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The average (arithmetic mean) of 5 distinct, single digit integers is [#permalink]
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The average (arithmetic mean) of 5 distinct, single digit integers is 5. If two of the integers are discarded, the new average is 4. What is the greatest of these integers? (1) Exactly 3 of the integers are consecutive primes. (2) The least integer is 3.
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Last edited by Bunuel on 24 Jul 2017, 05:30, edited 2 times in total.
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Re: The average (arithmetic mean) of 5 distinct, single digit integers is [#permalink]
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02 Sep 2012, 20:29
A: Let's say 2, 3, 5, 7, 8 and we discard 5 & 8, the new average is 4. So A is sufficient
B: 3, 4, 5, 6, 7 and we discard 6 & 7, the new average is 4, so B is also sufficient.
Why is the OA not D?



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Re: The average (arithmetic mean) of 5 distinct, single digit integers is [#permalink]
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02 Sep 2012, 22:46
The solution with statement (1) is actually 3, 4, 5, 6, 7 too. Your solution has 4 consectutive primes. But "exactly 3" implies that there are not 4, I think.
So choices are either 2, 3, 5 without the 7. To get to average 5, the remaining two numbers would be 6 and 9. But then it's not possible to get average 4 with 3 of those numbers. So it's 3, 5, 7 without the 2. To get to average 5, the remaining two numbers can be 4 and 6 or 1 and 9. If it's 1 and 9, again, average 4 with 3 of those numbers isn't possible. So 3, 4, 5, 6, 7 is the solution.
I agree with you that statement (2) alone is sufficient too. Because given this statement, we have to take the lowest possible combination to get to average 25. If we take the number 8 or 9, we will exceed the average of 5 for sure.



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Re: The average (arithmetic mean) of 5 distinct, single digit integers is [#permalink]
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03 Sep 2012, 02:45
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Correct answer must be D, not A. The average of 5 distinct single digit integers is 5. If two of the integers are discarded, the new average is 4. What is the largest of the 5 integers?From the stem: The sum of 5 distinct single digit integers is 5*5=25; The sum of 3 of the integers is 3*4=12; The sum of the other 2 of the integers is 2512=13. (1) Exactly 3 of the integers are consecutive primes. We can have two cases: Case 1The three consecutive primes are: 2, 3 and 5 > 2+3+5=10. The remaining 2 integers could be: 0, 1, 4, 6, 8 or 9 (notice that neither of the remaining 2 integer can be 7, since in this case we would have 4 consecutive primes, not 3). The sum of these two integers must be 2510=15. From the list, only two integers whose sum is 15 are 6 and 9 (6+9=15). Now, in this case the 5 integers would be {2, 3, 5, 6, 9}, but no 2 integers from the list give the sum of 13, thus the case when 3 consecutive primes are 2, 3 and 5 is not possible. Case 2The three consecutive primes are: 3, 5, and 7 > 3+5+7=15. The remaining 2 integers could be: 0, 1, 4, 6, 8 or 9 (notice that neither of the remaining 2 integer can be 2, since in this case we would have 4 consecutive primes, not 3). The sum of these two integers must be 2515=10. From the list, there are two pairs of integers whose sum is 10: (1, 9) and (4, 6). Now, in this case the 5 integers would be {1, 3, 5, 7, 9} or {3, 4, 5, 6, 7}. From the first list there are no 2 integers whose sum is 13. In the second list, such two integers are 6 and 7. Hence, the 5 integers are {3, 4, 5, 6, 7}. Sufficient. (2) The smallest integer is 3. The sum of the other 4 integers, each of which must be greater than 3, must be 253=22. Only 4+5+6+7=22 is possible (the sum of any other 4 integers will be more than 22), so the 5 integers are {3, 4, 5, 6, 7}. Sufficient. Answer: D. Hope it's clear.
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Re: The average (arithmetic mean) of 5 distinct, single digit integers is [#permalink]
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01 Oct 2012, 23:55
Bunuel wrote: Correct answer must be D, not A.
The average of 5 distinct single digit integers is 5. If two of the integers are discarded, the new average is 4. What is the largest of the 5 integers?
From the stem: The sum of 5 distinct single digit integers is 5*5=25; The sum of 3 of the integers is 3*4=12; The sum of the other 2 of the integers is 2512=13.
(1) Exactly 3 of the integers are consecutive primes. We can have two cases:
Case 1 The three consecutive primes are: 2, 3 and 5 > 2+3+5=10. The remaining 2 integers could be: 0, 1, 4, 6, 8 or 9 (notice that neither of the remaining 2 integer can be 7, since in this case we would have 4 consecutive primes, not 3). The sum of these two integers must be 2510=15. From the list, only two integers whose sum is 15 are 6 and 9 (6+9=15). Now, in this case the 5 integers would be {2, 3, 5, 6, 9}, but no 2 integers from the list give the sum of 13, thus the case when 3 consecutive primes are 2, 3 and 5 is not possible.
Case 2 The three consecutive primes are: 3, 5, and 7 > 3+5+7=15. The remaining 2 integers could be: 0, 1, 4, 6, 8 or 9 (notice that neither of the remaining 2 integer can be 2, since in this case we would have 4 consecutive primes, not 3). The sum of these two integers must be 2515=10. From the list, there are two pairs of integers whose sum is 10: (1, 9) and (4, 6). Now, in this case the 5 integers would be {1, 3, 5, 7, 9} or {3, 4, 5, 6, 7}. From the first list there are no 2 integers whose sum is 13. In the second list, such two integers are 6 and 7.
Hence, the 5 integers are {3, 4, 5, 6, 7}. Sufficient.
(2) The smallest integer is 3. The sum of the other 4 integers, each of which must be greater than 3, must be 253=22. Only 4+5+6+7=22 is possible (the sum of any other 4 integers will be more than 22), so the 5 integers are {3, 4, 5, 6, 7}. Sufficient.
Answer: D.
Hope it's clear. The question does not mention that the integers are positive. What about the case where integers are negative?



