Question 1192366: What is the smallest positive integer that leaves a remainder of 1 when divided by 12, 25 and 2020?
Answer by ikleyn(52887)   (Show Source):
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What is the smallest positive integer that leaves a remainder of 1 when divided by 12, 25 and 2020?
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Let N be this smallest positive integer number.
Then N-1 is the smallest positive integer number which is a multiple of 12, 25 and 2020.
12 = 2^2^3;
25 = 5^2;
2020 = 20*101 = 2^2*5*101.
Therefore N-1 = Least Common Multiple of the numbers 12, 25 and 2020 = LCM (12, 25, 2020) = 2^2*3*5^2*101 = 30300.
ANSWER. The smallest positive integer that leaves a remainder of 1 when divided by 12, 25 and 2020 is 30300+1 = 30301.
Solved.
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