Question
Find the smallest number of five digits which when
divided by 15, 30, 38, 40 respectively always leaves 2 as remainder.Solution
ATQ, Let the LCM of 15, 30, 38, and 40 is = 2280 (Least Common Multiple) TO find the smallest five-digit number in the form: N=2280k+2 We need N ≥ 10000. Solve for k: 2280k≥9998 k = 5 (smallest integer) To find N: N=2280×5+2=11402
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