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A set of five positive integers has an arithmetic mean of 150. A parti [#permalink]
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07 Apr 2015, 06:25
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Re: A set of five positive integers has an arithmetic mean of 150. A parti [#permalink]
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Hi, I was not able to solve but tried till some point and no idea how to proceed further. Assume numbers : X, Y, Y, Y, X+100 average (x+y+y+y+x+100) = 5*150 2x+3y = 650 y = (650 – 2x)/3 we know that x < y < x+100 x < (650 – 2x)/3 < x+100 3x < 650 – 2x < 3x + 300 split into two inequalities: 3x < 650 – 2x and 650 – 2x < 3x + 300 1: 3x < 650 – 2x 5x < 650 x < 130 2: 650 – 2x < 3x + 300 5x > 350 x > 70 So we get that 70 < X < 130. Number x should fall between 71 to 129. I'm lost here, how to proceed further.
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A set of five positive integers has an arithmetic mean of 150. A parti [#permalink]
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I think it would be 19. Once we have got the range for X, we need to find out the numbers between 70 and 130 that will be divisible by 3. From 70 to 130, there are 19 multiples of 3. So, every third value would give us an integer value for Y. Eg: X73 Y168 X76 Y166 Correct me if I am wrong.



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A set of five positive integers has an arithmetic mean of 150. A parti [#permalink]
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Lets say numbers are in this order: x < y = y = y < x+100. Question is, how many different values can x take. x+3y+x+100 = 750 2x+3y = 650 We have condition x < y < x + 100 so lets use it 2x + 3y < 2x + 3*(x+100) = 5x + 300 2x + 3y > 2x + 3x = 5x 2x + 3y = 650
So we got double inequality: 5x < 650 < 5x + 300 which turns into 350 < 5x < 650 70 < x < 130 x can take 130701 = 59 values which corresponds to answer E edit: forgot to take into account that 2x + 3y = 650 and both x and y = integers. y = (650  2x)/3 = 2/3*(325x)  integer, so 325  x has to bi divisable by 3 to be correct Now if I input all my available X (from 71 to 129) I'll notice that with a period of 3 my x values will yield me the result I need, I just need to carefully count them. x = 71: 325  71 = 254, not divisable by 3  fail x = 72: 325  72 = 253, not divisable by 3  fail x = 73: 325  73 = 252, divisable by 3  good same goes for 249, 246, 243 and etc. That means with x go with period of 3 starting from 73 and finishing with 73 + 3*18 = 127 making it total 19 numbers and corresponding to option B afterall. edit: zzz, it seems like one of the main problems on the GMAT is actually remembering what type of numbers you are working with.
Originally posted by Zhenek on 07 Apr 2015, 14:58.
Last edited by Zhenek on 08 Apr 2015, 00:12, edited 3 times in total.



