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The sum of n consecutive positive integers is 45

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Re: The sum of n consecutive positive integers is 45 [#permalink]

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New post 29 Jul 2012, 06:46
Q5. If a and b are integers, and a not= b, is |a|b > 0?
(1) |a^b| > 0
(2) |a|^b is a non-zero integer

The question can be rephrased as are a and b distinct non-zero integers and is b positive?
(1) We can deduce that \(a\neq0\), but b can be 0.
Take a=1, b=0, then \(|a^0|=1>0\) and |a|b=0.
But if we take a=2 and b=1, then \(|a^b|=2>0\) and |a|b=2>0.
Not sufficient.

(2) Again \(a\neq0\). But a can be 1, in which case b can be negative or 0.
Not sufficient.

(1) and (2) Since we cannot guarantee that \(b\neq0\), again not sufficient.

Answer E
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New post 29 Jul 2012, 06:56
Q6. If M and N are integers, is (10^M + N)/3 an integer?
1. N = 5
2. MN is even

(1) M can be negative, in which case \((10^M + N)/3\) is not an integer.
But for example, if M=0, N=5, \((10^M + N)/3=6/3=2\) is an integer.
Not sufficicent.

(2) MN even means at least one of the two numbers must be even.
We can take M=0 and N=3, then \((10^M + N)/3=4/3\) it's not an integer.
But for M=1 and N=2, \((10^M + N)/3=12/3=4\) it is an integer.
Not sufficient.

(1) and (2): N=5, but since MN is even, it means that M must be even.
Still, we cannot exclude the case M negative, for which the given expression is not an integer.
Therefore, not sufficient.

Answer E
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New post 29 Jul 2012, 07:04
Q8. If x and y are non-zero integers and |x| + |y| = 32, what is xy?
(1) -4x - 12y = 0
(2) |x| - |y| = 16

(1) From the given equality we get 4x=-12y, or x=-3y, which gives |x|=3|y|.
We can deduce that |y|=8, |x|=24, and |x||y|=|xy|=192.
Not sufficient, because xy=192 or -192.

(2) Since |x|=|y|+16, we find again that |y|=8, |x|=24, and |x||y|=|xy|=192.
Same situation as in (1), not sufficient.

(1) and (2) together cannot help, as seen above.

Answer E
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Re: The sum of n consecutive positive integers is 45 [#permalink]

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EvaJager wrote:
Q8. If x and y are non-zero integers and |x| + |y| = 32, what is xy?
(1) -4x - 12y = 0
(2) |x| - |y| = 16

(1) From the given equality we get 4x=-12y, or x=-3y, which gives |x|=3|y|.
We can deduce that |y|=8, |x|=24, and |x||y|=|xy|=192.
Not sufficient, because xy=192 or -192.

(2) Since |x|=|y|+16, we find again that |y|=8, |x|=24, and |x||y|=|xy|=192.
Same situation as in (1), not sufficient.

(1) and (2) together cannot help, as seen above.

Answer E


Answer to this question is A, not E.

If x and y are non-zero integers and |x| + |y| = 32, what is xy?

(1) \(-4x - 12y = 0\) --> \(x+3y=0\) --> \(x=-3y\) --> \(x\) and \(y\) have opposite signs --> so either \(|x|=x\) and \(|y|=-y\) OR \(|x|=-x\) and \(|y|=y\) --> either \(|x|+|y|=-x+y=3y+y=4y=32\): \(y=8\), \(x=-24\), \(xy=-24*8\) OR \(|x|+|y|=x-y=-3y-y=-4y=32\): \(y=-8\), \(x=24\), \(xy=-24*8\), same answer. Sufficient.

(2) \(|x| - |y| = 16\). Sum this one with th equations given in the stem --> \(2|x|=48\) --> \(|x|=24\), \(|y|=8\). \(xy=-24*8\) (x and y have opposite sign) or \(xy=24*8\) (x and y have the same sign). Multiple choices. Not sufficient.

Answer: A.

Hope it's clear.
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EvaJager wrote:
Q7

Since the given equality must hold for any value of x, if we substitute x = 0, we obtain \(c=d^2\).

Then, we can immediately see that (1) alone is sufficient, but (2) is not.

Answer A


The answer to this question is D, not A.

If b, c, and d are constants and x^2 + bx + c = (x + d)^2 for all values of x, what is the value of c?

