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The sum of n consecutive positive integers is 45 [#permalink]
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Please find below new set of DS problems: TIP: many of these problems act in GMAT zone, so beware of ZIP trap. 1. The sum of n consecutive positive integers is 45. What is the value of n? (1) n is even (2) n < 9 2. Is a product of three integers XYZ a prime? (1) X=Y (2) Z=1 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. 4. Is y – x positive? (1) y > 0 (2) x = 1 – y 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 6. If M and N are integers, is (10^M + N)/3 an integer? 1. N = 5 2. MN is even 7. 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? (1) d = 3 (2) b = 6 8. If x and y are nonzero integers and x + y = 32, what is xy? (1) 4x  12y = 0 (2) x  y = 16 9. Is the integer n odd (1) n is divisible by 3 (2) 2n is divisible by twice as many positive integers as n 10. The sum of n consecutive positive integers is 45. What is the value of n? (1) n is odd (2) n >= 9 Please share your way of thinking, not only post the answers. OA and explanations to follow. Also you can check new set of PS problems: goodsetofps85414.html
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Re: Good set of DS 3 [#permalink]
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ANSWERS: 1. The sum of n consecutive positive integers is 45. What is the value of n? (1) n is even (2) n < 9 (1) n=2 > 22+23=45, n=4 > n=6 x1+(x1+1)+(x1+2)+(x1+3)+(x1+4)+(x1+5)=45 x1=5. At least two options for n. Not sufficient. (2) n<9 same thing not sufficient. (1)+(2) No new info. Not sufficient. Answer: E. 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 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. 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. 4. Is y – x positive? (1) y > 0 (2) x = 1 – y Easy one even if y>0 and x+y=1, we can find the x,y when yx>0 and yx<0 Answer: E. 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 This is tricky ab > 0 to hold true: a#0 and b>0. (1) a^b>0 only says that a#0, because only way a^b not to be positive is when a=0. Not sufficient. NOTE having absolute value of variable a, doesn't mean it's positive. It's not negative > a>=0 (2) a^b is a nonzero integer. What is the difference between (1) and (2)? Well this is the tricky part: (2) says that a#0 and plus to this gives us two possibilities as it states that it's integer: A. 1>a>1 (a>1), on this case b can be any positive integer: because if b is negative a^b can not be integer. OR B. a=1 (a=1 or 1) and b can be any integer, positive or negative. So (2) also gives us two options for b. Not sufficient. (1)+(2) nothing new: a#0 and two options for b depending on a. Not sufficient. Answer: E. 6. If M and N are integers, is (10^M + N)/3 an integer? (1) N = 5 (2) MN is even Note: it's not given that M and N are positive. (1) N=5 > if M>0 (10^M + N)/3 is an integer ((1+5)/3), if M<0 (10^M + N)/3 is a fraction ((1/10^M+5)/3). Not sufficient. (2) MN is even > one of them or both positive/negative AND one of them or both even. Not sufficient (1)+(2) N=5 MN even > still M can be negative or positive. Not sufficient. Answer: E. 7. 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? (1) d = 3 (2) b = 6 Note this part: "for all values of x" So, it must be true for x=0 > c=d^2 > b=2d (1) d = 3 > c=9 Sufficient (2) b = 6 > b=2d, d=3 > c=9 Sufficient Answer: D. 8. 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. 9. Is the integer n odd (1) n is divisible by 3 (2) 2n is divisible by twice as many positive integers as n (1) 3 or 6. Clearly not sufficient. (2) 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. 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.
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Re: Good set of DS 3 [#permalink]
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Re: Good set of DS 3 [#permalink]
