Question

Positive integers x, y, z satisfy xy + z = 160. Compute the smallest possible value of x + yz

Answer 1

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158 + 2 = 160

158*1 + 2 = 160

x + yz = 158 + (1*2) = 160

79*2 + 2 = 160

x + yz = 79 + (2*2) = 83

156 + 4 = 160

156*1 + 4 = 160

x + yz = 156 + (1*4) = 160

78*2 + 4 = 160

x + yz = 78 + (2*4) = 86

52*3 + 4 = 160

x + yz =52 + (3*4) = 64

39*4 + 4 = 160

x + yz =39 + (4*4) = 55

26*6 + 4 = 160

x + yz =26 + (6*4) = 50

13*12 + 4 = 160

x + yz =13 + (12*4) = 61

the smallest possible value of x + yz. = 50

Answer 2

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158 + 2 = 160

158*1 + 2 = 160

x + yz = 158 + (1*2) = 160

79*2 + 2 = 160

x + yz = 79 + (2*2) = 83

156 + 4 = 160

156*1 + 4 = 160

x + yz = 156 + (1*4) = 160

78*2 + 4 = 160

x + yz = 78 + (2*4) = 86

52*3 + 4 = 160

x + yz =52 + (3*4) = 64

39*4 + 4 = 160

x + yz =39 + (4*4) = 55

26*6 + 4 = 160

x + yz =26 + (6*4) = 50

13*12 + 4 = 160

x + yz =13 + (12*4) = 61

the smallest possible value of x + yz. = 50

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