\(x^2 + bx + c = (x + d)^2\) --> \(x^2+bx+c=x^2+2dx+d^2\) --> \(bx+c=2dx+d^2\).

Now, as above expression is true "for ALL values of \(x\)" then it must hold true for \(x=0\) too --> \(c=d^2\).

Next, substitute \(c=d^2\) --> \(bx+d^2=2dx+d^2\) --> \(bx=2dx\) --> again it must be true for \(x=1\) too --> \(b=2d\).

So we have: \(c=d^2\) and \(b=2d\). Question: \(c=?\)

(1) \(d=3\) --> as \(c=d^2\), then \(c=9\). Sufficient
(2) \(b=6\) --> as \(b=2d\) then \(d=3\) --> as \(c=d^2\), then \(c=9\). Sufficient.

Answer: D.

Hope it's clear.
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Re: The sum of n consecutive positive integers is 45 [#permalink]

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New post 29 Jul 2012, 07:17
Q9. Is the integer n odd
(1) n is divisible by 3
(2) 2n is divisible by twice as many positive integers as n

(1) There are odd as well as even numbers, which are divisible by 3 (3 and 6, for example).
Not sufficient.

(2) The statement is true for any non-zero integer. But still, n can be even or odd.
Try n=3 and n=4, for example.
Remark: n cannot be 0, as zero has infinitely many divisor (any non-zero integer is a factor of 0).
Not sufficient.

(1) and (2): From the above, it is clear that not sufficient.

Answer E
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Re: The sum of n consecutive positive integers is 45 [#permalink]

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New post 29 Jul 2012, 07:25
EvaJager wrote:
Q9. Is the integer n odd
(1) n is divisible by 3
(2) 2n is divisible by twice as many positive integers as n

(1) There are odd as well as even numbers, which are divisible by 3 (3 and 6, for example).
Not sufficient.

(2) The statement is true for any non-zero integer. But still, n can be even or odd.
Try n=3 and n=4, for example.
Remark: n cannot be 0, as zero has infinitely many divisor (any non-zero integer is a factor of 0).
Not sufficient.

(1) and (2): From the above, it is clear that not sufficient.

Answer E


Answer to this question is B, not E.

Is the integer n odd ?

(1) n is divisible by 3. Clearly insufficient, consider n=3 and n=6.

(2) 2n is divisible by twice as many positive integers as n
TIP:
When odd number n is doubled, 2n has twice as many factors as n.
Thats because odd number has only odd factors and when we multiply n by two, we remain all these odd factors as divisors and adding exactly the same number of even divisors, which are odd*2.

Sufficient.

Answer: B.

For more on this topic check Number Theory chapter of Math Book: math-number-theory-88376.html

Hope it helps.
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Re: The sum of n consecutive positive integers is 45 [#permalink]

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New post 29 Jul 2012, 07:31
Q10. The sum of n consecutive positive integers is 45. What is the value of n?
(1) n is odd
(2) n >= 9

(1) When n is odd, the sum of the consecutive numbers is a multiple of the middle term.
For example, we can have 9 terms, with the middle term 5, so the numbers are 1,2,3,4,5,6,7,8,9 or we can have 5 terms, with the middle term 9 - 7,8,9,10,11.
Not sufficient.

(2) If n=9, the numbers are as above from 1 to 9.
If n is greater than 9, than the average of the numbers is less than 45/9 = 5, and necessarily the sequence must contain non-positive numbers, which is impossible. So, n must be 9, sufficient.

Answer B
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Re: The sum of n consecutive positive integers is 45 [#permalink]

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New post 29 Jul 2012, 07:38
Bunuel wrote:
EvaJager wrote:
Q8. If x and y are non-zero integers and |x| + |y| = 32, what is xy?
(1) -4x - 12y = 0
(2) |x| - |y| = 16

(1) From the given equality we get 4x=-12y, or x=-3y, which gives |x|=3|y|.
We can deduce that |y|=8, |x|=24, and |x||y|=|xy|=192.
Not sufficient, because xy=192 or -192.

(2) Since |x|=|y|+16, we find again that |y|=8, |x|=24, and |x||y|=|xy|=192.
Same situation as in (1), not sufficient.

(1) and (2) together cannot help, as seen above.

Answer E


Answer to this question is A, not E.

If x and y are non-zero integers and |x| + |y| = 32, what is xy?