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Re: A set of five positive integers has an arithmetic mean of 150. A parti [#permalink]
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07 Apr 2015, 18:02
Hi All, This question has some rather specific 'restrictions' that you have to pay attention to. 1) All 5 numbers are positive INTEGERS 2) The average of the 5 is 150 (so the sum of those 5 integers is 750) 3) The greatest of the 5 is 100 greater than the smallest of the 5 4) The remaining 3 integers are all the SAME and are somewhere between the biggest and smallest. The question asks for the number of different POSSIBLE values for the LARGEST integer. To start, I'm going to name the simplest example that fits the above information: 100, 150, 150, 150 and 200 To figure out the other options, we have to "slide" the numbers. There are some numbers that are IMPOSSIBLE though. For example..... 101 _ _ _ 201 101+201 = 302 So the sum of the other 3 integers would have to be 448. Since the 3 integers have to be the SAME number, there is no way for them to sum to 448. Thus, the largest number CANNOT be 201 This same "problem" happens with.... 102 _ _ _ 202 102+202 = 304 So the sum of the other 3 integers would have to be 446. Since the 3 integers have to be the SAME number, there is no way for them to sum to 446. Thus, the largest number CANNOT be 202 We DO come across another solution though with.... 103 _ _ _ 203 Here, the 3 other integers would be 148 103, 148, 148, 148, 203 This provides the basis for the pattern that we need to answer this question: We can increase/decrease the smallest and largest integers by 3 and decrease/increase the other integers by 2. The options that "fit" would start with.... 103 106 109 112 115 118 121 124 127, 130, 130, 130, 227 At this point, there are no other options, since the "middle 3" integers have to be BIGGER than the smallest integer and that would not happen beyond that last example. This proves that there are 9 options beyond the first one (the 100, 150,150,150, 200 option). We can use this same pattern to SUBTRACT 3 from the smallest and largest values and ADD 2 to the other values.... Those options would be 97 152 152 152 197 94 91 88 85 82 79 76 73 168 168 168 173 Just as in the prior list, there are no other options past the last one. This is another 9 options. In total, we have 1+9+9 = 19 options. Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: A set of five positive integers has an arithmetic mean of 150. A parti [#permalink]
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07 Apr 2015, 23:37
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Bunuel wrote: A set of five positive integers has an arithmetic mean of 150. A particular number among the five exceeds another by 100. The rest of the three numbers lie between these two numbers and are equal. How many different values can the largest number among the five take?
(A) 18 (B) 19 (C) 21 (D) 42 (E) 59
Kudos for a correct solution. numbers are a,b,b,b,100+a 2a+3b+100=150*5=750 \(a= \frac{650  3b}{2}\) , or \(a= 325  \frac{3b}{2}\) , note that 'b' has to be an even integer in order to get a an integer value . we know b>a so b> \(\frac{650  3b}{2}\) so b>130we know that b<100+aso b < \(100 + \frac{650  3b}{2}\) so b<170130<b<170 and since b is even , we will get only 19 such values. Answer 19.
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Re: A set of five positive integers has an arithmetic mean of 150. A parti [#permalink]
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07 Apr 2015, 23:42
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Lucky2783 wrote: Bunuel wrote: A set of five positive integers has an arithmetic mean of 150. A particular number among the five exceeds another by 100. The rest of the three numbers lie between these two numbers and are equal. How many different values can the largest number among the five take?
(A) 18 (B) 19 (C) 21 (D) 42 (E) 59
Kudos for a correct solution. numbers are a,b,b,b,100+a 2a+3b+100=150*5=750 \(a= \frac{650  3b}{2}\) , or \(a= 325  \frac{3b}{2}\) , note that 'b' has to be an even integer in order to get a an integer value . we know b>a so b> \(\frac{650  3b}{2}\) so b>130we know that b<100+aso b < \(100 + \frac{650  3b}{2}\) so b<170130<b<170 and since b is even , we will get only 19 such values. Answer 19. need not to mention that for each unique value of 'b' there will be a unique value of 'a' and '100+a' .
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Re: A set of five positive integers has an arithmetic mean of 150. A parti [#permalink]
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08 Apr 2015, 00:02