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Re: The average (arithmetic mean) of 5 distinct, single digit integers is [#permalink]
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02 Oct 2012, 00:22
Nowhere in the question is it mentioned that the integers are positive. I can get a combination which suffices the required rules set with inclusion of negative integers.
Consider { 8, 2, 7, 11, 13},
1. Avg. {8, 2, 7, 11, 13} = 5
2. Avg. {8, 7, 13} = 4
3. We have only three consecutive integers 7, 11 and 13
So condition 1. alone does not suffice but condition 2. alone will be sufficient as the smallest integer is given to be 3 and the only combination of numbers we can get will be {3, 4, 5, 6, 7} this way.



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Re: The average (arithmetic mean) of 5 distinct, single digit integers is [#permalink]
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02 Oct 2012, 01:38
raghupro wrote: Bunuel wrote: Correct answer must be D, not A.
The average of 5 distinct single digit integers is 5. If two of the integers are discarded, the new average is 4. What is the largest of the 5 integers?
From the stem: The sum of 5 distinct single digit integers is 5*5=25; The sum of 3 of the integers is 3*4=12; The sum of the other 2 of the integers is 2512=13.
(1) Exactly 3 of the integers are consecutive primes. We can have two cases:
Case 1 The three consecutive primes are: 2, 3 and 5 > 2+3+5=10. The remaining 2 integers could be: 0, 1, 4, 6, 8 or 9 (notice that neither of the remaining 2 integer can be 7, since in this case we would have 4 consecutive primes, not 3). The sum of these two integers must be 2510=15. From the list, only two integers whose sum is 15 are 6 and 9 (6+9=15). Now, in this case the 5 integers would be {2, 3, 5, 6, 9}, but no 2 integers from the list give the sum of 13, thus the case when 3 consecutive primes are 2, 3 and 5 is not possible.
Case 2 The three consecutive primes are: 3, 5, and 7 > 3+5+7=15. The remaining 2 integers could be: 0, 1, 4, 6, 8 or 9 (notice that neither of the remaining 2 integer can be 2, since in this case we would have 4 consecutive primes, not 3). The sum of these two integers must be 2515=10. From the list, there are two pairs of integers whose sum is 10: (1, 9) and (4, 6). Now, in this case the 5 integers would be {1, 3, 5, 7, 9} or {3, 4, 5, 6, 7}. From the first list there are no 2 integers whose sum is 13. In the second list, such two integers are 6 and 7.
Hence, the 5 integers are {3, 4, 5, 6, 7}. Sufficient.
(2) The smallest integer is 3. The sum of the other 4 integers, each of which must be greater than 3, must be 253=22. Only 4+5+6+7=22 is possible (the sum of any other 4 integers will be more than 22), so the 5 integers are {3, 4, 5, 6, 7}. Sufficient.
Answer: D.
Hope it's clear. The question does not mention that the integers are positive. What about the case where integers are negative? Single digit integers mean integers from 0 till 9, inclusive.
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Re: The average (arithmetic mean) of 5 distinct, single digit integers is [#permalink]
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02 Oct 2012, 03:42
Bunuel wrote: raghupro wrote: Bunuel wrote: Correct answer must be D, not A.
The average of 5 distinct single digit integers is 5. If two of the integers are discarded, the new average is 4. What is the largest of the 5 integers?
From the stem: The sum of 5 distinct single digit integers is 5*5=25; The sum of 3 of the integers is 3*4=12; The sum of the other 2 of the integers is 2512=13.
(1) Exactly 3 of the integers are consecutive primes. We can have two cases:
Case 1 The three consecutive primes are: 2, 3 and 5 > 2+3+5=10. The remaining 2 integers could be: 0, 1, 4, 6, 8 or 9 (notice that neither of the remaining 2 integer can be 7, since in this case we would have 4 consecutive primes, not 3). The sum of these two integers must be 2510=15. From the list, only two integers whose sum is 15 are 6 and 9 (6+9=15). Now, in this case the 5 integers would be {2, 3, 5, 6, 9}, but no 2 integers from the list give the sum of 13, thus the case when 3 consecutive primes are 2, 3 and 5 is not possible.
