If a, b and c are integers, is a even?

Question

(1)  15^a=9^b25^c

(2)  4^{a+5}=64^{b+1}

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av1901 · Answer

Given information: a, b, c are integers
To find: If a is even

(1)  15^a=9^b25^c

15 can be written as 3*5; 9 can be written as  3^2 ; 25 can be written as  5^2

3^a 5^a = 3^{2b} 5^{2c}

Since bases are same, we can safely say that:
 a = 2b  
 a = 2c 
Now, since both b and c are also integers, we know that even*(any integer odd or even) results in even so
a is even

SUFFICIENT

(2)  4^{a+5}=64^{b+1}

64 can be written as  4^3

4^{a+5}=4^{3b+3}

Since bases are same, we can safely say that:
 a + 5 = 3b + 3 
 a = 3b - 2

Now, here, we can have 2 cases:

Case 1: b is even => a = 3*even - 2 = even - 2 = even
Case 2: b is odd => a = 3*odd - 2 = odd - 2 = odd

So we DO NOT have a definite answer

NOT SUFFICIENT

Answer - A

gmatophobia · Answer

Question : is a even ?

Inference : a is even if its a multiple of 2.

Statement 1

(3 * 5) ^ a  =  3^{2b} * 5^{2c}

(3^a* 5^a)  =  3^{2b} * 5^{2c}

If we compare the powers of 3 we get -

a= 2b

We know that b is an integer, hence we can conclude that a is a multiple of 2.

Thus this statement is  sufficient .

We can eliminate B C E

Statement 2

4^{a+5}=64^{b+1}

2^{2*(a+5)}=2^{6*(b+1)}

2^{2a+10}=2^{6b+6}

Comparing the powers as the base is same -

2a + 10 = 6b + 6

Dividing both sides by 2

a + 5 = 3b + 3

a = 3b - 2

Now we know that that 3b is an integer, however we don't know if the integer is even or odd.

If 3b is even, a is even

If 3b is odd, a is odd.

Hence this statement by itself is not sufficient.

The correct answer is   A

tanyas311 · Answer

Given Information : a, b, c are integers.

Statement 1:    15^a=9^b25^c

Writing the prime factors of LHS and RHS:
LHS:  15^a=3^a*5^a 
RHS:  9^b25^c=3^{2b}*5^{2c}

Equating LHS & RHS:
 3^a*5^a=3^{2b}*5^{2c} 
Comparing the powers of the prime factors gives us:
a=2b (powers of 3)
a=2c (powers of 5)

Since a = 2*integer, a is an even integer.

Hence Statement 1 is Sufficient.

Statement 2:    4^{a+5}=64^{b+1}

Writing the prime factors of LHS and RHS:
LHS:  4^{a+5}=2^{2a+10} 
RHS:  64^{b+1}=2^{6b+6}

Equating LHS & RHS:
 2^{2a+10}=2^{6b+6} 
Comparing the powers of 2 gives us:
2a + 10 = 6b + 6
a= 3b - 2

Case 1: b is even
a = 3*Even - 2 will give us a as even

Case 2: b is odd
a = 3*Odd - 2 will give us a as odd

Two different answers hence   Statement 2 is Insufficient

Answer: Option A

tukaisg · Answer

This question tests the concepts of odd even , exponents and prime factorization :
question stem :

If a, b and c are integers, is a even?

Statement number 1 :

(1) 15^a=9^b25^c

We need to do a prime factorization and represent the two sides of the equation as power of prime numbers.

(3*5)^a = 3^2b * 5^2c
=> 3^a * 5^a = 3^2b * 5^2c

We compare RHS and LHS

a = 2b, a = 2c

Hence statement 1 is sufficient to conclude that a is even .We can eliminate options B,C,E . From this point the answer is either A or D.We have 50% probability to choosing the right option in case of time constraint .

Next , coming to statement number two 
(2) 4^(a+5)=64^(b+1)

=> 2 ^ (2a+10) = 2 ^ (6b+6)

comparing LHS and RHS

2a+10 = 6b+6

=> 2a = 6b -4
=> a = 3b -2

lets place some variables of b to check what we values we result we get for a

when b=1 , a=1
when b=2 , a=4
when b=3, a=7

Hence statement 2 is insufficient .

Finally , the correct option is A .

If a, b and c are integers, is a even?

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