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01 Jan 2021, 13:05
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E
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Difficulty:
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Question Stats:
36% (02:08) correct
64%
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wrong
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If a and b are positive integers and \(\frac{2019}{2020} < \frac{a}{b} < \frac{2020}{2021}\), what is the minimum possible value of a + b?
A. 4039
B. 4040
C. 4041
D. 8078
E. 8080
Are You Up For the Challenge: 700 Level Questions
A. 4039
B. 4040
C. 4041
D. 8078
E. 8080
Are You Up For the Challenge: 700 Level Questions
01 Jan 2021, 20:26
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Bunuel
We are looking for positive integer values for a and b, but \(\frac{2019}{2020} < \frac{a}{b} < \frac{2020}{2021}\) tells us that there are no such values if we take these values as it is because \(\frac{2019}{2020} = \frac{2020-1}{2021-1}\).
So we multiply the given fractions by 2/2 =>\(\frac{2019*2}{2020*2} < \frac{a}{b} < \frac{2020*2}{2021*2}\)
\(\frac{4038}{4040} < \frac{a}{b} < \frac{4040}{4042}\)
So \(\frac{a}{b}\) can be \(\frac{4039}{4041}\).
\(a+b=4039+4041=8080\)
E
General Discussion
29 May 2025, 02:57
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If a and b are positive integers and \(\frac{2019}{2020} < \frac{a}{b} < \frac{2020}{2021}\), what is the minimum possible value of a + b?
\(\frac{2019}{2020} < \frac{a}{b} < \frac{2020}{2021} \)
Since a & b are positive integers, let us create a gap in between to find a & b
\(\frac{4038}{4040} < \frac{a}{b} < \frac{4040}{4042}\)
Possible value of \(\frac{a}{b} = \frac{4039}{4041}\)
Minimum value of a+b = 4039+4041 = 8080
IMO E
\(\frac{2019}{2020} < \frac{a}{b} < \frac{2020}{2021} \)
Since a & b are positive integers, let us create a gap in between to find a & b
\(\frac{4038}{4040} < \frac{a}{b} < \frac{4040}{4042}\)
Possible value of \(\frac{a}{b} = \frac{4039}{4041}\)
Minimum value of a+b = 4039+4041 = 8080
IMO E