Zhenek wrote: Lets say numbers are in this order: x < y = y = y < x+100. Question is, how many different values can x take. x+3y+x+100 = 750 2x+3y = 650 We have condition x < y < x + 100 so lets use it 2x + 3y < 2x + 3*(x+100) = 5x + 300 2x + 3y > 2x + 3x = 5x 2x + 3y = 650
So we got double inequality: 5x < 650 < 5x + 300 which turns into 350 < 5x < 650 70 < x < 130 x can take 130701 = 59 values which corresponds to answer E note 2x+3y = 650 > \(y= \frac{650  2x}{3} > \frac{(650+x)  3x}{3}\) implies 650+x should be multiple of 3 . X is of the form (2k1) . have a look at it.
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Re: A set of five positive integers has an arithmetic mean of 150. A parti [#permalink]
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13 Apr 2015, 07:16
Bunuel wrote: A set of five positive integers has an arithmetic mean of 150. A particular number among the five exceeds another by 100. The rest of the three numbers lie between these two numbers and are equal. How many different values can the largest number among the five take?
(A) 18 (B) 19 (C) 21 (D) 42 (E) 59
Kudos for a correct solution. VERITAS PREP OFFICIAL SOLUTION:Let’s assume that the 5 natural numbers in increasing order are: a, b, b, b, a+100 We are given that a < b < a+100. Also, we are given that a and b are positive integers. This information is critical – we will see later why. The average of the 5 numbers is (a+b+b+b+a+100)/5 = 150 (a+b+b+b+a+100) = 5*150 2a+3b = 650 We need to find the number of distinct values that a can take because a+100 will also take the same number of distinct values. Now there are two methods to proceed. Let’s discuss both of them. Method 1: Pure Algebra – Write b in terms of a and plug it in the inequality b = (650 – 2a)/3 a < (650 – 2a)/3 < a+100 3a < 650 – 2a < 3a + 300 Now split it into two inequalities: 3a < 650 – 2a and 650 – 2a < 3a + 300 Inequality 1: 3a < 650 – 2a 5a < 650 a < 130 Inequality 2: 650 – 2a < 3a + 300 5a > 350 a > 70 So we get that 70 < a < 130. Since a is an integer, can we say that a can take all values from 71 to 129? No. What we are forgetting is that b is also an integer. We know that b = (650 – 2a)/3 For which values will be get b as an integer? Note that 650 is not divisible by 3. You need to add 1 to it or subtract 2 out of it to make it divisible by 3. So a should be of the form 3x+1. b = (650 – 2*(3x+1))/3 = (648 – 6x)/3 = 216 – 2x Here, for any positive integer x, b will be an integer. From 71 to 129, we have the following numbers which are of the form 3x+1: 73, 76, 79, 82, 85, … 127 This is an Arithmetic Progression. How many terms are there here? Last term = First term + (n – 1)*Common Difference 127 = 73 + (n – 1)*3 n = 19 a will take 19 distinct values so the last term i.e. (a+100) will also take 19 distinct values. Method 2: Using Transition Points Note that a < b < a+100 Since a < b, let’s find the point where a = b, i.e. the transition point 2a + 3a = 650 a = 130 = b But b must be greater than a. If we increase b by 1, we need to decrease a by 3 to keep the average same. But decreasing a by 3 decreases the largest number i.e. a+100 by 3 too; so we need to increase b by another 1. We get a = 127 and b = 132. This give us the numbers as 127, 132, 132, 132, 227. Here the average is 150 Since b < a+100, let’s find the point where b = a+100 2a + 3(a+100) = 650 a = 70, b = 170 But b must be less than a+100. If we decrease b by 1, we need to increase a by 3 to keep the average same. But increasing a by 3 increases the largest number, i.e. a+100 by 3 too, so we need to decrease b by another 1. We get a = 73 and b = 168. This gives us the numbers as 73, 168, 168, 168, 173. Here the average is 150 Values of a will be: 73, 76, 79, ….127 (Difference of 3 to make b an integer) This is an Arithmetic Progression. Last term = First term + (n – 1)*Common difference 127 = 73 + (n – 1)*3 n = 19 a will take 19 distinct values so the last term i.e. (a+100) will also take 19 distinct values. Answer (B)
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A set of five positive integers has an arithmetic mean of 150. A parti [#permalink]
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23 Sep 2016, 11:56
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Here is my solution:
n=5; average=150 => sum=750
we have a n n n a+100 => 2a+3n=750100=650
also we know that a<n<a+100. than means that n=a+T, where T is some number from 1 to 99
then we have 2a+3a+3*T and this equals to 650 5a+3*T=650; 5a=6503*T.
a=133*T/5. As "a" must be integer then T must be a multiple of 5. Thus, T can be 5,10,15,...,95.
Lets count this possibilities:
(955)/5+1=19
Thus T, and therefore a and a+100 have 19 different values.
Answer: (B)
Hope this helps



