# Find all pairs of positive integers $(x, y)$ for which $261x + 48y = 7881$ [closed]

How do you use the Euclidean Algorithm to solve the following: Find all pairs of positive integers $$(x, y)$$ for which $$261x + 48y = 7881$$

## closed as off-topic by T. Bongers, John Omielan, mrtaurho, GNUSupporter 8964民主女神 地下教會, Delta-uFeb 28 at 9:16

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• Look online for linear diophantine equations of degree 1... – Bruno Reis Feb 28 at 4:07
• $261x+48y=7881$ iff $87x+16y=2627$ – J. W. Tanner Feb 28 at 4:21
• solutions are (5,137) and (21,50) – J. W. Tanner Feb 28 at 4:33

Euclidean algorithm:

$$261=5\times48+21$$

$$48=2\times21+6$$

$$\color{blue}{21}=3\times6+3$$

so $$\color{blue}{(261-5\times48)}=3\times(48 - 2\color{blue}{(261-5\times48)})+3;$$

i.e., $$7\times261-38\times48=3.$$

Therefore, $$2627\times7\times261-2627\times38\times48=3\times2627=7881;$$

i.e., $$18389\times261-99826\times48=7881.$$

More generally $$(18389-16k)\times261+(87k-99826)\times48=7881.$$

$$18389-16k>0$$ and $$87k-99826 > 0$$ were requested.

Thus $$k < 1149.3125$$ and $$k > 1147.4...$$; i.e., $$k = 1148$$ or $$1149.$$

Solutions are therefore $$21\times261+50\times48=7881$$ and $$5\times261+137\times48=7881.$$

• You missed positive – Bill Dubuque Feb 28 at 4:17
• Oops.................... – J. W. Tanner Feb 28 at 4:18
• Corrected...... – J. W. Tanner Feb 28 at 4:46