How many triples (a,b,c) of even positive integers satisfy

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

Masterscorp · Answer

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.

WizakoBaskar · Answer

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.

MathRevolution · Answer

=>

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

JeffTargetTestPrep · Answer

The only positive even perfect cube less than 50 is 2^3 = 8.

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

Kinshook · Answer

How many triples (a,b,c) of even positive integers satisfy  a^3 + b^2 + c = 50? 
If a=4; a^3 = 64>50 ; Not possible
Since a , b, c are even positive integers, only a = 2 is possible since A^3 = 8 <50

a=2
a^3 = 8

b^2 + c = 50 -8 = 42
If b=8; b^2 = 64 > 42; Not possible
b = {2,4,6}

b = 2; b^2=4; c = 42-4 = 38
b = 4; b^2 = 16; c = 42 - 16 = 26
b = 6; b^2 = 36; c = 42 - 36 = 6

(a, b, c) = {(2,2,38),(2,4,26),(2,6,6)}

IMO C

HarshavardhanR · Answer

Given:  a, b, c are all even, positive integers

Let's first think about what values "a" can take?

- a cannot be 8 or 6. a^3 will already be > 50.
- a cannot even be 4. Even 4^3 >50.
- a cannot be 0. (a is a positive integer)

Thus, a = 2. Only possibility!

Then,

a^3 + b^2 + c = 50

a = 2

=> b^2 + c = 42

Now, let' focus on what values b can take.

- Can b be 2? Yes. b = 2 -> b^2 = 4 -> c = 38. Works.
- Can b be 4? Yes. b = 4 -> b^2 = 16 -> c = 26. Works.
- Can b be 6? Yes. b = 6 -> b^2 = 36 -> c = 6. Works.
- Can b be 8? No. b = 8 -> b^2 = 64 which is > 50. Not possible.

So, we have exactly three possible triplets for (a,b,c) ->

(1) (2,2,38)
(2) (2,4,26)
(3) (2,6,6)

Answer:  3 (Choice C).

---
Harsha
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How many triples (a,b,c) of even positive integers satisfy

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How many triples (a,b,c) of even positive integers satisfy

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