Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 49496

The remainder when positive integer N is divided by 2 is 1, when divid
[#permalink]
Show Tags
24 Apr 2018, 23:30
Question Stats:
65% (03:02) correct 35% (03:00) wrong based on 149 sessions
HideShow timer Statistics




eGMAT Representative
Joined: 04 Jan 2015
Posts: 2014

Re: The remainder when positive integer N is divided by 2 is 1, when divid
[#permalink]
Show Tags
24 Apr 2018, 23:49
Solution Given:• N is a positive integer • When N is divided by 2, 3, 4, and 5, the remainders are 1, 2, 3, and 4 respectively To find:• The sum of the digits of the least possible value of N Approach and Working: • The number N can be written as: o N = 2a + 1 = 2a + 2 – 1 = 2(a+1) – 1 = 2a’ – 1 • Therefore, the least value of number N = lcm (2, 3, 4, 5) – 1 Or, least (N) = 60 – 1 = 59 • Sum of digits of least (N) = 5 + 9 = 14 Hence, the correct answer is Option C. Answer: C
_________________
Register for free sessions Number Properties  Algebra Quant Workshop
Success Stories Guillermo's Success Story  Carrie's Success Story
Ace GMAT quant Articles and Question to reach Q51  Question of the week
Must Read Articles Number Properties – Even Odd  LCM GCD  Statistics1  Statistics2 Word Problems – Percentage 1  Percentage 2  Time and Work 1  Time and Work 2  Time, Speed and Distance 1  Time, Speed and Distance 2 Advanced Topics Permutation and Combination 1  Permutation and Combination 2  Permutation and Combination 3  Probability Geometry Triangles 1  Triangles 2  Triangles 3  Common Mistakes in Geometry Algebra Wavy line  Inequalities Practice Questions Number Properties 1  Number Properties 2  Algebra 1  Geometry  Prime Numbers  Absolute value equations  Sets
 '4 out of Top 5' Instructors on gmatclub  70 point improvement guarantee  www.egmat.com




Director
Joined: 02 Oct 2017
Posts: 616

Re: The remainder when positive integer N is divided by 2 is 1, when divid
[#permalink]
Show Tags
13 May 2018, 03:30
For remainder by 5 to be 4 we need to have unit digit either 4 or 9
If unit digit is 4 then that number is always divisible by 2 so this possibility is out
We need to find number with unit digit 9 Possible numbers are19,29,39,49,59,69,79,89,99 9 is discarded as sum in options is at least 11 39,69,99 are divisible by 3 so not possible 19,49,79 give remainder 1 by 3 so out Among Number(29,59,89)above only 59 fulfill criteria So sum is 14
Give kudos if it helps
Posted from my mobile device




Senior Manager
Joined: 21 Jan 2015
Posts: 347
Location: India
Concentration: Strategy, Marketing
GMAT 1: 620 Q48 V28 GMAT 2: 690 Q49 V35
WE: Sales (Consumer Products)

Re: The remainder when positive integer N is divided by 2 is 1, when divid
[#permalink]
Show Tags
24 Apr 2018, 23:56
Ans: C as it is given that number N leaves reminder of 4 when divided by 5 so : Number must be= 5i+4 where i is an int. now. we get values of possible N by putting i= 1,2,3,4,5.. and so on. Possibilities: N= 9, 14,19,24,29,34,39 and so on. but as given number is not divisible by 2 and 3 so we will lest out the even numbers and multiple of 3 from the list which will leave us possible N= 19, 29, 49, 59, 79.. and so on.. now just follow the conditions and divide these by 2,3,4 and check for reminders. 59 fulfill it. so sum of the digits is5+9=14. Ans : 14
_________________
 The Mind is Everything, What we Think we Become. Kudos will encourage many others, like me. Please Give Kudos !! Thanks



VP
Joined: 07 Dec 2014
Posts: 1091

The remainder when positive integer N is divided by 2 is 1, when divid
[#permalink]
Show Tags
13 May 2018, 13:02
Bunuel wrote: The remainder when positive integer N is divided by 2 is 1, when divided by 3 is 2, when divided by 4 is 3, and when divided by 5 is 4. What is the sum of the digits of the least possible value of N?
(A) 11 (B) 13 (C) 14 (D) 16 (E) 17 let x=difference between first two quotients: x=(n1)/2(n2)/3➡(n+1)/6 thus, n+1 is multiple of 6 let y=difference between last two quotients: y=(n3)/4(n4)/5➡(n+1)/20 thus, n+1 is a multiple of 20 if n+1 is a multiple of 20 and 6, then least possible value of n=59 5+9=14 C



Senior Manager
Joined: 31 Jul 2017
Posts: 464
Location: Malaysia
GPA: 3.95
WE: Consulting (Energy and Utilities)

Re: The remainder when positive integer N is divided by 2 is 1, when divid
[#permalink]
Show Tags
14 May 2018, 02:28
Bunuel wrote: The remainder when positive integer N is divided by 2 is 1, when divided by 3 is 2, when divided by 4 is 3, and when divided by 5 is 4. What is the sum of the digits of the least possible value of N?
(A) 11 (B) 13 (C) 14 (D) 16 (E) 17 The best way to solve this is  Any number divisible by 5 ends with 0 or 5. Numbers ending with 0 are divisible by 2, hence, we need to consider number ending with 5 + 4. For Ex  9, 19,29,39,49,59,69,79.. etc. So, minimum would be 59.
_________________
If my Post helps you in Gaining Knowledge, Help me with KUDOS.. !!



