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The least multiple of $$23$$ which when divided by $$18,21$$ and $$24$$ leaves the remainders $$7,10$$ and $$13$$ respectively is
Correct Answer: 3013
The conditions imply that the number is congruent to $$7$$ modulo $$18$$, $$10$$ modulo $$21$$, and $$13$$ modulo $$24$$. Solving these simultaneous congruences gives $$N=3013$$ as the smallest positive solution satisfying all three conditions. Also $$3013=23\times131$$, so it is a multiple of $$23$$.
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