solution to the equation.

We then notice that something interesting happens when we change just one of the $x_i$ . Specifically, any arrangement of six $1$ s and an integer $2 \leq n \leq 8$ satisfies the equation. This is because the six $1$ s add up to $6$ and have a product of $1$ , so the product of the $x_i$ is $n$ and their sum is $6 + n$ $(1 + 1 + 1 + 1 + 1 + 1 + n - 1^6\cdot{n} = 6 + n - n = 6)$ .

Now we must look at what happens when more than one of the $x_i$ is increased from $1$ . For example, can the 7-tuple $1, 1, 1, 1, 1, n, m$ with $2 \leq m, n \leq 8$ satisfy the equation? The answer is no. If we have less then six $1$ s in the 7-tuple, the equation cannot be satisfied. This is because $\forall m,n > 1, m\cdot{n} \geq m + n$ , which means that we will be subtracting from $5 + m + n$ at least $m + n$ , which clearly cannot give us a total of $6$ . A similar argument shows that the equation cannot be satisfied if more than two of the $x_i > 1$ .

Therefore, the only 7-tuples that satisfy the equation are $1, 1, 1, 1, 1, 1, 1$ and ones of the form $1, 1, 1, 1, 1, 1, n$ with $n \geq 2$ . There are $7$ ways to choose which of the $x_i$ to increase and $8 - 2 +1 = 7$ numbers to increase it to, so there are a total of $1 + 7\cdot{7} = 50$ 7-tuples that satisfy the equation.

I will divide it into $8$ case based on how many $1$ in the tuple, WLOG $x_1 \le x_2 \le x_3 \le x_4 \le x_6 \le x_7$

Case 1 : Assume $(1, 1, 1, 1, 1, 1, 1)$ is the solution. So, after we put it into the equation, we obtain : $7 - 1 = 6$ . So, satisfy the equation.

Case 2 : Assume $(1, 1, 1, 1, 1, 1, a)$ , for $a \ne 1$ . So, we obtain $(a + 6) - a = 6$ . Satisfy the equation

Case 3 : Assume $(1, 1, 1, 1, 1, a, b)$ , for $a, b \ne 1$ . So, we obtain $(a + b + 5) - ab = 6$ . So, $ab - a - b + 1 = 0 \implies (a - 1)(b-1) = 0$ . Not satisfy that $a, b \ne 1$

Case 3 : Assume $(1, 1, 1, 1, a, b, c)$ , for $a, b, c \ne 1$ . So, we obtain $(a + b + c + 4) - abc = 6$ . So, $abc - a - b - c + 2 = 0 \implies a(\frac{bc}{3} - 1) + b(\frac{ac}{3} - 1) + c(\frac{ab}{3} - 1) + 2 = 0$ . But, as we know that $ab, bc, ca \ge 4$ . So, $a(\frac{bc}{3} - 1) + b(\frac{ac}{3} - 1) + c(\frac{ab}{3} - 1) + 2 \ge 1 + 1 + 1 + 2 = 5$ . Contradictive with asumption. So, there is no solution for this case.

Case 4 : Assume $(1, 1, 1, a, b, c, d)$ , for $a, b, c \ne 1$ . So, we obtain $(a + b + c + d + 3) - abcd = 6$ . So, $abcd - a - b - c - d + 3 = 0 \implies 0 = a(\frac{bcd}{4} - 1) + b(\frac{acd}{4} - 1) + c(\frac{abc}{4} - 1) + d(\frac{abc}{4} - 1) + 2 \ge 1 + 1 + 1 + 1 + 2 = 5$ , since $abc, abd, acd, bcd \ge 8)$ . So, contradictive that $0 \ge 5$ . There is no solution

Case 5 : Assume $(1, 1, a, b, c, d, e)$ , for $a, b, c, d, e \ne 1$ . Analogue with case $3$ and $4$ . There is no solution.

Case 6 : Assume $(1, a, b, c, d, e, f)$ , for $a, b, c, d, e, f \ne 1$ . Analogue with case $3$ and $4$ . There is no solution.

Case 7 : Assume $(a, b, c, d, e, f, g)$ , for $a, b, c, d, e, f, g \ne 1$ . Analogue with case $3$ and $4$ . There is no solution.