Case 2 The three consecutive primes are: 3, 5, and 7 > 3+5+7=15. The remaining 2 integers could be: 0, 1, 4, 6, 8 or 9 (notice that neither of the remaining 2 integer can be 2, since in this case we would have 4 consecutive primes, not 3). The sum of these two integers must be 2515=10. From the list, there are two pairs of integers whose sum is 10: (1, 9) and (4, 6). Now, in this case the 5 integers would be {1, 3, 5, 7, 9} or {3, 4, 5, 6, 7}. From the first list there are no 2 integers whose sum is 13. In the second list, such two integers are 6 and 7.
Hence, the 5 integers are {3, 4, 5, 6, 7}. Sufficient.
(2) The smallest integer is 3. The sum of the other 4 integers, each of which must be greater than 3, must be 253=22. Only 4+5+6+7=22 is possible (the sum of any other 4 integers will be more than 22), so the 5 integers are {3, 4, 5, 6, 7}. Sufficient.
Answer: D.
Hope it's clear. The question does not mention that the integers are positive. What about the case where integers are negative? Single digit integers mean integers from 0 till 9, inclusive. Oops! Missed that part. Thanks for clarifying



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Re: The average (arithmetic mean) of 5 distinct, single digit integers is [#permalink]
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30 Aug 2013, 02:27
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Somehow I did not pay attention to the word "exactly" and that is why considered 2,3,5,7 a valid answer.



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Re: The average (arithmetic mean) of 5 distinct, single digit integers is [#permalink]
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10 Nov 2013, 09:13
Bunuel wrote: Correct answer must be D, not A.
The average of 5 distinct single digit integers is 5. If two of the integers are discarded, the new average is 4. What is the largest of the 5 integers?
From the stem: The sum of 5 distinct single digit integers is 5*5=25; The sum of 3 of the integers is 3*4=12; The sum of the other 2 of the integers is 2512=13.
(1) Exactly 3 of the integers are consecutive primes. We can have two cases:
Case 1 The three consecutive primes are: 2, 3 and 5 > 2+3+5=10. The remaining 2 integers could be: 0, 1, 4, 6, 8 or 9 (notice that neither of the remaining 2 integer can be 7, since in this case we would have 4 consecutive primes, not 3). The sum of these two integers must be 2510=15. From the list, only two integers whose sum is 15 are 6 and 9 (6+9=15). Now, in this case the 5 integers would be {2, 3, 5, 6, 9}, but no 2 integers from the list give the sum of 13, thus the case when 3 consecutive primes are 2, 3 and 5 is not possible.
Case 2 The three consecutive primes are: 3, 5, and 7 > 3+5+7=15. The remaining 2 integers could be: 0, 1, 4, 6, 8 or 9 (notice that neither of the remaining 2 integer can be 2, since in this case we would have 4 consecutive primes, not 3). The sum of these two integers must be 2515=10. From the list, there are two pairs of integers whose sum is 10: (1, 9) and (4, 6). Now, in this case the 5 integers would be {1, 3, 5, 7, 9} or {3, 4, 5, 6, 7}. From the first list there are no 2 integers whose sum is 13. In the second list, such two integers are 6 and 7.
Hence, the 5 integers are {3, 4, 5, 6, 7}. Sufficient.
(2) The smallest integer is 3. The sum of the other 4 integers, each of which must be greater than 3, must be 253=22. Only 4+5+6+7=22 is possible (the sum of any other 4 integers will be more than 22), so the 5 integers are {3, 4, 5, 6, 7}. Sufficient.
Answer: D.
Hope it's clear. Couldn't it still be 7 though, even though the hint tells you there's 3 consecutive primes...even with 7, you could still have 3 consecutive. Granted there would be 4 in a row, but WITHIN that series of 4, there are two different ways to have 3 consecutive primes. Or does the GMAT not play 'tricks' like that?



