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Re: The sum of n consecutive positive integers is 45 [#permalink]
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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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Re: The sum of n consecutive positive integers is 45 [#permalink]
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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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Re: The sum of n consecutive positive integers is 45 [#permalink]
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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]
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29 Jul 2012, 06:46
Q5. If a and b are integers, and a not= b, is ab > 0? (1) a^b > 0 (2) a^b is a nonzero integer The question can be rephrased as are a and b distinct nonzero 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 ab=0. But if we take a=2 and b=1, then \(a^b=2>0\) and ab=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]
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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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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 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
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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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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 nonzero 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 nonzero 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]
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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 nonzero 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 nonzero 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: mathnumbertheory88376.htmlHope it helps.
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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 nonpositive 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 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. 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]
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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]
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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 nonzero 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 nonzero 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: mathnumbertheory88376.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]
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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]
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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 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.
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]
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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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