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Re: The sum of n consecutive positive integers is 45 [#permalink]
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25 Oct 2009, 20:53
goldgoldandgold wrote: Bunuel wrote: ANSWERS:
10. The sum of n consecutive positive integers is 45. What is the value of n? (1) n is odd (2) n >= 9
Look at the Q 1 we changed even to odd and n<9 to n>=9
(1) not sufficient see Q1. (2) As we have consecutive positive integers max for n is 9: 1+2+3+...+9=45. (If n>9=10 first term must be zero. and we are given that all terms are positive) So only case n=9. Sufficient.
Answer: B. sum of n integers = (n*(n+1))/2 (n*(n+1))/2 = 45 which yields n = 9 could be the only answer Stmt 1 and 2 (D) are both sufficient. Does GMAT assume that we would not indulge in any formula? Just curious This is not correct. First of all the formula you referring n(1+n)/2 is the formula for counting the sum of n FIRST integers (meaning that starting from 1). Question stem does not give us that information: "The sum of n consecutive positive integers is 45." You can not assume anything. The formula for counting the sum of consecutive integers (not necessarily first n integers), which is in fact sum of AP (arithmetic progression) is: Sn=n*(a1+an)/2 or Sn=n*(2a1+d(n1))/2, you can substitute d=1, as the numbers are consecutive. If you further substitute a1 (the first term of the progression) with 1 you'll get exactly the formula you wrote: Sn=n(2+n1)/2=n(n+1)/2 Second, you can find the n consecutive odd positive integers (n=odd) to total 45: n=3 > 14, 15, 16 n=5 > 7, 8, 9, 10, 11. Hope it's clear.
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Re: The sum of n consecutive positive integers is 45 [#permalink]
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27 Oct 2009, 22:19
tejal777 wrote: Quote: 3. Multiplication of the two digit numbers wx and cx, where w,x and c are unique nonzero digits, the product is a three digit number. What is w+cx?
(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. >>What about 8?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+cx is the different even number.
Answer: A.
Could somebody please elaborate the solution..i don't understand:( 8 is also not possible. ........................................................................ 1. wx * cx = aaa (aaa can only be 111 (for 9), 444 (for 2), 666 (for 8), and 999 (for 3 and 7)). 222, 333, 555, 777, and 888 are not possible as noneo f the unit difgits of the 3 digit numbers is square of an integer. wx * cx = 111 = 3x37. Not possible as there is only one 2 digit number. Needs 2 two digit numbers. wx * cx = 444 = 2x3x37 = 6x37. Not possible as there is only one 2 digit number. Needs 2 two digit numbers. wx * cx = 666 = 2x3x3x37 = 18x37. Not possible as the unit digit of these 2 digit numbers are not the same. wx * cx = 999 = 3x3x3x37 = 27x37. Possible as the unit digit of these 2 digit numbers are same. 2. is self explanatory..
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Re: The sum of n consecutive positive integers is 45 [#permalink]
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26 Dec 2009, 03:47
please explain 2nd Q.
i want an example where XYZ can be prime using STATEMENT 1 ALONE



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Re: The sum of n consecutive positive integers is 45 [#permalink]
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26 Dec 2009, 09:04
jan4dday wrote: please explain 2nd Q.
i want an example where XYZ can be prime using STATEMENT 1 ALONE First note prime numbers are only positive. (Also note that \(x\), \(y\) and \(z\) are integers) Q: \(xyz=p\), is \(p\) prime? (1) \(x=y\) > \(p=x^2z\). Let's check when this expression gives a prime number: Well first of all \(p\) to be prime \(z\) MUST be negative, as \(p\) MUST be positive to be a prime. Next if \(x>1\), (eg \(2\), \(3\), ...) OR equals to zero, \(p\) won't be prime. So \(x\) must be equal to \(1\). But it's not enough. We'll have \(p=x^2z=z\), so \(p\) to be a prime number \(z\) must be equal to \(prime\). You are asking how using statement (1) \(p\) could be a prime: according to above, when \(x=1\) and \(z=p\). eg.: \(x=1\) > \(y=1\) > z\(=7\) > \(p=(1)*1*(7)=7\), which is prime. Statement (1) may or may not give the prime number for \(xyz\). Not sufficient. (2) \(z=1\) > \(p=xy\). Again for \(p\) to be a prime number \(xy\) must be \(>0\) (both positive or both negative). Then if \(x=prime\) and \(y=1\), OR \(y=prime\) and \(x=1\), so that \(xy>0\), then \(xy\) is a prime number. For any other values or combinations of \(x\) and \(y\), \(p\) won't be a prime. Not sufficient. (1)+(2) \(p=xyz=x^2\) (as \(x=y\) and \(z=1\)). \(x^2\) is never positive, hence \(p\) is not a prime. Sufficient. Answer: C. Hope it's clear.