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Re: The average (arithmetic mean) of 5 distinct, single digit integers is [#permalink]
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10 Nov 2013, 10:48
AccipiterQ wrote: Bunuel wrote: Correct answer must be D, not A.
The average of 5 distinct single digit integers is 5. If two of the integers are discarded, the new average is 4. What is the largest of the 5 integers?
From the stem: The sum of 5 distinct single digit integers is 5*5=25; The sum of 3 of the integers is 3*4=12; The sum of the other 2 of the integers is 2512=13.
(1) Exactly 3 of the integers are consecutive primes. We can have two cases:
Case 1 The three consecutive primes are: 2, 3 and 5 > 2+3+5=10. The remaining 2 integers could be: 0, 1, 4, 6, 8 or 9 (notice that neither of the remaining 2 integer can be 7, since in this case we would have 4 consecutive primes, not 3). The sum of these two integers must be 2510=15. From the list, only two integers whose sum is 15 are 6 and 9 (6+9=15). Now, in this case the 5 integers would be {2, 3, 5, 6, 9}, but no 2 integers from the list give the sum of 13, thus the case when 3 consecutive primes are 2, 3 and 5 is not possible.
Case 2 The three consecutive primes are: 3, 5, and 7 > 3+5+7=15. The remaining 2 integers could be: 0, 1, 4, 6, 8 or 9 (notice that neither of the remaining 2 integer can be 2, since in this case we would have 4 consecutive primes, not 3). The sum of these two integers must be 2515=10. From the list, there are two pairs of integers whose sum is 10: (1, 9) and (4, 6). Now, in this case the 5 integers would be {1, 3, 5, 7, 9} or {3, 4, 5, 6, 7}. From the first list there are no 2 integers whose sum is 13. In the second list, such two integers are 6 and 7.
Hence, the 5 integers are {3, 4, 5, 6, 7}. Sufficient.
(2) The smallest integer is 3. The sum of the other 4 integers, each of which must be greater than 3, must be 253=22. Only 4+5+6+7=22 is possible (the sum of any other 4 integers will be more than 22), so the 5 integers are {3, 4, 5, 6, 7}. Sufficient.
Answer: D.
Hope it's clear. Couldn't it still be 7 though, even though the hint tells you there's 3 consecutive primes...even with 7, you could still have 3 consecutive. Granted there would be 4 in a row, but WITHIN that series of 4, there are two different ways to have 3 consecutive primes. Or does the GMAT not play 'tricks' like that? I guess you are talking about case 1 in the first statement. Please be more specific when asking a question. Now, if there is 7 and 4 then the set is {x, 2, 3, 4, 5, 7}. Ask yourself: how many consecutive primes are there?
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Re: The average (arithmetic mean) of 5 distinct, single digit integers is [#permalink]
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10 Nov 2013, 15:15
Bunuel wrote: AccipiterQ wrote: Bunuel wrote: Correct answer must be D, not A.
The average of 5 distinct single digit integers is 5. If two of the integers are discarded, the new average is 4. What is the largest of the 5 integers?
From the stem: The sum of 5 distinct single digit integers is 5*5=25; The sum of 3 of the integers is 3*4=12; The sum of the other 2 of the integers is 2512=13.
(1) Exactly 3 of the integers are consecutive primes. We can have two cases:
Case 1 The three consecutive primes are: 2, 3 and 5 > 2+3+5=10. The remaining 2 integers could be: 0, 1, 4, 6, 8 or 9 (notice that neither of the remaining 2 integer can be 7, since in this case we would have 4 consecutive primes, not 3). The sum of these two integers must be 2510=15. From the list, only two integers whose sum is 15 are 6 and 9 (6+9=15). Now, in this case the 5 integers would be {2, 3, 5, 6, 9}, but no 2 integers from the list give the sum of 13, thus the case when 3 consecutive primes are 2, 3 and 5 is not possible.
Case 2 The three consecutive primes are: 3, 5, and 7 > 3+5+7=15. The remaining 2 integers could be: 0, 1, 4, 6, 8 or 9 (notice that neither of the remaining 2 integer can be 2, since in this case we would have 4 consecutive primes, not 3). The sum of these two integers must be 2515=10. From the list, there are two pairs of integers whose sum is 10: (1, 9) and (4, 6). Now, in this case the 5 integers would be {1, 3, 5, 7, 9} or {3, 4, 5, 6, 7}. From the first list there are no 2 integers whose sum is 13. In the second list, such two integers are 6 and 7.
Hence, the 5 integers are {3, 4, 5, 6, 7}. Sufficient.
(2) The smallest integer is 3. The sum of the other 4 integers, each of which must be greater than 3, must be 253=22. Only 4+5+6+7=22 is possible (the sum of any other 4 integers will be more than 22), so the 5 integers are {3, 4, 5, 6, 7}. Sufficient.
Answer: D.
Hope it's clear. Couldn't it still be 7 though, even though the hint tells you there's 3 consecutive primes...even with 7, you could still have 3 consecutive. Granted there would be 4 in a row, but WITHIN that series of 4, there are two different ways to have 3 consecutive primes. Or does the GMAT not play 'tricks' like that? I guess you are talking about case 1 in the first statement. Please be more specific when asking a question. Now, if there is 7 and 4 then the set is {x, 2, 3, 4, 5, 7}. Ask yourself: how many consecutive primes are there? ah ok, my question was more in regards to whether the GMAT would ever state that there were 3 consecutive primes in a set, but that in the solution you find out there were actually more than 3, and they just happened to tell you about 3 of them.



