This is #1 on pg 81 of the MGMAT Number Properties book:
If x, y, and z are integers, is x even?
1) 10^x = (4^y)(5^z)
2) 3^x+5 = 27^y+1
Can someone do this problem for me with a step by step process?
Thanks.
MGMAT Number Properties Problem
This topic has expert replies
-
- Master | Next Rank: 500 Posts
- Posts: 118
- Joined: Mon May 21, 2012 10:07 pm
- Thanked: 23 times
- Followed by:4 members
Hi rastis
1) 10^x = (4^y)(5^z)
=>(2*5)^x = 2^(2y) * 5^z
=>2^x * 5^x = 2^(2y) * 5^z
Comparing the powers of 2 we get x = 2y
Therefore x is even. Suff
2) 3^x+5 = 27^y+1
I am assuming it means 3^(x+5) = 27^ (y+1)
=> 3^(x+5) = 3^3(y+1)
Comparing powers of 3
x+5 = 3y + 3
=>x= 3y-2
For y = 0 , x= -2 .... Even
For y = 1, x= 1...... Odd
THerefore Insuff
ANS A
hope it helps
1) 10^x = (4^y)(5^z)
=>(2*5)^x = 2^(2y) * 5^z
=>2^x * 5^x = 2^(2y) * 5^z
Comparing the powers of 2 we get x = 2y
Therefore x is even. Suff
2) 3^x+5 = 27^y+1
I am assuming it means 3^(x+5) = 27^ (y+1)
=> 3^(x+5) = 3^3(y+1)
Comparing powers of 3
x+5 = 3y + 3
=>x= 3y-2
For y = 0 , x= -2 .... Even
For y = 1, x= 1...... Odd
THerefore Insuff
ANS A
hope it helps
-
- Master | Next Rank: 500 Posts
- Posts: 118
- Joined: Mon May 21, 2012 10:07 pm
- Thanked: 23 times
- Followed by:4 members
Hi rastis
1) 10^x = (4^y)(5^z)
=>(2*5)^x = 2^(2y) * 5^z
=>2^x * 5^x = 2^(2y) * 5^z
Comparing the powers of 2 we get x = 2y
Therefore x is even. Suff
2) 3^x+5 = 27^y+1
I am assuming it means 3^(x+5) = 27^ (y+1)
=> 3^(x+5) = 3^3(y+1)
Comparing powers of 3
x+5 = 3y + 3
=>x= 3y-2
For y = 0 , x= -2 .... Even
For y = 1, x= 1...... Odd
THerefore Insuff
ANS A
hope it helps
1) 10^x = (4^y)(5^z)
=>(2*5)^x = 2^(2y) * 5^z
=>2^x * 5^x = 2^(2y) * 5^z
Comparing the powers of 2 we get x = 2y
Therefore x is even. Suff
2) 3^x+5 = 27^y+1
I am assuming it means 3^(x+5) = 27^ (y+1)
=> 3^(x+5) = 3^3(y+1)
Comparing powers of 3
x+5 = 3y + 3
=>x= 3y-2
For y = 0 , x= -2 .... Even
For y = 1, x= 1...... Odd
THerefore Insuff
ANS A
hope it helps
-
- Master | Next Rank: 500 Posts
- Posts: 118
- Joined: Mon May 21, 2012 10:07 pm
- Thanked: 23 times
- Followed by:4 members
Hi rastis
Let me start from the basic. In a question like this,
1. You need to reduce the terms on both side of the equation in terms of prime factors.
That is what I have done here.
10^x = (4^y)(5^z) =>(2*5)^x = 2^(2y) * 5^z
2. You need to compare the powers/ exponent of the common bases.
In this case, we have both sides in powers of 2 and 5
Comparing powers of 2 , we get x=2y
Comparing powers of 5 , we get x=z
Now, the question asks us whether x is even or not.
From x= z, we can't say as we don't know the value of z
But, from x=2y , x will always be even irrespective of the value of y
Hope it's clear now
Let me start from the basic. In a question like this,
1. You need to reduce the terms on both side of the equation in terms of prime factors.
That is what I have done here.
10^x = (4^y)(5^z) =>(2*5)^x = 2^(2y) * 5^z
2. You need to compare the powers/ exponent of the common bases.
In this case, we have both sides in powers of 2 and 5
Comparing powers of 2 , we get x=2y
Comparing powers of 5 , we get x=z
Now, the question asks us whether x is even or not.
From x= z, we can't say as we don't know the value of z
But, from x=2y , x will always be even irrespective of the value of y
Hope it's clear now