So, all of the solution is, $(1, 1, 1, 1, 1, 1, 1)$ , and there is only $1$ permutation. And also, $(1, 1, 1, 1, 1, 1, a)$ , and there are $7$ permutation for each $a \in \{ 2, 3, 4, 5, 6, 7, 8 \}$ . So, total there are $1 + 7 \times 7 = \boxed{50}$ ordered tuple

Let $f(x_1,x_2,x_3,x_4,x_5,x_6,x_7)=\sum_{i=1}^7 x_i - \prod_{i=1}^7 x_i$ .

We can write the equation as $x_1(1-x_2x_3x_4x_5x_6x_7)+x_2+x_3+x_4+x_5+x_6+x_7=6,$ and conclude that the LHS is increasing in $x_1$ unless $x_2x_3\cdots x_7=1.$ The same conclusion can be reached for each $x_i$ . However, $f(1,1,1,1,1,1,1)=6$ . Suppose that there are two $x_i$ which are not $1$ , which we assume WLOG to be $x_1=a$ and $x_2=b$ . Then $f(a,b,1,1,1,1,1)<f(a,1,1,1,1,1)=6$ , which means that $(a,b,1,1,1,1,1)$ is not a solution. It is easy to see that the same argument extends to any number of $x_i\ne 1$ other than $0$ and $1$ . There is exactly one solution with all ones. For solutions with one non-one, we first choose a non-one entry, which can be done in 7 ways, and then choose a number other than one, which can also be done in 7 ways. Thus, there are 49 solutions with one non-one, and 50 solutions in total.

Well, this is not at all rigorous, but... Note that plugging in 1 for everything works. Yay!

Note that replacing one of the x's with another number also works, since it cancels with the product and the remaining 6*1 carries you to success. There are 7 choices for the index and 7 numbers (not =1) to choose. This gives you another 49, so 50 solutions so far.

If you put in a second number>1, the product grows faster than the sum, or x+y<xy. Let y be the larger number; adding a smaller number results in a number less than adding a bunch of larger numbers. So there are no more solutions. I wonder if there's a more formal way to state this intuitive result?

for the number of non-1 variables are more than one, assume five of the variables is 1, the others are n and k, with n has non-1 value 5+n+k-nk=6 nk-n-k+1=0 (n-1)(k-1)=0 n=1 or k=1 since n is non-1, that mean k=1 and that makes the number of non-1 variable is one. if the number of non-1 variables are two or more, when you try to put a fixed value on one of them, it will make the value of the other become 1. That mean the number of non-1 variables cannot be more than 1 therefore, it is safe to conclude that the number of non-1 variable in one or less.

for the number of non-1 variable is 1 or less, assume 6 of the variables are 1 and the other is k, then it will result, 6+k-k=6 (true) for k=1, it has 1 permutation for k=2, it has 7 permutation for k=3, it has 7 permutation for k=4, it has 7 permutation for k=5, it has 7 permutation for k=6, it has 7 permutation for k=7, it has 7 permutation for k=8, it has 7 permutation 50 solution set in total

NB : It's possible that there are some misswrite in my solution. Sorry for that. And, sorry for my bad English. ^_^

we can have every term 1 and get the solution, we can have 6 1's , we can have one 2 and rest 1, ( 7 ordered triplets), similarly for one 3 and rest 1,, and so on till 7 and rest 1, so we have 50 solutions till now for any value more, the difference becomes less than 6... hence 50is the answer

There is one trivial solution, all $x$ are one. It's obvious that only one $x$ can be another number than $1$ because the product grows faster than the sum. Hence, there are seven other solutions, if one counts the permutation this gives $49$ . Adding the trivial one gives $50$ .

The first possibility is if all numbers are 1. 1 permutation.

The next possibility is all 1's and one 2. Since we're talking about ordered pairs, order is important, so 2 could be in any of 7 places. 7 permutations.

The next several possibilities are all 1's and one 3, all 1's and one 4, etc. up to all 1's and one 8. As the non-1 number increases by 1, so does the product increase by 1. In each, the non-1 digit could be in any place. 7 permutations each.

