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18 May 2018, 01:49
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
50% (02:11) correct
50%
(02:09)
wrong
based on 201
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[GMAT math practice question]
How many triples (a,b,c) of even positive integers satisfy \(a^3 + b^2 + c = 50?\)
A. one
B. two
C. three
D. four
E. five
How many triples (a,b,c) of even positive integers satisfy \(a^3 + b^2 + c = 50?\)
A. one
B. two
C. three
D. four
E. five
18 May 2018, 04:28
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We should check the biggest number (\(a^3\) first because it will exceed 50 before the other numbers do.
If a=2 then \(a^3=8\)
If a=3 then \(a^3=64\). 64 > 50, which can't be valid in this equation.
Hence, a must be 2 and \(a^3=8\).
Let's proceed by checking the next biggest number, \(b^2\).
If b=2, then \(b^2=4\). \(a^3 + b^2=8+4=12\). Hence, \(c=50-12=48\) Option 1
If b=4, then \(b^2=16.\) \(a^3 + b^2=8+16=24\). Hence, \(c=50-24=16\) Option 2
If b=6, then \(b^2=36\). \(a^3+b^2=8+36=44\). Hence, \(c=50-36=24\) Option 3
If b > 8 the equation doesn't work anymore. Therefore we can conclude that we have three different combinations of variables to solve the equation.
Correct answer is C.
Kudos if it helps.
If a=2 then \(a^3=8\)
If a=3 then \(a^3=64\). 64 > 50, which can't be valid in this equation.
Hence, a must be 2 and \(a^3=8\).
Let's proceed by checking the next biggest number, \(b^2\).
If b=2, then \(b^2=4\). \(a^3 + b^2=8+4=12\). Hence, \(c=50-12=48\) Option 1
If b=4, then \(b^2=16.\) \(a^3 + b^2=8+16=24\). Hence, \(c=50-24=16\) Option 2
If b=6, then \(b^2=36\). \(a^3+b^2=8+36=44\). Hence, \(c=50-36=24\) Option 3
If b > 8 the equation doesn't work anymore. Therefore we can conclude that we have three different combinations of variables to solve the equation.
Correct answer is C.
Kudos if it helps.
18 May 2018, 06:12
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MathRevolution
Key point
All a, b, and c are even positive integers.
\(a^3 + b^2 + c = 50?\)
The easiest part is finding possible values for 'a' as it has the highest power among the 3 unknowns. Let us start with the smallest possible even number that a can take.
If a = 2, \(a^3\) = 8. Value of \(a^3\) < 50. So, a = 2 is a possible value for a.
If a = 4, the next even number, \(a^3\) = 64 > 50. So, a = 4 is not a possible value of a.
Any other even positive integer value for a will meet the same result. So, the only value possible for a is 2.
Let us find values that b can take starting with the least even integer value for b. The question does not state a, b, and c are distinct. So, b can start with 2
If b = 2, \(b^2\) = 4.
When a = 2, b = 2, c = 50 - \(2^3 - 2^2\) = 50 - 8 - 4 = 38. Possibility 1 {2, 2, 38}
If b = 4, the next value that b can take, \(b^2\) = 16.
When a = 2, b = 4, c = 50 - \(2^3 - 4^2\) = 50 - 8 - 16 = 26. Possibility 2 {2, 4, 26}
If b = 6, \(b^2\) = 36.
When a = 2, b = 6, c = 50 - \(2^3 - 6^2\) = 50 - 8 - 36 = 6. Possibility 3 {2, 6, 6}
If b = 8, \(b^2\) = 64. \(b^2\) > 50. So, b = 8 or values greater than 8 are not possible values for b.
Therefore, 3 triples exist that satisfy the information given in the stem.
Choice C is the answer.
