Product of Differences

Because $a, b, c, d,$ and $e$ are integers, our factors $5-a, 5-b, 5-c, 5-d,$ and $5-e$ are also integers. The prime factors of $28$ are $2 \times 2 \times 7$ . In order for five integers to multiply to be $28$ , at least two of them must be $1$ or $-1$ . Because our numbers are distinct and therefore our factors are distinct, we may not use $2$ twice as a factor. We may however, use $2$ and $-2$ once each as factors. If we use $-1$ and $-2$ as our only negative factor, our product will be positive, which we need. Therefore we have $-2, -1, 1, 2,$ and $7$ as our factors, which are our factors (5-a, 5-b, 5-c, 5-d,) and $5-e$ in no particular order. So $a, b, c, d,$ and $e$ are $7, 6, 4, 3,$ and $-2$ in no particular order. So our sum is $18$ .

Given that a, b, c, d, and e are five distinct integers, then (5-a), (5-b), (5-c), (5-d), and (5-e) are also 5 distinct integers, whose product (we are told) is 28. If we want to write 28 as a product of distinct positive integers, then we can use at most three integers. (1x4x7 = 28, or 1x2x14 = 28 are the only two ways to do this.) Therefore two or four of the five factors (5-a), (5-b), (5-c), (5-d), and (5-e) must be negative since their product is positive. A quick check verifies that the only option for these integers is -2, -1, 1, 2, and 7.That is,

(-2) + (-1) + 1 + 2 + 7 = (5-a) + (5-b) + (5-c) + (5-d) + (5-e), which simplifies to

7 = 25 - a - b - c - d - e, which gives us

a + b + c + d + e = 25 - 7 = 18.

Since a, b, c, d, e (call this set X) are distinct integers then 5-a, 5-b, 5-c, 5-d, 5-e (call this set Y) will be distinct integers too. We know that 28=2 2 7. So there must be at least 2 numbers in Y that the result of multiplication between them may be 1. But these 2 numbers are supposed to be distinct integers, so we can choose 1 of them is -1 and the other is 1. Since 1*(-1)=-1, we must have 1 negative integers among 3 numbers left of set Y.. With 3 integers left, we can't use some choice such as 1, 4, 7 or 1, 2, 14, because 1 will appear again and set Y won't be a set of distinct integers. So we can only have 1 choice is -2, 2, 7 for 3 integers left of set Y that meet both the distinct requirement and having a negative integers. So the answer for set Y: -1, 1, -2, 2, 7 => answer for set X: 6, 4, 7, 3, -2 Then the value of a+b+c+d+e must be equal to 18

We must express 28 as a product of 5 distinct integers 28 = 1 x 7 x (-2) x (2) x (-1)

Thus a, b, c, d, e are 4, -2, 7, 3, 6 a + b + c + d + e = 18

First note that $28=7\cdot 2\cdot 2$. Since we have five integral numbers multiplying up to 28, two of them must be of absolute value one. Now, since each integer is distinct, we need to find five different numbers whose product is 28. This has an integral solution of $-1\cdot 1\cdot -2\cdot 2\cdot 7=28$. Solving for the variables and adding, we get $6+4+7+3-2=18$.

28 = 7 x 2^2 = 7 x 2 x 2 x 1 x 1 As all a,b,c,d,e are integers all the brackets are integers. For the values to be distinct two of the brackets must be negative to account for needing to have two brackets equaling + or - 1 and + or - 2. This leads to a,b,c,d,e being equal to 4, 6, -2, 3, 7.

4 + 6 - 2 + 3 + 7 =18

Given the factor of 28 = 1,2,4,7,14,28. And what we looking for is 28 then we focusing in 2,1, and 7. then it will be -1 x 1 x -2 x 2 x 7 = 28 5-a = -1 a = 6

and so forth until you achieve: 6, 4, 7, 3, and -2 6 + 4 + 7 + 3 + (-2) = 18

28=2 2 7 =(1)(-1)(2)(-2)(7) =(5-4)(5-6)(5-3)(5-7)(5-(-2)) comparing with (5−a)(5−b)(5−c)(5−d)(5−e) a=4,b=6,c=3,d=7,e=-2 a+b+c+d+e=4+6+3+7-2=22

first: find 5 factors of 28

28 = 1 x (-1) x (-2) x 2 x 7

second: equate the 5 factors of the problem to the 5 factors you got.

(5 - a) = 1 ; (5 - b) = -1 ; (5 - c) = -2 ; (5 - d) = 2 ; (5 - e) = 7

third: evaluate the missing variables. a = 4 ; b = 6 ; c = 7 ; d = 3 ; e = -2

last: 4 + 6 + 7 + 3 + (-2) = 18

28=1 2 2 7 to get 7 by subtracting from 5 we have to subtract -2 i.e 5-(-2)=7 and here as there are 5 variables and only one equation so we have to compare the digits.There are 4 factors and 5 variables.So we have to break a factor.So ,1=1 1.but again no variables are equal,so = 1 (-1) 7 2 (-2)=28.so the values of the variables are 4,6,3,7,(-2),and there sum =18

Write 28 = 2 2 7 = (-2) (-1) 1 2 7. Comparing with 28 = (5-a)(5-b)(5-c)(5-d)(5-e), we can write a = 7, b = 6, c = 4, d = 3, and e = -2. The sum a+b+c+d+e is then 18.

We see that 28 can be written as a product of 5 distinct integers only in the way shown below: 7 * 2 * - 2 *1 * - 1. Expressing the above integers in the 5-x form we get a=-2 , b=3 , c=7, d=4 and e=6 and hence a+b+c+d+e=18

The prime factorization of 28 is 2 * 2 * 7. This factorization is also maximal in terms of the number of factors. Since we need five factors, the other two must be either 1 or -1. Since {a,b,c,d,e} must all be distinct we need to make one of the 2's a -2. Putting this together we get -1 * 1 * 2 * -2 * 7 = 28, or {a,b,c,d,e} = {6,4,3,7,-2}. Adding these we get 18.

Factorize 28 in 5 terms as:

28=1 * 1 * 2 * 2 * 7

Now try to make the product terms as each of these.

Since the answer i.e 28 is positive, we can use even number of negative terms also to get our required values. So,

5-4=1

5-6=-1

5-3=2

5-7=-2

5-(-2)=7

Therefore, (1)(-1)(2)(-2)(7)=28

and 4+6+3+7+(-2)=18

Since $a$ , $b$ , $c$ , $d$ and $e$ are distinct integers, thus each of $(5 - a), (5 - b), (5 - c), (5 - d)$ and $(5 - e)$ are distinct. So we look to split $28$ into a product of $5$ distinct integers.

We have that $28 = 2 \times 2 \times 7 = -1 \times 1 \times -2 \times 2 \times 7$ is the only possibility.

Thus $5 - a + 5 - b + 5 - c + 5 - d + 5 - e = -1 + 1 - 2 + 2 + 7$ $\Rightarrow a + b + c + d + e = 25 - 7 = 18$ .