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Re: The sum of n consecutive positive integers is 45 [#permalink]
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02 Jan 2010, 10:28
Hi Bunuel..... Would appreciate if you could explain this in detail as I am confused... Thanks! JT
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Re: The sum of n consecutive positive integers is 45 [#permalink]
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02 Jan 2010, 11:58
jeeteshsingh wrote: Hi Bunuel..... Would appreciate if you could explain this in detail as I am confused... Thanks! JT If a and b are integers, and a not= b, is ab > 0? (1) a^b > 0 (2) a^b is a nonzero integer. \(ab > 0\) is true when \(b>0\) and \(a\) does not equal to zero. (1) \(a^b > 0\) > \(a\) does not equal to zero, but we don't know about \(b\), it can be any value, positive or negative. Not sufficient. (2) \(a^b\) is a nonzero integer > \(a\) can be 1 and \(b\) any integer, positive or negative. Not sufficient. (1)+(2) If a=1 and b=2, then ab > 0, but if a=1 and b=2, then ab <0. Not sufficient. Answer: E. This is the easiest way to solve this problem. In my previous solution I just tried to show what each statement means algebraically. Please tell which part needs more clarification.
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Re: The sum of n consecutive positive integers is 45 [#permalink]
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03 Jan 2010, 03:14
Bunuel wrote: If a and b are integers, and a not= b, is ab > 0? (1) a^b > 0 (2) a^b is a nonzero integer.
\(ab > 0\) is true when \(b>0\) and \(a\) does not equal to zero.
(1) \(a^b > 0\) > \(a\) does not equal to zero, but we don't know about \(b\), it can be any value, positive or negative. Not sufficient.
(2) \(a^b\) is a nonzero integer > \(a\) can be 1 and \(b\) any integer, positive or negative. Not sufficient.
(1)+(2) If a=1 and b=2, then ab > 0, but if a=1 and b=2, then ab <0. Not sufficient.
Answer: E.
This is the easiest way to solve this problem. In my previous solution I just tried to show what each statement means algebraically. Please tell which part needs more clarification.
Not able to get why do you say this (marked in red)... I know that \(a\) would always give a positive value, but why cant we consider \(a\) as 2 or 3...? Any reason for this?
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Re: The sum of n consecutive positive integers is 45 [#permalink]
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03 Jan 2010, 09:23
jeeteshsingh wrote: Bunuel wrote: If a and b are integers, and a not= b, is ab > 0? (1) a^b > 0 (2) a^b is a nonzero integer.
\(ab > 0\) is true when \(b>0\) and \(a\) does not equal to zero.
(1) \(a^b > 0\) > \(a\) does not equal to zero, but we don't know about \(b\), it can be any value, positive or negative. Not sufficient.
(2) \(a^b\) is a nonzero integer > \(a\) can be 1 and \(b\) any integer, positive or negative. Not sufficient.
(1)+(2) If a=1 and b=2, then ab > 0, but if a=1 and b=2, then ab <0. Not sufficient.
Answer: E.
This is the easiest way to solve this problem. In my previous solution I just tried to show what each statement means algebraically. Please tell which part needs more clarification.
Not able to get why do you say this (marked in red)... I know that \(a\) would always give a positive value, but why cant we consider \(a\) as 2 or 3...? Any reason for this? OK. For statement (2): First of all I'm not saying that \(a\) can ONLY be 1. I'm saying that it MAY be 1 and in this case \(b\) can take any value statement (2): " \(a^b\) is a nonzero integer " to hold true. So \(b\) can take positive as well as negative values, hence \(ab > 0\) may or may not be true. That's why (2) is not sufficient. To elaborate further: we are told that \(a^b\) is A. not zero and B. it's an integer. A. Means that \(a\) is not zero, as the only way \(a^b\) to be zero, is \(a\) to be zero. B. \(a^b\) is an integer (non=zero). What does that mean? When \(a\) is any integer so that \(a>1\) (eg 2, 2, 3, 3, ...), \(b\) can take any value but negative. Example: \(a=3\) > \(3=3\), \(3\) in integer power (given \(a\) and \(b\) are integers) to be an integer, power (\(b\)), must be positive or zero, as when \(b\) is negative, let's say \(2\), we'll have: \(3^{2}=\frac{1}{3^2}\) which is not integer. But when \(a=1\) (\(1\) or \(1\)), then \(b\) can take ANY integer value as 1 in any power equals to 1 which is integer. So from statement (2) we'll have:\(a\) does not equals to 0. Plus, \(b\) can be ANY integer (positive or negative or zero) when \(a=1\) and \(b\) must be positive or zero when \(a>1\). Hope it's clear.
