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thanatoz
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If x and y are positive integers, is x^16 - y^8 + 345y^2 div 29 Apr 2010, 00:29
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67% (02:16) correct 33% (02:17) wrong based on 334 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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B
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If x and y are positive integers, is x^16 - y^8 + 345y^2 div 29 Apr 2010, 08:58
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thanatoz
So I know the basic rule is that, if a is a factor of both b and c, then a is a factor of (b + c)

But anything else?

Specifically, for a question like this on the math challenge. Is there a clear rule that will show quickly that since x and y are not multiples of 3, their addition cannot have it either? Thanks.

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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a is a factor of both b and c then a is a factor of b-c, b*c...
if x and y are not multiples of 3, when x^2n and y^2n are divided by 3, the remainder is 1.
1. That x is a multiple of 25 and y is a multiple of 20 doesn't give any clue of remainder of x,y when divided by 3 so insuf
2. suf
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If x and y are positive integers, is x^16 - y^8 + 345y^2 div 04 May 2010, 14:32
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thanatoz
So I know the basic rule is that, if a is a factor of both b and c, then a is a factor of (b + c)

But anything else?

Specifically, for a question like this on the math challenge. Is there a clear rule that will show quickly that since x and y are not multiples of 3, their addition cannot have it either? Thanks.

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. \(y = x^2\)
Show more

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

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 --> they both 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 (eg x=25*15 and y=20) and in this case \(x^{16}-y^8\) won't be divisible by 15 (as we can not factor out 15 from \(x^{16}-y^8\)). Not sufficient.

(2) \(y = x^2\) --> \(x^{16}-y^8=x^{16}-(x^2)^8=x^{16}-x^{16}=0\). 0 is divisible by 15. Sufficient.

Answer: B.

thanatoz
So I know the basic rule is that, if a is a factor of both b and c, then a is a factor of (b + c)

But anything else?

Specifically, for a question like this on the math challenge. Is there a clear rule that will show quickly that since x and y are not multiples of 3, their addition cannot have it either? Thanks.
Show more

If x and y both are not multiples of 3, then their sum or difference may or may not be multiple of 3:

x=2 and y=1 --> x+y=3 --> sum is multiple of 3;
x=2 and y=2 --> x-y=0 --> difference is multiple of 3;
x=2 and y=3 --> x+y=5 --> sum is not multiple of 3;
x=2 and y=0 --> x-y=2 --> difference is not multiple of 3.

BUT, if x is multiple of 3 and y is not (or vise-versa), then their sum or difference won't be multiple of 3.

Hope it helps.
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If x and y are positive integers, is x^16 - y^8 + 345y^2 div 15 Aug 2013, 22:39
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REM(x^16-y^8 +345*y^2)/15

REM(345/15)=0

So REM(x^16-y^8/15)

(1).

It may have different remainders for different values of x/y.
Insufficient

(2).

y=x^2
=> REM(x^16-x^16)/15
=>REM=0

=>Sufficient

(B) it is !
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If x and y are positive integers, is x^16 - y^8 + 345y^2 div 07 May 2023, 03:39
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Official Solution:


If \(x\) and \(y\) are positive integers, is \(x^{16} - y^8 + 345y^2\) divisible by 15?

Firstly, note that since \(345y^2\) is divisible by 15, we can ignore it, as it won't affect the remainder.

The question then becomes: is \(x^{16} - y^8\) divisible by 15?

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

In this case, both \(x\) and \(y\) could be multiples of 15 as well (e.g., \(x = 25*15\) and \(y = 20*5\)) and in this case, \(x^{16} - y^8 = 15 \cdot (...)\). This would be divisible by 15. However, one could be a multiple of 15 and the other not (e.g., \(x = 25* 15\) and \(y = 20\)), and in this case, \(x^{16} - y^8\) would not be divisible by 15, as we cannot factor out 15 from \(x^{16} - y^8\). Not sufficient.

(2) \(y = x^2\).

Substitute \(y\) with \(x^2\): \(x^{16} - y^8 = x^{16} - (x^2)^8 = x^{16} - x^{16} = 0\). Zero 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 \(k > 1\) (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 \(k > 1\) and the other 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 \(k > 1\) (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.

Thus, according to the above, the fact that \(x\) is a multiple of 25 and \(y\) is a multiple of 20 is insufficient to determine whether \(x^{16} - y^8\) is divisible by 15.


Answer: B
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If x and y are positive integers, is x^16 - y^8 + 345y^2 div 25 Dec 2025, 22:11
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