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Positive integers a, b, c, d, and e such that
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22 Jun 2026, 05:12
22 Jun 2026, 05:12
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Expert Reply
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Question Stats:
45% (03:26) correct
54% (02:46) wrong based on 11 sessions
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Positive integers \(\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}\) and e are such that \(\mathrm{a}<\mathrm{b}<\mathrm{c}<\mathrm{d}< e\). If the average (arithmetic mean) of the five numbers is 6 and \(\mathrm{d}-\mathrm{b}=3\) , then what is the greatest possible range of the five numbers?
A. 12
B. 17
C. 18
D. 19
E. 20
A. 12
B. 17
C. 18
D. 19
E. 20
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Re: Positive integers a, b, c, d, and e such that
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24 Jun 2026, 07:04
24 Jun 2026, 07:04
1
Expert Reply
Explanation
Let the five integers be \(a<b<c<d<e.\)
Since their average is 6,
\(a+b+c+d+e=5*6=30.\)
Also, \(d-b=3.\) We want to maximize the range: \(e−a\).
Since \(d=b+3,\) \(a+b+c+(b+3)+e=30,\) so
\(a+2b+c+e=27.\)
To make \(e-a\) as large as possible, we want a as small as possible and e as large as possible. Because the numbers are positive integers and strictly increasing:
smallest possible a=1,
then \(b≥2,\)
and since \(d=b+3,\) we need a c with \(b<c<b+3.\)
Thus \(c\) can only be \(b+1\) or \(b+2.\)
Substitute \(a=1:\)
\(2b+c+e=26.\)
To maximize e, minimize \(2b+c.\)
Take the smallest possible \(b=2.\) Then c must be 3 (since 2<c<5).
Thus \(e=26-(2⋅2+3)=26-7=19.\)
The numbers are \(1,2,3,5,19,\) which satisfy all conditions and have sum 30. Therefore the greatest possible range is
\(e−a=19−1=18.\)
Answer: C
Let the five integers be \(a<b<c<d<e.\)
Since their average is 6,
\(a+b+c+d+e=5*6=30.\)
Also, \(d-b=3.\) We want to maximize the range: \(e−a\).
Since \(d=b+3,\) \(a+b+c+(b+3)+e=30,\) so
\(a+2b+c+e=27.\)
To make \(e-a\) as large as possible, we want a as small as possible and e as large as possible. Because the numbers are positive integers and strictly increasing:
smallest possible a=1,
then \(b≥2,\)
and since \(d=b+3,\) we need a c with \(b<c<b+3.\)
Thus \(c\) can only be \(b+1\) or \(b+2.\)
Substitute \(a=1:\)
\(2b+c+e=26.\)
To maximize e, minimize \(2b+c.\)
Take the smallest possible \(b=2.\) Then c must be 3 (since 2<c<5).
Thus \(e=26-(2⋅2+3)=26-7=19.\)
The numbers are \(1,2,3,5,19,\) which satisfy all conditions and have sum 30. Therefore the greatest possible range is
\(e−a=19−1=18.\)
Answer: C
Positive integers a, b, c, d, and e such that
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08 Sep 2026, 00:28
08 Sep 2026, 00:28
1
minimize a and maximize e
minimum value a i 1s
now to maximize e, minimize all others
different integers so b will be 2 , c will be 3 , d has to be b+3 = 5
total is 30, so e will be 19 after subsituting
range 18
minimum value a i 1s
now to maximize e, minimize all others
different integers so b will be 2 , c will be 3 , d has to be b+3 = 5
total is 30, so e will be 19 after subsituting
range 18
