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Updated on: 29 Mar 2014, 14:55
Originally posted by jokkemauritzen on 29 Mar 2014, 14:23.
Last edited by Bunuel on 29 Mar 2014, 14:55, edited 1 time in total.
Last edited by Bunuel on 29 Mar 2014, 14:55, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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73% (01:37) correct
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If (x # y) represents the remainder that results when the positive integer x is divided by the positive integer y, what is the sum of all the possible values of y such that (16 # y) = 1?
A. 8
B. 9
C. 16
D. 23
E. 24
Is there a quick and easy way to find divisors to an integer/expression when we are looking for a distinct remainders?
correct answer is, I found the answer by calculating and doing several divisions, is there an easier way?
A. 8
B. 9
C. 16
D. 23
E. 24
Is there a quick and easy way to find divisors to an integer/expression when we are looking for a distinct remainders?
correct answer is
23
29 Mar 2014, 15:02
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jokkemauritzen
(x # y) represents the remainder that results when the positive integer x is divided by the positive integer y.
Thus (16 # y) = 1 implies that \(16=yq+1\) --> \(15=yq\) --> y is a factor of 15. The factors of 15 are 1, 3, 5, and 15. Now, y cannot be 1, since 16 divided by 1 yields the remainder of 0 not 1.
Therefore the sum of all the possible values of y is 3+5+15=23.
Answer: D.
P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention to rules 3, and 8. Thank you.
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Updated on: 17 May 2021, 07:16
Originally posted by BrentGMATPrepNow on 19 Apr 2019, 06:07.
Last edited by BrentGMATPrepNow on 17 May 2021, 07:16, edited 1 time in total.
Last edited by BrentGMATPrepNow on 17 May 2021, 07:16, edited 1 time in total.
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jokkemauritzen
If y > 16, (16 # y) = 16, so we need only check the values from 1 to 15
Also, we need not check the FACTORS of 16, since they will all yield a remainder of 0
We're left with:
(16 # 3) = 1 KEEP!
(16 # 5) = 1 KEEP
(16 # 7) = 2
(16 # 9) = 7
(16 # 10) = 6
(16 # 11) = 5
(16 # 12) = 4
(16 # 13) = 3
(16 # 14) = 2
(16 # 15) = 1 KEEP
3 + 5 + 15 = 23
Answer: D
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