Four positive integers A, B, C and D have a sum of 36. If A+2 = B-2 = C+2 = D-2, what is the value of the product ABCD?
A + B + C + D = 36
A + 2 = B - 2 = C x 2 = D ÷ 2
A = B - 4
B = 2C + 2
C = D ÷ 4
D = 2A + 4
Let's find A first.
D = 2A + 4
C = (2A + 4) ÷ 4
= 0.5A + 1
B = A + 4
A + B + C + D = 36
A + A + 4 + 0.5A + 1 + 2A + 4 = 36
4.5A + 9 = 36
4.5 A = 27
A = 6
Since A = 6,
B = A + 4
= (6) + 4
= 10
C = 0.5A + 1
= 0.5(6) + 1
= 3 + 1
= 4
D = 2A + 4
= 2(6) + 4
= 12 + 4
= 16
Which means the product of the four, ABCD = 6 * 10 * 4 * 16 = 3840, which is your answer
+36916 A = C and B = D
a+ a + b + b = 36
a+b =18
a+2 = b-2
then a = 7 b = 11 c = 11 and d= 7
7 * 11*7*11 = 170
+36916 *** edit *** ( the last line is incorrect ...I added instead of multiplying)
A = C and B = D
a+ a + b + b = 36
a+b =18
a+2 = b-2
then a = 7 b = 11 c = 11 and d= 7
7 * 11*7*11 = 170
7 * 11 *7 * 11 = 5929