Senior Manager
Joined: 31 May 2017
Posts: 325

Re: The remainder when positive integer N is divided by 2 is 1, when divid
[#permalink]
Show Tags
14 May 2018, 20:10
Solved this using substituting values Constraint 1: The number should not be divisible by 2,3 or 4. So this removes any even number from contention. Constraint: When divided by 5, the remainder should be 4. So the number cannot end in 1, 6, 7. Also number cannot end in 0. From the above 2 constraints , we get that number should end in 9. But at the same time cannot be divisible by 3. The options are 19,29,39,49,59,69,79,89,99. Out of these numbers 39,69 and 99 are divisible by 3. 19, 49,79  gives remainder of 1 when divided by 3. 29  Remainder 1 when divided by 4 59 is the option left which satisfied all constraints. When the digits are added , gives 5+9=14 Ans : C
_________________
Please give kudos if it helps
Resources Ultimate GMAT Quantitative Megathread  ALL YOU NEED FOR QUANT ! ! !  SC Blogs by Magoosh  How to improve your verbal score  Things i wish i could've done earlier  Ultimate Q51 Guide



Target Test Prep Representative
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2835

Re: The remainder when positive integer N is divided by 2 is 1, when divid
[#permalink]
Show Tags
15 May 2018, 16:14
Bunuel wrote: The remainder when positive integer N is divided by 2 is 1, when divided by 3 is 2, when divided by 4 is 3, and when divided by 5 is 4. What is the sum of the digits of the least possible value of N?
(A) 11 (B) 13 (C) 14 (D) 16 (E) 17 Let’s use the last condition first: when N is divided by 5, the remainder is 4. The reason is: 5 is the largest divisor here, so if we list the numbers that satisfies this condition, the numbers will increase faster than those of the other three conditions. Recall that if the remainder is 4 when a number is divided by 5, the units digit must be either 4 or 9. However, it cannot be 4 since we are told that there is a remainder of 1 when the N is divided by 2. Now let’s list the numbers: 9, 19, 29, 39, 49, 59, 69, … We can also omit the numbers 9, 39, 69, etc. since these numbers are divisible by 3 (we are given that N is not divisible by 3). So we are left with: 19, 29, 49, 59, 79, 89, … We see that all of these numbers have a remainder of 1 when divided by 2, so let’s check division by 3 and 4: 19/3 = 6 R 1…. No 29/3 = 9 R 2…. Yes; 29/4 = 7 R 1 …. No 49/3 = 16 R 1 …. No 59/3 = 19 R 2…. Yes; 59/4 = 14 R …. Yes Thus 59 is the least possible value of N and 5 + 9 = 14. Alternate solution: Notice that each remainder is the largest possible remainder of their respective divisor (for example, 3 is the largest possible remainder when a number is divided by 4). Therefore, if we can the find a number that has remainder of 0 when it’s divided by 2, 3, 4 and 5, and then subtract 1 from it, we will have a suitable value for N. Since we want the least possible value of N, we want to find the smallest positive integer that is divisible by 2, 3, 4 and 5. This number is, of course, the least common multiple of the 4 numbers and it is: LCM(2, 3, 4, 5) = LCM(2, 3, 2^2, 5) = 2^2 x 3 x 5 = 60. Since 60 is the LCM of (2, 3, 4, 5), 60  1 = 59 is the least possible value of N. Let’s verify it: 59/2 = 29 R 1; 59/3 = 19 R 2; 59/4 = 14 R 3 and 59/5 = 11 R 4. Thus the sum of the digits of N = 5 + 9 = 14. Answer: C
_________________
Jeffery Miller
Head of GMAT Instruction
GMAT Quant SelfStudy Course
500+ lessons 3000+ practice problems 800+ HD solutions



Intern
Joined: 07 Jul 2018
Posts: 7

Re: The remainder when positive integer N is divided by 2 is 1, when divid
[#permalink]
Show Tags
08 Jul 2018, 10:25
EgmatQuantExpert wrote: Solution Given:• N is a positive integer • When N is divided by 2, 3, 4, and 5, the remainders are 1, 2, 3, and 4 respectively To find:• The sum of the digits of the least possible value of N Approach and Working: • The number N can be written as: o N = 2a + 1 = 2a + 2 – 1 = 2(a+1) – 1 = 2a’ – 1 • Therefore, the least value of number N = lcm (2, 3, 4, 5) – 1 Or, least (N) = 60 – 1 = 59 • Sum of digits of least (N) = 5 + 9 = 14 Hence, the correct answer is Option C. Answer: CHello, Would you please tell me where I can find this rule: N= lcm Q + Constant R? Thank you very much Jetmat
_________________
Welcoming critics is my way to improvement. So do not hesitate, tell me how I can improve. Thx



