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Re: The sum of n consecutive positive integers is 45 [#permalink]
29 Jul 2012, 04:10

Expert's post

1

This post was BOOKMARKED

dvinoth86 wrote:

1. The sum of n consecutive positive integers is 45. What is the value of n? (1) n is even (2) n < 9

Cant we use the formula for sum of n consecutive +ve integers (n)*(n+1)/2 = 45 n^2 +n -90 = 0 n=9 or -10

As stated in the previous post \(\frac{n(n+1)}{2}\) gives the sum of first \(n\) positive integers: \(1+2+3+...+n=\frac{n(n+1)}{2}\) and we cannot use that formula since we are not told that we have this case. Check this for more: math-number-theory-88376.html (Evenly spaced set chapter).

Solution of this problem is as follows:

The sum of n consecutive positive integers is 45. What is the value of n?

(1) n is even --> n can be 2: 22+23=45. But it also can be 6 --> x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)=45 --> x=5. At least two values of n are possible. Not sufficient.

(2) n<9 --> the above example is also valid for this statement, hence not sufficient.

(1)+(2) Still at least two values of n are possible. Not sufficient.

Re: The sum of n consecutive positive integers is 45 [#permalink]
29 Jul 2012, 04:32

Q1

The sum of n consecutive integers is either a multiple of the middle term, in case n is odd, or a multiple of the sum of the two middle terms, in case n is even.

(1) Since n is even and 45 is odd, we must have an odd number of pairs in our sequence, such that the sum of each pair is a factor of 45, and it is the same as the sum of the two middle terms. For example, we can have just one pair (22, 23) - or three pairs, each sum being 45/3 = 15 - 5, 6, 7, 8, 9, 10, with 5+10=6+9=7+8=15. So, (1) is not sufficient.

(2) From what we have seen above, (2) is not sufficient either. Just as an example, if n is odd and less than 9, we can have the sequence 14, 15, 16.

(1) and (2) taken together is obviously not sufficient.

Answer E _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: The sum of n consecutive positive integers is 45 [#permalink]
29 Jul 2012, 04:49

Q2. Is a product of three integers XYZ a prime? (1) X=-Y (2) Z=1

(1) We can take X=1, then Y=-1, and taking Z any non-negative integer will get a non-prime, as the product will be either negative or 0. But if we take X=1, Y=-1 and for example Z=-2, than XYZ=2, which is a prime. Therefore, (1) is not sufficient.

(2) If Z=1, we can take Y=1 as well, and then either X is a prime or not, so the final product XYZ=X, can be or not a prime. Not sufficient.

(1) and (2) together: If Z=1 and X=-Y, then \(XYZ=-X^2\), which can be either 0 (when X=0) or a negative integer. In either case, we won't get a prime number (by definition, a prime must be a positive integer). So, the answer is a definite NO, therefore sufficient.

Answer C _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: The sum of n consecutive positive integers is 45 [#permalink]
29 Jul 2012, 05:18

Q3. 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) If the product is a three digit number, all digits identical, then the number must be of the form AAA=A*111=A*3*37, where A is a non-zero digit. It means that x = 7, and 3A must be a two digit number, which also ends in 7. It follows that 3A = 27, so the numbers are 37 and 27, 37 * 27 = 999, and w+c-x=5-2, sufficient (it doesn't matter who is w and who is c).

(2) Since w+c is odd, one of them must be even and the other one odd. Fro example, we can take w=2, c=1, x=3 or w=3, c=2, x=1. 23*13=299, 31*21=651, are both three digit numbers, and w+c-x is 0 and 4, respectively. Not sufficient.

Answer A.

Remark: In the body of the question, w,x and c are unique non-zero digits, shouldn't be "distinct non-zero digits"? _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: The sum of n consecutive positive integers is 45 [#permalink]
29 Jul 2012, 05:27

Q4. Is y – x positive? (1) y > 0 (2) x = 1 – y

The question we are asked is in fact is y>x?

(1) Not sufficient, as we don't know anything about x. (2) We have x+y=1, again not sufficient. We can have x=y=0.5, or x=0.25 < y=0.75, or x=0.75 > y=0.25

(1) and (2) not sufficient, we can use the examples from the analysis for (2) above.

Answer E _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: The sum of n consecutive positive integers is 45 [#permalink]
29 Jul 2012, 05:46

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 _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: The sum of n consecutive positive integers is 45 [#permalink]
29 Jul 2012, 05:56

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 _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: The sum of n consecutive positive integers is 45 [#permalink]
29 Jul 2012, 06:04

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 _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: The sum of n consecutive positive integers is 45 [#permalink]
29 Jul 2012, 06:10

1

This post received KUDOS

Expert's post

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.

Re: The sum of n consecutive positive integers is 45 [#permalink]
29 Jul 2012, 06:17

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 _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: The sum of n consecutive positive integers is 45 [#permalink]
29 Jul 2012, 06:25

Expert's post

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.

Re: The sum of n consecutive positive integers is 45 [#permalink]
29 Jul 2012, 06:31

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 _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: The sum of n consecutive positive integers is 45 [#permalink]
29 Jul 2012, 06:38

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]
29 Jul 2012, 06:41

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.

Re: Good set of DS 3 [#permalink]
14 Nov 2012, 18:26

Bunuel wrote:

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

Hello Bunuel, I did not understand this explanation. A prime number is one which is divisible by 1 and itself, right. So when you are multiplying two or three numbers.... even if it is prime or not.... the result is not only divisible by 1 and itself but also by that prime number. So in any case multiplication of any 3 numbers cannot be prime?!!

Re: Good set of DS 3 [#permalink]
14 Nov 2012, 19:15

Bunuel wrote:

ANSWERS: 3. 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.

Did not understand a bit of it! Is there something I need to study? Can you Please explain clearly. Thanks.

Re: Good set of DS 3 [#permalink]
25 Nov 2012, 10:49

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?

gmatclubot

Re: Good set of DS 3
[#permalink]
25 Nov 2012, 10:49

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