x and y are positive integers. Is x^16 - y^8 + 345y^2 divisible by 15?
1. x is a multiple of 25 and y is a multiple of 20
2. y = x^2
[spoiler]OA: B[/spoiler]
I took more than 2 mins to solve this. I am interested in different/ faster approaches to this problem.
thanks
is x^16 - y^8 + 345y^2 divisible by 15?
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Vital,vittalgmat wrote:x and y are positive integers. Is x^16 - y^8 + 345y^2 divisible by 15?
1. x is a multiple of 25 and y is a multiple of 20
2. y = x^2
[spoiler]OA: B[/spoiler]
I took more than 2 mins to solve this. I am interested in different/ faster approaches to this problem.
thanks
Start with the easy statement,
ST2: y = x^2
This will cancel the X^16 and you will have
345something/15 = 23 something SUF
ST1: This is where you will kill some time but if you insert
x =25k
y = 20 t
YOu will go NOWHERE.
So B
LGTCH
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Vittal,
On tough DS questions or in general start off with the easier statement since it increases your probability of answering the question correct(Strictly my opinion)
Stmt II
y=x^2
y^8 = x^16
x^16 cancels leaving 345y^2 which is always divisible by 15 not matter what y is since 345 is divisible by 15
SUFF
Now its down to B) or D) (50% chance of answering this question correct).
Stmt I
If individually x^16, y^8 are both divisible by 15 then the combination x^16-y^8+345y^2 is divisible by 15. (someone correct me here if I am mistaken)
We can see that there are multiple of 25(x) and 20(y) that are and are not divisible by 15 which leads us to the fact that x^16 - y^8 + 345y^2 may or may not be divisible by 15
INSUFF
Choose B)
I am sure there may be better approaches!!!!
On tough DS questions or in general start off with the easier statement since it increases your probability of answering the question correct(Strictly my opinion)
Stmt II
y=x^2
y^8 = x^16
x^16 cancels leaving 345y^2 which is always divisible by 15 not matter what y is since 345 is divisible by 15
SUFF
Now its down to B) or D) (50% chance of answering this question correct).
Stmt I
If individually x^16, y^8 are both divisible by 15 then the combination x^16-y^8+345y^2 is divisible by 15. (someone correct me here if I am mistaken)
We can see that there are multiple of 25(x) and 20(y) that are and are not divisible by 15 which leads us to the fact that x^16 - y^8 + 345y^2 may or may not be divisible by 15
INSUFF
Choose B)
I am sure there may be better approaches!!!!
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Cramya - QUIT MATH! Move to VERBAL!cramya wrote:
I am sure there may be better approaches!!!!
I repeat one more time:
MOVE TO VERBAL!!!
LGTCH
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Is it divisible by 15? This is just asking: is it divisible by both 3 and 5. Those are easy things to check, at least when given a number.vittalgmat wrote:x and y are positive integers. Is x^16 - y^8 + 345y^2 divisible by 15?
1. x is a multiple of 25 and y is a multiple of 20
2. y = x^2
[spoiler]OA: B[/spoiler]
I took more than 2 mins to solve this. I am interested in different/ faster approaches to this problem.
thanks
345 is divisible by 3 and 5 (add the digits to check for 3), so 345y^2 is certainly divisible by 15. So all we need to know is whether x^16 - y^8 is divisible by 15 (since, if you add multiples of 15, you're guaranteed to get another multiple of 15, and if you add a multiple of 15 to something which is not a multiple of 15, you are guaranteed not to get a multiple of 15 - this is an elementary fact of number theory, but a very useful one to understand well for many questions, and of course this is true for any number, not only 15).
Statement 2 is quick to check, as pointed out above- x^16 - y^8 is equal to zero, so sufficient.
As for Statement 1, x and y could each be multiples of 15 (x could be 3*25 and y could be 3*20, for example), so it's certainly possible that x^16 - y^8 is a multiple of 15. But, x could be a multiple of 15 and y might not be (x could be 3*25 and y could be 20), in which case x^16 - y^8 will not be a multiple of 15. So insufficient.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com
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Hi Ian,
Thank you!
Question:
If individually x^16, y^8 are both divisible by 15 then the combination x^16-y^8+345y^2 is divisible by 15. (someone correct me here if I am mistaken)
Is this correct or am I mistaken?
Also lets say for some reason we find y^8 to be never be divisible by 15 can we for sure say the expression will not be divisible by 15?
What are the norms/rules when there are individual components like these in an expression and the question of divisibility by a number comes in to play?
Please advice.
Thank you!
Question:
If individually x^16, y^8 are both divisible by 15 then the combination x^16-y^8+345y^2 is divisible by 15. (someone correct me here if I am mistaken)
Is this correct or am I mistaken?
Also lets say for some reason we find y^8 to be never be divisible by 15 can we for sure say the expression will not be divisible by 15?
What are the norms/rules when there are individual components like these in an expression and the question of divisibility by a number comes in to play?
Please advice.
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- Ian Stewart
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If you add two (or more) multiples of 15, you'll always get a multiple of 15. Say x and y are multiples of 15. Then we can write:cramya wrote:Hi Ian,
Thank you!
Question:
If individually x^16, y^8 are both divisible by 15 then the combination x^16-y^8+345y^2 is divisible by 15. (someone correct me here if I am mistaken)
Is this correct or am I mistaken?
Also lets say for some reason we find y^8 to be never be divisible by 15 can we for sure say the expression will not be divisible by 15?
What are the norms/rules when there are individual components like these in an expression and the question of divisibility by a number comes in to play?
Please advice.
x = 15a for some integer a
y = 15b for some integer b
so x+y = 15a + 15b = 15(a+b), and we see that x+y is a multiple of 15.
On the other hand, say x is a multiple of 15, and z is not - say z gives a remainder of r when we divide by 15. Then we can write:
x = 15a
z = 15q + r
so x+z = 15a + 15q + r = 15(a+q) + r, and we see that we will get a remainder of r when we divide x+z by 15.
What if we add two numbers, neither of which is a multiple of 15? Here you'd need more information to say anything. We can, as above, write our numbers:
w = 15Q + R
z = 15q + r
and w+z = 15(Q + q) + R + r, but here, R+r might be a multiple of 15, so it's possible that w+z is a multiple of 15, possible that it's not.
And of course you can replace 15 throughout the above with any positive integer 2 or larger.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com
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