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Re: A set of five positive integers has an arithmetic mean of 150. A parti [#permalink]
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I used Diophantine equations for solving this question. May be somebody will find this approach easier. One integer greater by 100, rest are the same. So we have: a+b+b+b+a+100=750 2a+3b=650 Now let’s solve this Diophantine equation taking into consideration following restriction: a<b<a+100 3b=6502a => 3b=2(325a) b=2n, a=3253n a should be a positive integer, so 3253n>0, solving this we have first inequality n<108 Now we apply second restriction – a<b 3253n<2n => 325<5n => n>65 And final restriction b<a+100 2n<3253n+100 => 5n<425 => n<85 Combining above inequalities we have 65<n<85 => 66≤n≤84 from where number of possible values including the boundaries is 8466+1=19 Answer B Hope this will help. Regards



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A set of five positive integers has an arithmetic mean of 150. A parti [#permalink]
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29 Oct 2016, 15:14
Bunuel wrote: A set of five positive integers has an arithmetic mean of 150. A particular number among the five exceeds another by 100. The rest of the three numbers lie between these two numbers and are equal. How many different values can the largest number among the five take?
(A) 18 (B) 19 (C) 21 (D) 42 (E) 59
Kudos for a correct solution. let the 5 integers be a,b,b,b,a+100 to find lowest values that work, let a+100=b thus, 5b=850➡b=170 70+170+170+170+170=750 this won't work, but if we add 3 to a and subtract 2 from b we get 73+168+168+168+173=750, the lowest set, and 76+166+166+166+176=750, the next lowest set let x=the number of different values the largest number may take 73+3x=1682x 5x=95 x=19 B



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Re: A set of five positive integers has an arithmetic mean of 150. A parti [#permalink]
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10 Aug 2017, 20:31
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Bunuel wrote: A set of five positive integers has an arithmetic mean of 150. A particular number among the five exceeds another by 100. The rest of the three numbers lie between these two numbers and are equal. How many different values can the largest number among the five take?
(A) 18 (B) 19 (C) 21 (D) 42 (E) 59
Kudos for a correct solution. 5 numbers will be a, a+100, a+x, a+x, a+x (x can be between 1 and 99) 5a + 100 + 3x = 750 5a = 6503x 6503x needs to be a multiple of 5. This happens when 3x is a multiple of 5 => x is a list of multiples of 5 (between 1,99) 5,10,15, 95 19 values (B)



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Re: A set of five positive integers has an arithmetic mean of 150. A parti [#permalink]
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12 Aug 2017, 04:22
I used this approach and felt its the easiest:
x+y+y+y+(x+100) = 5*150 = 750  (1)
2x + 3y = 650  (2)
now we want to higher and lower limit, so we take, x=y :
5x = 650 x = 130
this implies, the set will be  130,130,130,130,230
lets try, 129  129+3y+229 = 750 3y = 750358  y will not be an integer as 352 is not divisible by 3 so, if you start thinking, even without doing the substitution and all... straight away analyse whether x + (x+100) is divisible by 3 or not
you will get 127 + 227 =354  divisible by 3
so our one limit is 127,132,132,132,227
similarly by taking y = x + 100
we will get the other limit > 73,168,168,168,173
the highest value has decreased from 227 to 173, with a jump of 3 units  227173 = 57  57/3=19
hence A=19 is the right answer..
Hope this helped Kudos if it did..



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Re: A set of five positive integers has an arithmetic mean of 150. A parti [#permalink]
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04 Jan 2018, 08:48
Mamyan94 wrote: Here is my solution:
n=5; average=150 => sum=750
we have a n n n a+100 => 2a+3n=750100=650
also we know that a<n<a+100. than means that n=a+T, where T is some number from 1 to 99
then we have 2a+3a+3*T and this equals to 650 5a+3*T=650; 5a=6503*T.
a=133*T/5. As "a" must be integer then T must be a multiple of 5. Thus, T can be 5,10,15,...,95.
Lets count this possibilities:
(955)/5+1=19
Thus T, and therefore a and a+100 have 19 different values.
Answer: (B)
Hope this helps I feel a becomes negative after certain values of T hence may be a problem




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