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Bunuel wrote: 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.
WX x CX = IJK 1.) I,J,K are the same and not equal to W,C or X. so 3 digit numbers with all digit same are 111,222,...., 999. basically multiples of 111 (37x3). so we get 1 number = 37 conditions the second number has to meet = last digit = 7, multiple of 3, double digit. so we get 27. 27 x 37 = 999 So suff. 2.) x and w+c are odd. this gives multiple values. So A.
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Re: Good set of DS 3 [#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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01 May 2014, 22:29
qlx wrote: So the terms " Integer" and " Digit " makes all the difference ?They do not mean the same thing, right?
Integer =both negative , positive or zero Digit = only 0 or positive.
I thought ( 2) was a digit too! a negative digit? Can we not have a negative digit?
Or is the Digit term in GMAT only used for non negative integers.
I think this is important info in terms of Gmat.
Really appreciate your help.Thank you so much . Responding to a pm: Here goes my honest opinion you asked for! An integer can be positive, negative or 0. A single digit, on the other hand always implies a positive single digit from 0 to 9. Given ab where a and b are single digits, ab is a positive integer. In the question, you have "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". This implies that w, c and x are distinct digits from 1 to 9 and wx and cx are positive integers. The author did not forget to mention "positive digits". He/she did not need to mention it because digits imply positive digits only. Also, in case of confusion, you can always search on Google. Say, put "digit in Math" and check out the various write ups. If there are multiple usages, the net will tell you that too.
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Re: Good set of DS 3 [#permalink]
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Bunuel wrote: 1. The sum of n consecutive positive integers is 45. What is the value of n? (1) n is even (2) n < 9 1. n could be 2 or 6 or 10 a + a+1 = 45 a = 22 n = 2 a + a+1 + a+2 + a+3 + a+4 + a+5 = 45 a = 5 n = 6 2. n could be 2, 3, 5 or 6 1&2: n could be 2 or 6. E. Bunuel wrote: 10. The sum of n consecutive positive integers is 45. What is the value of n? (1) n is even (2) n >= 9 1. n could be 2 or 6 or 10 n = 2: a + a+1 = 45 a = 22 n = 6: a + a+1 + a+2 + a+3 + a+4 + a+5 = 45 a = 5 2. n could be 9 or 10 or 14 or 15 or 18 & so on... 1&2: n could be 10 or 14 or 18. E.
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Re: Good set of DS 3 [#permalink]
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Bunuel wrote: TIP: many of these problems act in GMAT zone, so beware of ZIP trap.
1. The sum of n consecutive positive integers is 45. What is the value of n? (1) n is even (2) n < 9
2. Is a product of three integers XYZ a prime? (1) X=Y (2) Z=1
What is the ZIP trap? Q1) Statement 1) n = 2,4,6 etc n = 2 => x+(x+1)=45 => x=22 (works) n = 4 => x+(x+1)+(x+2)+(x+3)=45 => 4x+6=45 => x=39/4 (doesn't work) n = 6 => Take above equation+(x+4)+(x+5) => 6x+15=45 => x=5 (works) Not suff. Statement 2) n < 9. This is proven insufficient from the working above since both n=2 and n=6 n<9. 1 and 2 together still prove insufficient due to above working. ANS = E. Q2) Statement 1) X=Y This means Z needs to be negative and for XYZ to have a chance of being prime. Z can be anything. Insufficient. Statement 2) Z=1 X and Y could be anything such as 2 and 3 (non prime multiple) or 1 and 2 (prime). Insufficient. 1 and 2 Together) Z = 1. X=Y 1*Y*(Y) = Y^2 which cannot be prime as it is negative. ANS = C Edited: Got the right working but wrote E instead of C. I gotta stop doing that
Last edited by yangsta8 on 16 Oct 2009, 23:03, edited 1 time in total.