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Re: The sum of n consecutive positive integers is 45 [#permalink]
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18 Feb 2010, 13:38
jeeteshsingh,
To further clarify what Bunuel was saying. If b was negative and a was not 1 or 1, then a^b would be a fraction due to the properties of negative exponents. If a=1 (meaning a=1 or a=1) then the denominator of the fraction would be some power of 1 which is also one. In this case the fraction would be 1 over 1 which is again 1 and satisfies the conditions.
Also, Bunuel, for the second portion I noticed that you didn't include in your listed possibilities that for a^b is a non zero integer, b=0 is possible as it would create an answer of 1 for any value of a including 0.



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Re: The sum of n consecutive positive integers is 45 [#permalink]
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18 Feb 2010, 14:07
theturk123 wrote: jeeteshsingh,
To further clarify what Bunuel was saying. If b was negative and a was not 1 or 1, then a^b would be a fraction due to the properties of negative exponents. If a=1 (meaning a=1 or a=1) then the denominator of the fraction would be some power of 1 which is also one. In this case the fraction would be 1 over 1 which is again 1 and satisfies the conditions.
Also, Bunuel, for the second portion I noticed that you didn't include in your listed possibilities that for a^b is a non zero integer, b=0 is possible as it would create an answer of 1 for any value of a including 0. Welcome to the Gmat Club. Your logic is correct. Though b=0 IS included as one of the possibilities. See the bolded part from the explanation: Bunuel wrote: So from statement (2) we'll have:
\(a\) does not equals to 0. Plus, \(b\) can be ANY integer (positive or negative or zero) when \(a=1\) and \(b\) must be positive or zero when \(a>1\). You also mention \(0^0\) issue, this not the case for statement 2 (as \(a\) cannot be zero), but still as you brought this up: \(0^0\), in some sources equals to 1, some mathematicians say it's undefined. Anyway you won't need this for GMAT: "During the past decade, mathematicians argued extensively about the value of 0^0. Some answer that 0^0 = 1, while others answer that 0^0 is undefined. In the unlikely event that this question appears in some format or is a required intermediary calculation, the correct answer is more likely that 0^0 = 1." http://www.platinumgmat.com/gmat_study_ ... ial_powersand: "Note: the case of 0^0 is not tested on the GMAT." http://www.manhattangmat.com/npexponents.cfm
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Re: The sum of n consecutive positive integers is 45 [#permalink]
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15 Sep 2010, 10:06
jax91 wrote: Bunuel wrote: 5. If a and b are integers, and a not= b, is ab > 0? (1) a^b > 0 (2) a^b is a nonzero integer
ab > 0? a is always +ve. So we need to know if b is +ve or ve. 1.) mod of any number is +ve. Insuff. 2.) a^b is an integer. we know a and b are integers. so a is a +ve integer. any +ve integer raised to a ve integer will give us a fraction. e.g. 4 ^ 3 = 1/ (4^3) which will never be an integer. so for a^b to be an integer b has to be +ve. So its suff. So B. Hi,Can anyone explain me how a is positive?We generally look at the sign of any number before deciding its abs value..Right?But how come that is not followed here? Generally we take x=x (x <0) x=x (x>0) but y is a taken positive directly?In the question,it is just mentioned as an integer Explanation plz..



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Re: The sum of n consecutive positive integers is 45 [#permalink]
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15 Sep 2010, 10:16
ravitejapandiri wrote: Hi,Can anyone explain me how a is positive?We generally look at the sign of any number before deciding its abs value..Right?But how come that is not followed here? Generally we take x=x (x <0) x=x (x>0) but y is a taken positive directly?In the question,it is just mentioned as an integer Explanation plz.. Seems that you need to brush up on absolute value. Please check Walker's post on Absolute Value at: mathabsolutevaluemodulus86462.html\(x=x\), when \(x<0\) and \(x=x\) when \(x>0\) > CORRECT. But when \(x=negative<0\) then \(x=x=(negative)=positive\) and when \(x=positive>0\) then \(x=x=positive\), so in any case when \(x\) is either positive or negative \(x\) is still positive. There is one more case though: when \(x=0\) then \(x=0\). So generally we can say that absolute value of an expression is alway nonnegative: \(some \ expression\geq{0}\) > \(x\geq{0}\). As for this particular question: see my posts on it on pages 1, 2, and 3. Hope it helps.
