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# M02-02

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 00:16
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45% (medium)

Question Stats:

66% (01:50) correct 34% (02:06) wrong based on 235 sessions

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If $$x$$ 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) $$y = x^2$$

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16 Sep 2014, 00:16
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Official Solution:

First of all notice that since $$345y^2$$ is divisible by 15, we can drop it (this term won't affect the remainder).

The question becomes: is $$x^{16}-y^8$$ divisible by 15?

(1) $$x$$ is a multiple of 25, and $$y$$ is a multiple of 20. Now, both $$x$$ and $$y$$ could be multiples of 15 as well (eg $$x=25*15$$ and $$y=20*15$$) and in this case $$x^{16}-y^8=15*(...)$$ will be divisible by 15 OR one could be multiple of 15 and another not (e.g. $$x=25*15$$ and $$y=20$$) and in this case $$x^{16}-y^8$$ won't be divisible by 15 (as we cannot factor out 15 from $$x^{16}-y^8$$). Not sufficient.

(2) $$y = x^2$$. Substitute $$y$$ with $$x^2$$: $$x^{16}-y^8=x^{16}-(x^2)^8=x^{16}-x^{16}=0$$. 0 is divisible by 15 (zero is divisible by every integer except zero itself). Sufficient.

Notes for statement (1):

If integers $$a$$ and $$b$$ are both multiples of some integer $$k \gt 1$$ (divisible by $$k$$), then their sum and difference will also be a multiple of $$k$$ (divisible by $$k$$):

Example: $$a=6$$ and $$b=9$$, both divisible by 3: $$a+b=15$$ and $$a-b=-3$$, again both divisible by 3.

If out of integers $$a$$ and $$b$$ one is a multiple of some integer $$k \gt 1$$ and another is not, then their sum and difference will NOT be a multiple of $$k$$ (divisible by $$k$$):

Example: $$a=6$$, divisible by 3 and $$b=5$$, not divisible by 3: $$a+b=11$$ and $$a-b=1$$, neither is divisible by 3.

If integers $$a$$ and $$b$$ both are NOT multiples of some integer $$k \gt 1$$ (divisible by $$k$$), then their sum and difference may or may not be a multiple of $$k$$ (divisible by $$k$$):

Example: $$a=5$$ and $$b=4$$, neither is divisible by 3: $$a+b=9$$, is divisible by 3 and $$a-b=1$$, is not divisible by 3

OR: $$a=6$$ and $$b=3$$, neither is divisible by 5: $$a+b=9$$ and $$a-b=3$$, neither is divisible by 5

OR: $$a=2$$ and $$b=2$$, neither is divisible by 4: $$a+b=4$$ and $$a-b=0$$, both are divisible by 4.

So according to above info that $$x$$ is a multiple of 25, and $$y$$ is a multiple of 20 tells us nothing whether $$x^{16}-y^8$$ is divisible by 15.

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07 Jan 2015, 09:28
1
I think this question is good and helpful.
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10 Feb 2016, 05:55
I think this is a high-quality question and I agree with explanation.
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09 Mar 2016, 00:45
This is an Excellent Question
here is my approach
here in statement 1 => 1st is a non multiple of 15 and the rest two terms are
hence the result is a non multiple of 15 as we know that a multiple + non multiple is a non multiple.
statement 2 is sufficient as 345 is a multiple of 15 and rest two terms cancel out
So B
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23 May 2016, 18:08
Thanks for the question, got caught by it - was struggling with the (1) condition

stonecold:
stonecold wrote:
This is an Excellent Question
here is my approach
here in statement 1 => 1st is a non multiple of 15 and the rest two terms are
hence the result is a non multiple of 15 as we know that a multiple + non multiple is a non multiple.
statement 2 is sufficient as 345 is a multiple of 15 and rest two terms cancel out
So B

I quite not understand your reason for y^8 is divisible by 15 - the statement only says it's a multiple of 20 - could you explain a little bit more?
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09 Jan 2018, 22:24
I think the right answer is D.
x16−y8 has the x2+y as a factor which is 645 and divisible by 15 ==> D is correct.
x16−y8 = (x8+y4)(x8-y4)
(x8-y4)=(x4+y2)(x4-y2)
(x4-y2)=(x2+y)(x2-y)
you see my point?
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09 Jan 2018, 23:10
Java85 wrote:
I think the right answer is D.
x16−y8 has the x2+y as a factor which is 645 and divisible by 15 ==> D is correct.
x16−y8 = (x8+y4)(x8-y4)
(x8-y4)=(x4+y2)(x4-y2)
(x4-y2)=(x2+y)(x2-y)
you see my point?

You are not right. x is a multiple of 25, and y is a multiple of 20 does NOT mean that x^2 + y is a multiple of 645. You could check it VERY easily by picking some numbers: x = 25 any y = 40 --> x^2 + y = 665, which is NOT a multiple of 645. Please re-read the solution above carefully, it explains in detail (with examples) why the first statement is not sufficient.
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12 May 2018, 15:54
Hello Bunuel,

I approached this question using the remainder theorem. According to the theorem, the remainder of the expression [AxBxC]/M is the same as the remainder of expression [ArxBrxCr]/M (where Ar, Br, and Cr are the respective remainders when A, B, and C are divided by M. (For example, remainder of [16x19]/15 == remainder [1x3]/15 = 3).
Please let me know if you agree with this.