Trying all 1's and two 2's does not yield a solution, and any numbers higher than that give a much larger product than sum.

Total permutations = 1 + 7(7) = 50

The important thing to see is that given the parameters, the function is monotonic for an increase in any of the $x_i$ 's. The lowest value is therefore given by $x_i=1$ for all $i$ . That value is 6. The only way that it will remain the same for an increase in an $x_i$ is if the product of the other $x_i$ 's is 1. Therefore the answers are : one of the $x_i$ is 2 through 8 and all the others are one, or all of them are 1. There are 7 ways to choose which one is different and 7 ways to choose how, plus one for the ${1,1,1,1,1,1,1}$ solution. That equals 50.

We first seek to find how many ones are there among the $x_i$ s. Note that if all $x_i = 1$ it clearly satisfies, so we start by considering the case where there are 6 ones. If $x_1, ..., x_6 = 1$ Then $x_7$ can be anything and it's still satisfied. Since $x_7$ has to be greater than one, there are seven solutions. Now, the number that's not one can be any of the $x_i$ s, so there are 49 solutions altogether.

If there are five ones, say $x_1, ..., x_5$ then we have $x_6 + x_7 - x_6x_7 = 1$ which is impossible since $x_6,x_7 > 1$ . Similarly, cases with less than five ones are all impossible because with numbers greater than one, the products will outgrow the sum much faster.

Let us examine $x_7$ suppose at least one of $x_1,x_2,x_3,x_4,x_5,x_6$ is not equal to 1. So, as we increase $x_7$ we see that the value of expression decreases. As 6 is the maximum value of this expression (for natural numbers), so the only possibility is that six of them are equal to 1 and the remaining equal to any natural number. therefore $no. of ways=7*7+1=\boxed{50}$

For positive integers, x1 x2 .....*xn >= x1 + x2 + ....+ xn except when one or more of those variables are 1. Here the difference is exactly 6 and the number of variables is 7. Each 1 in the set (x1, ...x7) will contribute to exactly one 1 in the sum (x1,...x7). So there are exactly 6 1s. Using this argument and using basic permutation anc combination, you will get the answer 50.

After some observation, we see that the unordered set of integers. $(1,1,1,1,1,1,1), (1,1,1,1,1,1,2), (1,1,1,1,1,1,3), (1,1,1,1,1,1,4),$ $(1,1,1,1,1,1,5), (1,1,1,1,1,1,6), (1,1,1,1,1,1,7), (1,1,1,1,1,1,8)$ are solutions to the above Diophantine equation. since there are no restrictions on $x_1, x_2, x_3, x_4, x_5, x_6, x_7$

Since when $x_i \neq 1$ , the sum of ${x_i}$ is certainly smaller than the product of ${x_i}$ , then the only possible condition is when 6 of ${x_i}$ is equal to 1, and the remainder $x_j$ fulfills $1 \leq x_j \leq 8$ .

Hence, for each value of $2 \leq x_j \leq 8$ , there are 7 arrangements of the ordered tuples. When ${x_j = 1}$ , there are 1 arrangement of the ordered tuples.

Thus, there are $7 \times 7 + 1 = \boxed {50}$ ordered tuples.

Let's consider the case when all x are equal to 1. We have as only tuple

$\left( {1,1,1,1,1,1,1} \right)$

Then let's consider the case when all the x equals 1 except 1 that could be any value. We have :

${x_1} - {x_1} = 0$

Because all the 1 cancel out. There are 7 possible numbers for x and 7 way of ordering it so we have for now 50 possible tuples.

If 2 x are different, let's see what happens :

${x_1} + {x_2} - {x_1}{x_2} = 1$

If the 2 x are equal we have :

$\begin{array}{l} {x^2} - 2x + 1 = 0\\ {\left( {x - 1} \right)^2} = 0\\ x = 1 \end{array}$

Which is one of the solutions we already found. If x is different by only 1 we have :

$x\left( {1 - x} \right) = 0$

Which says that one of the values must be x=1. If we continue on we always get that one of the x is 1.

Other cases like with 3 x will get use the same result.

The final answer is 50 tuples.