Question

if A is the hcf of 408 and 1032,then find a pair of integral values of x and y such that

A= 408x + 1032y

Answer 1

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Answer 2

408 = 8*51 = (2^3)*3*17

1032 = 8*129 = (2^3)*3*43

Therefore HCF is (2^3)*3 = 24

The (Diophantine) eqn is: 408X + 1032Y = 24

i.e. 17X + 43Y = 1 (there shouild be no common factors here)   ---------------- (1)

Taking the factors 43 and17,

43 = 2 * 17 + 9  ---- (2)

17 = 1 * 9  + 8  -----(3)

9 = 1 * 8 + 1   ------(4)

From (4), 1 = 9 - 1 * 8

now substitute for 8 from (3), giving

1 = 9 - 1 * (17 - 1 * 9)

1 = 2 * 9 - 17

now substitute for 9 from (2), giving

1 = 2 * (43 - 2 * 17) - 17

1 = -5 * 17 + 2 * 43   ------------------------(5)

Comparing (5) with (1), a pair of values for X and Y are (X,Y) = (-5, 2)

Although you don't need it, the general solution would be,

X = -5 + a*K

Y = 2 - b*K

where a is the coefft of Y, and b is the coefft of X, in (1). i.e. a = 43, b = 17