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

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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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PhD in Applied Mathematics Love GMAT Quant questions and running.

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

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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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PhD in Applied Mathematics Love GMAT Quant questions and running.

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

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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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PhD in Applied Mathematics Love GMAT Quant questions and running.

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.

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

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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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PhD in Applied Mathematics Love GMAT Quant questions and running.

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.

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

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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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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).
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

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

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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.

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?

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).

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

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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.

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 \(xyz=(-1)*1*(-7)=7=prime\).

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

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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

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

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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.

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.

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

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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,

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.

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