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Re: The average (arithmetic mean) of 5 distinct, single digit integers is [#permalink]
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11 Jan 2014, 02:39
Bunuel wrote: Correct answer must be D, not A.
The average of 5 distinct single digit integers is 5. If two of the integers are discarded, the new average is 4. What is the largest of the 5 integers?
From the stem: The sum of 5 distinct single digit integers is 5*5=25; The sum of 3 of the integers is 3*4=12; The sum of the other 2 of the integers is 2512=13.
(1) Exactly 3 of the integers are consecutive primes. We can have two cases:
Case 1 The three consecutive primes are: 2, 3 and 5 > 2+3+5=10. The remaining 2 integers could be: 0, 1, 4, 6, 8 or 9 (notice that neither of the remaining 2 integer can be 7, since in this case we would have 4 consecutive primes, not 3). The sum of these two integers must be 2510=15. From the list, only two integers whose sum is 15 are 6 and 9 (6+9=15). Now, in this case the 5 integers would be {2, 3, 5, 6, 9}, but no 2 integers from the list give the sum of 13, thus the case when 3 consecutive primes are 2, 3 and 5 is not possible.
Case 2 The three consecutive primes are: 3, 5, and 7 > 3+5+7=15. The remaining 2 integers could be: 0, 1, 4, 6, 8 or 9 (notice that neither of the remaining 2 integer can be 2, since in this case we would have 4 consecutive primes, not 3). The sum of these two integers must be 2515=10. From the list, there are two pairs of integers whose sum is 10: (1, 9) and (4, 6). Now, in this case the 5 integers would be {1, 3, 5, 7, 9} or {3, 4, 5, 6, 7}. From the first list there are no 2 integers whose sum is 13. In the second list, such two integers are 6 and 7.
Hence, the 5 integers are {3, 4, 5, 6, 7}. Sufficient.
(2) The smallest integer is 3. The sum of the other 4 integers, each of which must be greater than 3, must be 253=22. Only 4+5+6+7=22 is possible (the sum of any other 4 integers will be more than 22), so the 5 integers are {3, 4, 5, 6, 7}. Sufficient.
Answer: D.
Hope it's clear. Can such questions, which involve quite good amount of "trial and error" technique, be expected on GMAT?