(1) \(-4x - 12y = 0\) --> \(x+3y=0\) --> \(x=-3y\) --> \(x\) and \(y\) have opposite signs --> so either \(|x|=x\) and \(|y|=-y\) OR \(|x|=-x\) and \(|y|=y\) --> either \(|x|+|y|=-x+y=3y+y=4y=32\): \(y=8\), \(x=-24\), \(xy=-24*8\) OR \(|x|+|y|=x-y=-3y-y=-4y=32\): \(y=-8\), \(x=24\), \(xy=-24*8\), same answer. Sufficient.

(2) \(|x| - |y| = 16\). Sum this one with th equations given in the stem --> \(2|x|=48\) --> \(|x|=24\), \(|y|=8\). \(xy=-24*8\) (x and y have opposite sign) or \(xy=24*8\) (x and y have the same sign). Multiple choices. Not sufficient.

Answer: A.

Hope it's clear.


Yes, you're right. completely forgot about x and y having opposite signs in (1).
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Re: The sum of n consecutive positive integers is 45 [#permalink]

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New post 29 Jul 2012, 07:39
Bunuel wrote:
EvaJager wrote:
Q7

Since the given equality must hold for any value of x, if we substitute x = 0, we obtain \(c=d^2\).

Then, we can immediately see that (1) alone is sufficient, but (2) is not.

Answer A


The answer to this question is D, not A.

If b, c, and d are constants and x^2 + bx + c = (x + d)^2 for all values of x, what is the value of c?

\(x^2 + bx + c = (x + d)^2\) --> \(x^2+bx+c=x^2+2dx+d^2\) --> \(bx+c=2dx+d^2\).

Now, as above expression is true "for ALL values of \(x\)" then it must hold true for \(x=0\) too --> \(c=d^2\).

Next, substitute \(c=d^2\) --> \(bx+d^2=2dx+d^2\) --> \(bx=2dx\) --> again it must be true for \(x=1\) too --> \(b=2d\).

So we have: \(c=d^2\) and \(b=2d\). Question: \(c=?\)

(1) \(d=3\) --> as \(c=d^2\), then \(c=9\). Sufficient
(2) \(b=6\) --> as \(b=2d\) then \(d=3\) --> as \(c=d^2\), then \(c=9\). Sufficient.

Answer: D.

Hope it's clear.


Again, you are right. Did not check further that given b, one can find d, so c...Not worth speeding!
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Re: The sum of n consecutive positive integers is 45 [#permalink]

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New post 29 Jul 2012, 07:41
Bunuel wrote:
EvaJager wrote:
Q9. Is the integer n odd
(1) n is divisible by 3
(2) 2n is divisible by twice as many positive integers as n

(1) There are odd as well as even numbers, which are divisible by 3 (3 and 6, for example).
Not sufficient.

(2) The statement is true for any non-zero integer. But still, n can be even or odd.
Try n=3 and n=4, for example.
Remark: n cannot be 0, as zero has infinitely many divisor (any non-zero integer is a factor of 0).
Not sufficient.

(1) and (2): From the above, it is clear that not sufficient.

Answer E


Answer to this question is B, not E.

Is the integer n odd ?

(1) n is divisible by 3. Clearly insufficient, consider n=3 and n=6.

(2) 2n is divisible by twice as many positive integers as n
TIP:
When odd number n is doubled, 2n has twice as many factors as n.
Thats because odd number has only odd factors and when we multiply n by two, we remain all these odd factors as divisors and adding exactly the same number of even divisors, which are odd*2.

Sufficient.

Answer: B.

For more on this topic check Number Theory chapter of Math Book: math-number-theory-88376.html

Hope it helps.


Oooops! Terrible miss, as the factor 2 for an even number is not counted twice.
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New post 25 Nov 2012, 11:49
Bunuel wrote:
ANSWERS:
2. Is a product of three integers XYZ a prime?
(1) X=-Y
(2) Z=1

(1) x=-y --> for xyz to be a prime z must be -p AND x=-y shouldn't be zero. Not sufficient.
(2) z=1 --> Not sufficient.
(1)+(2) x=-y and z=1 --> x and y can be zero, xyz=0 not prime OR xyz is negative, so not prime. In either case we know xyz not prime.

