# Prove that if either of 2a + 3b and 9a + 5b is divisible by 17, so is the other. a, b E N (Hint: 4(2a + 3b) + 9a + 5b = 17a + 17b)

Let 2a + 3b be divisible by 17.

Therefore, for some integer k, 2a + 3b = 17k….. eq (1)

Now,

9a + 5b = 17a + 17b – 4(2a + 3b)

= 17(a + b) – 4(17k) (from eq (1))

= 17(a + b – 4k)

Therefore, we can say that 9a + 5b is divisible by 17.

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