Let a, b, c, d, and e be distinct integers such that (6 − a)(6 − b)(6

45=3*3*5

Since a, b, c, d and e are distinct integers, then (6-a), (6-b), (6-c), (6-d), and (6-e) must also be distinct integers, thus (6-a), (6-b), (6-c), (6-d), and (6-e) must be -1, 1, -3, 3, and 5.

(6-a) + (6-b) + (6-c) + (6-d) + (6-e) = 30- (a+b+c+d+e) = -1+1-3+3+5 = 5
a+b+c+d+e = 30 - 5 = 25.

Answer: C.
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Since 45 = 3 * 3* 5. The whole product should result into 3, 3, and 5. as 6-1= 5 we assume that a=1, also 6-3 = 3 so let say b=3, now we need one more 3 which can be 6-9 = -3 (as a, b,c,d,e are distinct we can not have c = 3), so c= 9. the whole product is now negative, so we have to balance it, so we say d= 7 (resulting into 6-7 = -1) and last must be 5 so that we stick to our product. so e= 5 (6-5 =1).

The total = 1+3+9+7+5= 25. Answer is c

Since the prime factorization is 3*3*5

The constraint in the question is that they are distinct so the only possible way is 3,-3,5,1,-1

6-a = 3 => a=3
6-b = -3 => b=9
6-c = 5 => c=1
6-d = 1 => c=5
6-e = -1 => e = 7

Answer is C = 25

The question is exclusively asking for value of a+b+c+d+e so that should be 5.
why are we subtracting it from 30.

Check here: https://gmatclub.com/forum/let-a-b-c-d- ... l#p1274086
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Prime factorization of 45 = (3) (3) (5)

However, we have 5 factors that multiply to 45 ———> the other two factors must be 1

(1) (1) (3) (3) (5) = 45

However, each of the variables must be Distinct. Therefore, since 6 is being subtracted by each variable, we can make some of the factors Negative.

Rule: an EVEN count of (-)Negative factors will yield a (+)positive product.

We either need to have 2 of the factors be negative or 4 of the factors be negative.

Since we need each variable to be distinct, it is necessary that the factor produce different signs for (1) and (1) ——- and (3) and (3)

Change (1) (1) (3) (3) (5) ———> to ———-> (-1) (1) (-3) (3) (5)


Furthermore, since the answer asks for the addition of each variable and each factor involves 6 minus a different variable , it does not matter which variable we choose to make which factor. The addition in the end will still be the same.


(6 - a) = -1 ———-> a = 7

(6 - b) = 1 ———-> b = 5

(6 - c) = 3 ———-> c = 3

(6 - d) = -3 ————> d = 9

(6 - e) = 5 ———-> e = 1


And

( a + b + c + d + e ) = ( 7 + 5 + 3 + 9 + 1) =


25

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