Answer: C

i did not understand the explanation you gave..... a prime is a number which is divisible by 1 and itself right? if x,y,z are three integers..... and for it to be prime.... two for those three integers should be 1 or -1 or 1,-1.... so the third one be prime number or negative prime number....
(1) says two of them are equal in magnitude... so z can be -p to be prime or negative composite number or positive non prime in either case not sufficient...
(2) z=1 nothing said about x,y..... not sufficient

(1) + (2)
product will be a positive or negative composite number or 1..... so not a prime which is sufficient....
am i thinking correctly?

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New post 26 Nov 2012, 02:40
Amateur wrote:
Bunuel wrote:
ANSWERS:
2. Is a product of three integers XYZ a prime?
(1) X=-Y
(2) Z=1

(1) x=-y --> for xyz to be a prime z must be -p AND x=-y shouldn't be zero. Not sufficient.
(2) z=1 --> Not sufficient.
(1)+(2) x=-y and z=1 --> x and y can be zero, xyz=0 not prime OR xyz is negative, so not prime. In either case we know xyz not prime.

Answer: C

i did not understand the explanation you gave..... a prime is a number which is divisible by 1 and itself right? if x,y,z are three integers..... and for it to be prime.... two for those three integers should be 1 or -1 or 1,-1.... so the third one be prime number or negative prime number....
(1) says two of them are equal in magnitude... so z can be -p to be prime or negative composite number or positive non prime in either case not sufficient...
(2) z=1 nothing said about x,y..... not sufficient

(1) + (2)
product will be a positive or negative composite number or 1..... so not a prime which is sufficient....
am i thinking correctly?


We have that \(x=-y\) and \(z=1\), thus \(xyz=-x^2\). Now, \(-x^2\leq{0}\), thus it cannot be a prime number (only positive numbers are primes).

Hope it's clear.
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Re: The sum of n consecutive positive integers is 45 [#permalink]

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New post 06 Jan 2013, 04:24
Q Is a product of three integers XYZ a prime?
(1) X=-Y
(2) Z=1

I'm unable to understand why (1) X=-Y is not sufficient to answer the question?

In all cases if (1) X=-Y, XYZ can not be a prime number, whether X, Y being 0 or Z being negative. I may be missing out something very basic, please help.

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Re: The sum of n consecutive positive integers is 45 [#permalink]

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New post 07 Jan 2013, 04:20
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pashraddha wrote:
Q Is a product of three integers XYZ a prime?
(1) X=-Y
(2) Z=1

I'm unable to understand why (1) X=-Y is not sufficient to answer the question?

In all cases if (1) X=-Y, XYZ can not be a prime number, whether X, Y being 0 or Z being negative. I may be missing out something very basic, please help.


If \(x=-1\), \(y=1\), \(z=-7\), then \(xyz=(-1)*1*(-7)=7=prime\).

Check here: the-sum-of-n-consecutive-positive-integers-is-85413-20.html#p667155 and here: the-sum-of-n-consecutive-positive-integers-is-85413.html#p640133

Hope it helps.
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Re: The sum of n consecutive positive integers is 45 [#permalink]

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New post 13 Sep 2013, 10:13
Bunuel wrote:
GMAT40 wrote:
Hi Bunnel,
Can you please explain Q3

(1) wx+cx=aaa (111, 222, ... 999=37*k) --> As x is the units digit in both numbers, a can be 1,4,6 or 9 (2,3,7 out because x^2 can not end with 2,3, or 7. 5 is out because in that case x also should be 5 and we know that x and a are distinct numbers).
1 is also out because 111=37*3 and we need 2 two digit numbers.
444=37*12 no good we need units digit to be the same.
666=37*18 no good we need units digit to be the same.
999=37*27 is the only possibility all digits are distinct except the unit digits of multiples.


Can you please elaborate your question? Thank you.


My question was how did you arrive at 37 * K
and how did you rule out 2, 3, 7

going thru your solution once again i could understand this but had already posted my query

Thanks

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Re: The sum of n consecutive positive integers is 45 [#permalink]

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New post 22 Nov 2013, 10:20
I had a problem with number 8 and 9
8
Why is statement 2 not sufficient? I mean |x| means positive x so cant we then arrange it as |x|-|y|=16
Same as x-y=16 then use substitution method with the equation |x|+|y|=32 above?
9
You mentioned that “when odd number n is doubleb, 2n has twice as many factors as n”
Is this always the case? Let’s say our odd number is 15 ,it has four factors 5 ,1,15and 3.when doubled it becomes 30.30 has 30,1,2,3,5 factors. Just one more factor than 15.
My understanding for YES/NO DS question is that a statement is sufficient only if it satisfies the question always.