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Re: The sum of n consecutive positive integers is 45 [#permalink]
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19 Dec 2010, 17:33
gettinit wrote: yangsta8 wrote: Bunuel wrote: 4. Is y – x positive? (1) y > 0 (2) x = 1 – y
Statement 1) y>0 Not suff, X could be anything larger or smaller than X. Statement 2) x=1y x+y=1 Let x=3 and y=2 then yx < 0. But if x=1/4 and y=3/4 then yx >0 Not suff. 1 and 2 together) From the example above we have: if x=1/4 and y=3/4 then yx >0 but if we flip it around: if x=3/4 and y=1/4 then yx <0 not suff. ANS = E I selected that both stmts were sufficient together because the examples I chose worked both ways so if x=2 y=3 you get 3(2)=5 or if y=2 than x = (1) so 2(1)=3. Can the insufficiency only be seen with fractional numbers? Thanks. Yes, in order to get NO answer you should choose values of x and y from (0, 1), note that this range can give you the YES answer as well. Is \(yx>0\)? > is \(y>x\)?(1) y > 0, not sufficient as no info about \(x\). (2) x = 1  y > \(x+y=1\) > the sum of 2 numbers equal to 1 > we can not say which one is greater. Not sufficient. (1)+(2) \(x+y=1\) and \(y>0\), still can not determine which one is greater: if \(y=0.1>0\) and \(x=0.9\) then \(y<x\) but if \(y=0.9>0\) and \(x=0.1\) then \(y>x\). Not sufficient. Answer: E.
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Re: The sum of n consecutive positive integers is 45 [#permalink]
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31 Jan 2011, 06:48
subhashghosh wrote: Hi Bunuel
For question # 8, please explain how this is true :
either 4y=32 y=8 x=3, xy=24 OR 4y=32
Regards, Subhash If x and y are nonzero integers and x + y = 32, what is xy? (1) \(4x12y=0\) > \(x=3y\) > \(x\) and \(y\) have opposite signs. So either: \(x=x\) and \(y=y\) > in this case \(x+y=xy=3yy=4y=32\): \(y=8\), \(x=24\), \(xy=24*8\); OR: \(x=x\) and \(y=y\) > \(x+y=x+y=3y+y=4y=32\) > \(y=8\) and \(x=24\) > \(xy=24*8\), the same answer. Sufficient. (2) \(x  y = 16\). Sum this one with th equations given in the stem > \(2x=48\) > \(x=24\), \(y=8\). \(xy=24*8\) (x and y have opposite sign) or \(xy=24*8\) (x and y have the same sign). Multiple choices. Not sufficient. Answer: A.
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Re: The sum of n consecutive positive integers is 45 [#permalink]
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29 Jul 2012, 05:02
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.
[/color] n^2 +n 90 = 0 n=9 or 10 Cant we use the formula for sum of n consecutive +ve integers (n)*(n+1)/2 = 45. This formula gives the sum of the first n consecutive positive integers: 1,2 3, ..., n.Nowhere is stated that our sequence starts with 1.
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29 Jul 2012, 05:10
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: mathnumbertheory88376.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. Answer: E. Hope it's clear.
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Re: The sum of n consecutive positive integers is 45 [#permalink]
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29 Jul 2012, 05:32
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
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29 Jul 2012, 05:38
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
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Re: The sum of n consecutive positive integers is 45 [#permalink]
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29 Jul 2012, 05:49
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 nonnegative integer will get a nonprime, 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
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Re: The sum of n consecutive positive integers is 45 [#permalink]
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29 Jul 2012, 06:18
Q3. Multiplication of the two digit numbers wx and cx, where w,x and c are unique nonzero digits, the product is a three digit number. What is w+cx? (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 nonzero 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+cx=52, 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+cx is 0 and 4, respectively. Not sufficient. Answer A. Remark: In the body of the question, w,x and c are unique nonzero digits, shouldn't be "distinct nonzero digits"?
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