If so, then taking x=25 and y=40, we get :

Rem [(25^16 – 40^8)]/15
== [10^16 – 10^8]/15 (because 25/15 gives a remainder of 10, and 40/15 gives a remainder of 10)
= [100^8 – 10^8]/15 (rewriting 10^16 as 100^8 )
= [10^8 – 10^8]/15 (remainder of 100/15 = 10)
= 0

Therefore, if we take x=25, and y =40, the expression IS divisible by 15.

Similarly, I checked the value of expression $$x^16 - y^8$$ for different values of (x,y) such as (50,40), (25,20), (50,20), etc. and got the remainder to be 0.

Please let me know where am I going wrong.

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17 May 2018, 05:49
Hello Bunuel, or other experts, any thoughts on my post above?

Regards,
Paresh
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05 Jun 2018, 19:23
prajpurohit wrote:

I approached this question using the remainder theorem. According to the theorem, the remainder of the expression [AxBxC]/M is the same as the remainder of expression [ArxBrxCr]/M (where Ar, Br, and Cr are the respective remainders when A, B, and C are divided by M. (For example, remainder of [16x19]/15 == remainder [1x3]/15 = 3).
Please let me know if you agree with this.

If so, then taking x=25 and y=40, we get :

Rem [(25^16 – 40^8)]/15
== [10^16 – 10^8]/15 (because 25/15 gives a remainder of 10, and 40/15 gives a remainder of 10)
= [100^8 – 10^8]/15 (rewriting 10^16 as 100^8 )
= [10^8 – 10^8]/15 (remainder of 100/15 = 10)
= 0

Therefore, if we take x=25, and y =40, the expression IS divisible by 15.

Similarly, I checked the value of expression $$x^16 - y^8$$ for different values of (x,y) such as (50,40), (25,20), (50,20), etc. and got the remainder to be 0.

Please let me know where am I going wrong.

prajpurohit I think this is an interesting thought. To use the theorem as you outlined, however, wouldn't the expression need to be in the form A*B*C instead of the form A-B-C? It does not seem that the theorem would be applicable to the expression in the question because it is subtraction, not multiplication.
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04 Jul 2018, 02:55
Bunuel wrote:
If $$x$$ 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) $$y = x^2$$

Another way of solving this question.

Following are a few divisibility rules.

1 a^n - b^n is divisible by (a+b) if n = even

2 a^n - b^n is divisible by (a-b) if n = even/odd

3 a^n + b^n is divisible by (a+b) if n = odd

Now, $$x^{16} - y^8 + 345y^2$$ can be written as (x^2)^8 - y^8. Im not including 345y^2 as it is divisible by 15.

(1) $$x$$ is a multiple of 25, and $$y$$ is a multiple of 20

If a= 25; b= 20. Then a+b = 45. Then x^{16} - y^8 is divisible by 15

But if a = 50; b = 60, Then a+b = 110, So x^{16} - y^8 is NOT divisible by 15.

Hence, NOT SUFFICIENT

(2)

Substituting y = x^2. We get x^16 - x^16 = 0. 0 is divisible by 15.

Hope this helps.
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28 Sep 2018, 21:25
I think this is a high-quality question and I don't agree with the explanation. using x^2-Y^2 = (x+y)(x-y) multiple times we get 25^32-2^16 as 5^8((5^3+2^2)(5^3-2^2)(5^6+2^4)(5^12+2^8))

that gives us a multiple of 5 multipled with 129 which is a multiple of 3

hence a multiple of 15. thus 1 is also sufficient
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29 Sep 2018, 00:14
Chinmoy25 wrote:
I think this is a high-quality question and I don't agree with the explanation. using x^2-Y^2 = (x+y)(x-y) multiple times we get 25^32-2^16 as 5^8((5^3+2^2)(5^3-2^2)(5^6+2^4)(5^12+2^8))

that gives us a multiple of 5 multipled with 129 which is a multiple of 3

hence a multiple of 15. thus 1 is also sufficient

If x=25∗15 and y=20, then x^16 −y^8 + 345y^2 = 4304672099999999999999999999974400138000, which is NOT a multiple of 15. So, (1) is not sufficient.
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01 Jul 2019, 10:29
Bunuel is this sound logic?

(1) $$x$$ is a multiple of 25, and $$y$$ is a multiple of 20
If we rewrite ... x=25^n and y=20^m
(25^n)^16 - (20^m)^8 + 345y*(20^m)^2
We know all 3 are divisible by 5.
We only know for sure the 3rd term is divisible by 3, so it's not sufficient.
This depends on n and m (e.g. n=1, m=1 is no but n=3, m=3 is yes)

(2) $$y = x^2$$
x^16 - x^16 + 345x^4
Since 345 is divisible by 15, the product of 345 and x^4 is as well; sufficient.
M02-02   [#permalink] 01 Jul 2019, 10:29
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