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Re: The sum of n consecutive positive integers is 45 [#permalink]
 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: 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. Answer: E. Hope it's clear.
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Re: The sum of n consecutive positive integers is 45 [#permalink]
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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Re: The sum of n consecutive positive integers is 45 [#permalink]
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]
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 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
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Re: The sum of n consecutive positive integers is 45 [#permalink]
29 Jul 2012, 06:18
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"?
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Re: The sum of n consecutive positive integers is 45 [#permalink]
29 Jul 2012, 06:27
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
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Re: The sum of n consecutive positive integers is 45 [#permalink]
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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Re: The sum of n consecutive positive integers is 45 [#permalink]
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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Re: The sum of n consecutive positive integers is 45 [#permalink]
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]
29 Jul 2012, 07:10
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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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Re: The sum of n consecutive positive integers is 45 [#permalink]
29 Jul 2012, 07:14
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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]
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]
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.htmlHope it helps.
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Re: The sum of n consecutive positive integers is 45 [#permalink]
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]
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]
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]
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.htmlHope it helps. Oooops! Terrible miss, as the factor 2 for an even number is not counted twice.
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Re: Good set of DS 3 [#permalink]
14 Nov 2012, 19:26
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?!!
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Re: Good set of DS 3 [#permalink]
14 Nov 2012, 20:15
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.
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Re: Good set of DS 3 [#permalink]
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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Re: Good set of DS 3
[#permalink]
25 Nov 2012, 11:49
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