[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
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- Max@Math Revolution
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A
B
C
D
E
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We should check the biggest number (a^3 because it will exceed 50 before the other number do.)
If a = 2 then a^3 = 8
If a = 3 then a^3 = 27
If a = 4 the 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. 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.
The correct answer is C.
Regards!
If a = 2 then a^3 = 8
If a = 3 then a^3 = 27
If a = 4 the 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. 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.
The correct answer is C.
Regards!
- Max@Math Revolution
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Consider the variable a first.
Since 4^3 > 50, we can only have a = 2.
If a = 2, then b^2 + c = 42 since a^3 = 2^3 = 8.
Since 8^2 = 64 > 42, we can only have b = 2, 4 or 6.
If b = 2, then b^2 + c = 2^2 + c = 4 + c = 42 and c = 38.
If b = 4, then b^2 + c = 4^2 + c = 16 + c = 42 and c = 26.
If b = 6, then b^2 + c = 6^2 + c = 36 + c = 42 and c = 6.
Thus, there are three possible triples: ( 2, 2, 38 ), ( 2, 4, 26 ) and ( 2, 6, 6 ).
Therefore, the answer is C.
Answer: C
Consider the variable a first.
Since 4^3 > 50, we can only have a = 2.
If a = 2, then b^2 + c = 42 since a^3 = 2^3 = 8.
Since 8^2 = 64 > 42, we can only have b = 2, 4 or 6.
If b = 2, then b^2 + c = 2^2 + c = 4 + c = 42 and c = 38.
If b = 4, then b^2 + c = 4^2 + c = 16 + c = 42 and c = 26.
If b = 6, then b^2 + c = 6^2 + c = 36 + c = 42 and c = 6.
Thus, there are three possible triples: ( 2, 2, 38 ), ( 2, 4, 26 ) and ( 2, 6, 6 ).
Therefore, the answer is C.
Answer: C
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The only positive even perfect cube less than 50 is 2^3 = 8.Max@Math Revolution wrote:
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
The positive even perfect squares less than 50 are 2^2 = 4, 4^2 = 16 and 6^2 = 36,
So we can have the following triples for a, b and c such that a^3 + b^2 + c = 50 :
(2, 2, 38): 2^3 + 2^2 + 38 = 8 + 4 + 38 = 50
(2, 4, 26): 2^3 + 4^2 + 26 = 8 + 16 + 26 = 50
(2, 6, 6): 2^3 + 6^2 + 6 = 8 + 36 + 6 = 50
Answer: C
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