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# The sum of n consecutive positive integers is 45

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Re: The sum of n consecutive positive integers is 45 [#permalink]

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29 Jul 2012, 05:10
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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.

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.

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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.

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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 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.

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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 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.

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]

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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.

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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 |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.

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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.

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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 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.

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Re: The sum of n consecutive positive integers is 45 [#permalink]

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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 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.

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
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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.

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.

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 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.

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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 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 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.

For more on this topic check Number Theory chapter of Math Book: math-number-theory-88376.html

Hope 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 non-positive numbers, which is impossible. So, n must be 9, sufficient.

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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 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 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.

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.

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.

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 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 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.

For more on this topic check Number Theory chapter of Math Book: math-number-theory-88376.html

Hope 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.

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:
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.

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:
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.

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?
Re: Good set of DS 3   [#permalink] 25 Nov 2012, 11:49

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