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The numbers a, b, c and d satisfy the following equations: a + b + 3c + 4d = k; 5a = 3b = 2c = d. What is the smallest value for k for which a, b, c and d are all positive integers

  1. 5a = 3b = 2c = d. d must be a multiple of 5, 3 and 2, therefore the smallest possible value for d is 30. This sets a = 6, b = 10 and c = 152) a + b + 3c + 4d = 6 + 10 + 3x15 + 4x30 = 181 k = 181
MH

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