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Re: Good set of DS 3 [#permalink]
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Bunuel wrote: 6. If M and N are integers, is (10^M + N)/3 an integer? 1. N = 5 2. MN is even
Statement 1) N=5 If M>=0 then it is always divisible by 3. Since the number will always consist of 1, trailing 0's and a 5. Of which the sum of digits =6 which is the rule for divisibility by 3. If M<0 then the equation is not divisble by 3. For example if M=1. Insufficient Statement 2) MN is even. Again this means M could still be negative so insufficient. For example M could be 1 and N could be 2 which is not divisible by 3. Or n=5 but m=2 which is not. Statements together) Still insuff. m=2 n=5 works. But m=2 n=5 doesn't work. ANS = E



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Re: Good set of DS 3 [#permalink]
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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.
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Re: Good set of DS 3 [#permalink]
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17 Oct 2009, 20:43
GMAT TIGER wrote: Bunuel wrote: Yes and about the ZIP trap:
GMAT likes to act in the zone 1<=x<=1. So I always ask myself:
Did I assumed, with no ground for it, that variable can not be Zero? Check 0! Did I assumed, with no ground for it, that variable is an Integer? Check fractions! Did I assumed, with no ground for it, that variable is Positive? Check negative values!
I called it ZIP trap. Helps me a lot especially with number property problems. Thats cool. You can say PINZF (or better) trap as well: P = positive I = integer N = negative Z = zero F = fraction Sure you can call whatever suits you, no copyright on that term, for me ZIP sounds good.)))
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Re: The sum of n consecutive positive integers is 45 [#permalink]
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29 Jul 2012, 07:10
EvaJager wrote: Q8. If x and y are nonzero 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=3y. We can deduce that y=8, x=24, and xy=xy=192. Not sufficient, because xy=192 or 192.
(2) Since x=y+16, we find again that y=8, x=24, and xy=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 nonzero 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 \(x+y=xy=3yy=4y=32\): \(y=8\), \(x=24\), \(xy=24*8\), same answer. Sufficient. (2) \(x  y = 16\). Sum this one with th equations given in the stem > \(2x=48\) > \(x=24\), \(y=8\). \(xy=24*8\) (x and y have opposite sign) or \(xy=24*8\) (x and y have the same sign). Multiple choices. Not sufficient. Answer: A. 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, 07:14
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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07 Jan 2013, 04:20
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: thesumofnconsecutivepositiveintegersis8541320.html#p667155 and here: thesumofnconsecutivepositiveintegersis85413.html#p640133Hope it helps.
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Re: The sum of n consecutive positive integers is 45 [#permalink]
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24 Apr 2014, 03:57
qlx wrote: Bunuel wrote: Twodigit numbers wx means positive integer, where w is the tens digit and x is the units digit: 10, 11, 12, ..., 99.
Twodigit numbers wx means negative integer, where w is the tens digit and x is the units digit: 10, 11, 12, ..., 99.
Thank you I finally got what you are saying. I think a variable representing an integer can be positive or negative , a variable without a minus sign does not mean that the variable is positive does it . using the same logic if a particular sum says if a and b are integers what is b ? 1) a+b = 1 2) a= 2 From here can we say that a is positive ,because it is not a ? so a means = a is a positive integer and a means = a is a negative integer In the same way when it says wx is a two digit integer , how can we say because it is not wx hence wx is positive. a is an integer does not mean that a is positive but a is a digit means that a is positive: 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9. Sorry cannot explain any better.
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Re: Good set of DS 3 [#permalink]
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16 Oct 2009, 23:01
Bunuel wrote: 2. Is a product of three integers XYZ a prime?
(1) X=Y (2) Z=1 (1) If x=y = 2 and z = 1, yes. Otherwise, no.. (2) If z = 1, x could be 2 and y = 1. xyz is a price. If something else, no. From 1 and 2: x = y and z = 1, xyz is always a ve integer, which cannot be a prime....C.
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Re: Good set of DS 3 [#permalink]
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16 Oct 2009, 23:03
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



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Re: Good set of DS 3 [#permalink]
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16 Oct 2009, 23:17
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Bunuel wrote: 7. 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? (1) d = 3 (2) b = 6
Expanding it out we get : x^2 + bx + c = x^2 + 2dx + d^2 Statement 1) d = 3 d^2 = c = 9 Suff. Statement 2) b=6 b=6=2d d=3 d^2=c=9 Suff ANS = D



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Re: Good set of DS 3 [#permalink]
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16 Oct 2009, 23:41
Bunuel wrote: 9. Is the integer n odd (1) n is divisible by 3 (2) 2n is divisible by twice as many positive integers as n
10. The sum of n consecutive positive integers is 45. What is the value of n? (1) n is odd (2) n >= 9
Q9) Statement 1) N is a multiple of 3. N could be 3 or 6. Insufficient. Statement 2) I am not sure how to prove this except by examples: Example 1:n=9 factors={1,3,9}, 2n=18 factors={1,2,3,6,9,18} N is odd is true. Example 2:n=6 factors={1,2,3,6} 2n=12 factors={1,2,3,4,6,12} Does not have twice as many factors. Example 3: n=3 factors={1,3} 2n=6 factors={1,2,3,6} N is odd is true. ANS = B Q10) Statement 1) N is odd. N could be 1. 45 N could also be 3. x+(x+1)+(x+2)=45 => 3x=42 x=14 Insufficient. Statement 2) N>=9 Let n=9. 9x+8+7+6+5+4+3+2+1=45 => 9x+36=45 => 9x=9 x=1 we cannot use n>10 because adding anymore positive integers means sum > 45. Sufficient. ANS = B




Re: Good set of DS 3
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