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It's because in this case you know B is already mod C, so you calculate (A mod C * B) mod C. In fact, mike's code started with 1, so it's actually ((1 * A) mod C * B) mod C
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It's because in this case you know B is already mod C, so you calculate (A mod C * B) mod C. In fact, mike's code started with 1, so it's actually ((1 * A) mod C * B) mod C