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Re: The average (arithmetic mean) of 5 distinct, single digit integers is [#permalink]
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14 Aug 2014, 22:56
What about 2+3+5+11+4=25, it has 4 primes but 3 consecutive primes??



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Re: The average (arithmetic mean) of 5 distinct, single digit integers is [#permalink]
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Re: The average (arithmetic mean) of 5 distinct, single digit integers is [#permalink]
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31 Aug 2014, 04:49
Bunuel wrote: Single digit integers mean integers from 0 till 9, inclusive.
Hi Bunuel, Do you have a link to some article or post that explains these phrases? By default, for me at least, "Single digit integers mean integers from 0 till 9, inclusive." means that these are the number from 9 ... 9.



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Re: The average (arithmetic mean) of 5 distinct, single digit integers is [#permalink]
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15 Dec 2015, 19:53
the solution to (i) is 3,4,5,6 and 7 only and from (ii) also, the largest number is 7.



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Re: The average (arithmetic mean) of 5 distinct, single digit integers is [#permalink]
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15 Aug 2016, 21:32
I'm confused about part (1)
I get 2,3,5,6,9 AND 1,3,5,7,9
Both of which average to 25, and can remove two numbers to average 4. The only issue I can see is that it doesn't follow "three consecutive prime numbers" to which I would say I don't have three consecutive primes. The first set everyone I'm sure agrees with, the second set 1,3,5,7 aren't consecutive primes because there is a 2 between 1 and 3. I would be frustrated if someone said the second set is NOT "consecutive primes" because then that would mean i got this problem wrong because of a vague interpretation of what "consecutive primes" means.



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Re: The average (arithmetic mean) of 5 distinct, single digit integers is [#permalink]
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03 Jan 2017, 19:49
I don't understand why the answer is D. I think the answer should be B.
I evaluated it as follows: avg of 5 distinct integers = 5 => sum of 5 distinct integers = 25 avg of 3 distinct integers = 4 => sum of 3 distinct integers = 12, when 2 of the above integers are discarded.
(1) Exactly 3 of the integers are consecutive primes. I got 3 solutions for this case: 2,3,5,7,8 or 2,3,4,7,8 or 3,4,5,6,7 therefore the largest integer can be either 8 or 9 => insufficient.
Unless, because (1) says exactly 3 integers are consecutive primes, meaning that only 3 of the nos. are prime, that eliminates 2 scenarios I mentioned and the solution can only be 2,3,4,7,8. In this case the largest integer can only be 8 => sufficient!
(2) The smallest integer is 3. The solution is 3,4,5,6,7 to give 25, and then drop 6,7 to give 12. Largest integer is 7 => sufficient!
Sorry, I guess I answered my own question, the answer is D.



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Re: The average (arithmetic mean) of 5 distinct, single digit integers is [#permalink]
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04 Jan 2017, 01:48
AnitaBhatt wrote: I don't understand why the answer is D. I think the answer should be B.
I evaluated it as follows: avg of 5 distinct integers = 5 => sum of 5 distinct integers = 25 avg of 3 distinct integers = 4 => sum of 3 distinct integers = 12, when 2 of the above integers are discarded.
(1) Exactly 3 of the integers are consecutive primes. I got 3 solutions for this case: 2,3,5,7,8 or 2,3,4,7,8 or 3,4,5,6,7 therefore the largest integer can be either 8 or 9 => insufficient.
Unless, because (1) says exactly 3 integers are consecutive primes, meaning that only 3 of the nos. are prime, that eliminates 2 scenarios I mentioned and the solution can only be 2,3,4,7,8. In this case the largest integer can only be 8 => sufficient!
(2) The smallest integer is 3. The solution is 3,4,5,6,7 to give 25, and then drop 6,7 to give 12. Largest integer is 7 => sufficient!
Sorry, I guess I answered my own question, the answer is D. For more you can check the following posts: theaverageof5distinctsingledigitintegersis5if138274.html#p1118237theaverageof5distinctsingledigitintegersis5if138274.html#p1290690They address the doubt you had.
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Re: The average (arithmetic mean) of 5 distinct, single digit integers is
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