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Re: The sum of n consecutive positive integers is 45 [#permalink]

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New post 26 Nov 2013, 08:35
mumbijoh wrote:
I had a problem with number 8 and 9
8
Why is statement 2 not sufficient? I mean |x| means positive x so cant we then arrange it as |x|-|y|=16
Same as x-y=16 then use substitution method with the equation |x|+|y|=32 above?
9
You mentioned that “when odd number n is doubleb, 2n has twice as many factors as n”
Is this always the case? Let’s say our odd number is 15 ,it has four factors 5 ,1,15and 3.when doubled it becomes 30.30 has 30,1,2,3,5 factors. Just one more factor than 15.
My understanding for YES/NO DS question is that a statement is sufficient only if it satisfies the question always.



Question 8 is discussed here: if-x-and-y-are-non-zero-integers-and-x-y-32-what-is-128845.html

Question 9 is discussed here: is-the-integer-n-odd-1-n-is-divisible-by-3-2-2n-is-91399.html

As for your question:
15 has 4 factors: 1, 3, 5, and 15.
30 has 8 factors: 1, 2, 3, 5, 6, 10, 15, and 30.

Hope this helps.
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Re: The sum of n consecutive positive integers is 45 [#permalink]

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New post 18 Apr 2014, 07:16
Bunuel wrote:
Multiplication of the two digit numbers wx and cx, where w,x and c are unique non-zero digits, the product is a three digit number. What is w+c-x?
(1) The three digits of the product are all the same and different from w c and x.
(2) x and w+c are odd numbers.

(1) wx+cx=aaa (111, 222, ... 999=37*k) --> As x is the units digit in both numbers, a can be 1,4,6 or 9 (2,3,7 out because x^2 can not end with 2,3, or 7. 5 is out because in that case x also should be 5 and we know that x and a are distinct numbers).
1 is also out because 111=37*3 and we need 2 two digit numbers.
444=37*12 no good we need units digit to be the same.
666=37*18 no good we need units digit to be the same.
999=37*27 is the only possibility all digits are distinct except the unit digits of multiples.
Sufficient
(2) x and w+c are odd numbers.
Number of choices: 13 and 23 or 19 and 29 and w+c-x is the different even number.

Answer: A.


am I missing anything?
it does not say that w x and c are positive, does it?
w= 3, c= 2 and x= 7
37*27 = 999, here w+c -7 = -2

but we also can have
w= -3 and c= -2 and x= 7
-37*-27 = 999, here w+c - 7 = -12

Both of these sets satisfy both the conditions , hence I am getting E,

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Re: The sum of n consecutive positive integers is 45 [#permalink]

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New post 18 Apr 2014, 11:12
qlx wrote:
Bunuel wrote:
Multiplication of the two digit numbers wx and cx, where w,x and c are unique non-zero digits, the product is a three digit number. What is w+c-x?
(1) The three digits of the product are all the same and different from w c and x.
(2) x and w+c are odd numbers.

(1) wx+cx=aaa (111, 222, ... 999=37*k) --> As x is the units digit in both numbers, a can be 1,4,6 or 9 (2,3,7 out because x^2 can not end with 2,3, or 7. 5 is out because in that case x also should be 5 and we know that x and a are distinct numbers).
1 is also out because 111=37*3 and we need 2 two digit numbers.
444=37*12 no good we need units digit to be the same.
666=37*18 no good we need units digit to be the same.
999=37*27 is the only possibility all digits are distinct except the unit digits of multiples.
Sufficient
(2) x and w+c are odd numbers.
Number of choices: 13 and 23 or 19 and 29 and w+c-x is the different even number.

Answer: A.


am I missing anything?
it does not say that w x and c are positive, does it?
w= 3, c= 2 and x= 7
37*27 = 999, here w+c -7 = -2

but we also can have
w= -3 and c= -2 and x= 7
-37*-27 = 999, here w+c - 7 = -12

Both of these sets satisfy both the conditions , hence I am getting E,


w, x and c are unique non-zero digits of the two digit numbers wx and cx means that w, x, and c are 1, 2, 3, 4, 5, 6, 7, 8, or 9.

Hope it's clear.
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Re: The sum of n consecutive positive integers is 45   [#permalink] 18 Apr 2014, 11:12

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