Director
Joined: 09 Mar 2016
Posts: 886

Re: The remainder when positive integer N is divided by 2 is 1, when divid
[#permalink]
Show Tags
20 Sep 2018, 09:57
JeffTargetTestPrep wrote: Bunuel wrote: The remainder when positive integer N is divided by 2 is 1, when divided by 3 is 2, when divided by 4 is 3, and when divided by 5 is 4. What is the sum of the digits of the least possible value of N?
(A) 11 (B) 13 (C) 14 (D) 16 (E) 17 Let’s use the last condition first: when N is divided by 5, the remainder is 4. The reason is: 5 is the largest divisor here, so if we list the numbers that satisfies this condition, the numbers will increase faster than those of the other three conditions. Recall that if the remainder is 4 when a number is divided by 5, the units digit must be either 4 or 9. However, it cannot be 4 since we are told that there is a remainder of 1 when the N is divided by 2. Now let’s list the numbers: 9, 19, 29, 39, 49, 59, 69, … We can also omit the numbers 9, 39, 69, etc. since these numbers are divisible by 3 (we are given that N is not divisible by 3). So we are left with: 19, 29, 49, 59, 79, 89, … We see that all of these numbers have a remainder of 1 when divided by 2, so let’s check division by 3 and 4: 19/3 = 6 R 1…. No 29/3 = 9 R 2…. Yes; 29/4 = 7 R 1 …. No 49/3 = 16 R 1 …. No 59/3 = 19 R 2…. Yes; 59/4 = 14 R …. Yes Thus 59 is the least possible value of N and 5 + 9 = 14. Alternate solution: Notice that each remainder is the largest possible remainder of their respective divisor (for example, 3 is the largest possible remainder when a number is divided by 4). Therefore, if we can the find a number that has remainder of 0 when it’s divided by 2, 3, 4 and 5, and then subtract 1 from it, we will have a suitable value for N. Since we want the least possible value of N, we want to find the smallest positive integer that is divisible by 2, 3, 4 and 5. This number is, of course, the least common multiple of the 4 numbers and it is: LCM(2, 3, 4, 5) = LCM(2, 3, 2^2, 5) = 2^2 x 3 x 5 = 60. Since 60 is the LCM of (2, 3, 4, 5), 60  1 = 59 is the least possible value of N. Let’s verify it: 59/2 = 29 R 1; 59/3 = 19 R 2; 59/4 = 14 R 3 and 59/5 = 11 R 4. Thus the sum of the digits of N = 5 + 9 = 14. Answer: C pushpitkc hello there on what rule is this statement based ? can you pls explain ? Recall that if the remainder is 4 when a number is divided by 5, the units digit must be either 4 or 9.
_________________
In English I speak with a dictionary, and with people I am shy.



Director
Joined: 09 Mar 2016
Posts: 886

The remainder when positive integer N is divided by 2 is 1, when divid
[#permalink]
Show Tags
20 Sep 2018, 10:34
JeffTargetTestPrep wrote: Bunuel wrote: The remainder when positive integer N is divided by 2 is 1, when divided by 3 is 2, when divided by 4 is 3, and when divided by 5 is 4. What is the sum of the digits of the least possible value of N?
(A) 11 (B) 13 (C) 14 (D) 16 (E) 17 Let’s use the last condition first: when N is divided by 5, the remainder is 4. The reason is: 5 is the largest divisor here, so if we list the numbers that satisfies this condition, the numbers will increase faster than those of the other three conditions. Recall that if the remainder is 4 when a number is divided by 5, the units digit must be either 4 or 9. However, it cannot be 4 since we are told that there is a remainder of 1 when the N is divided by 2. Now let’s list the numbers: 9, 19, 29, 39, 49, 59, 69, … We can also omit the numbers 9, 39, 69, etc. since these numbers are divisible by 3 (we are given that N is not divisible by 3). So we are left with: 19, 29, 49, 59, 79, 89, … We see that all of these numbers have a remainder of 1 when divided by 2, so let’s check division by 3 and 4: 19/3 = 6 R 1…. No 29/3 = 9 R 2…. Yes; 29/4 = 7 R 1 …. No 49/3 = 16 R 1 …. No 59/3 = 19 R 2…. Yes; 59/4 = 14 R …. Yes Thus 59 is the least possible value of N and 5 + 9 = 14. Alternate solution: Notice that each remainder is the largest possible remainder of their respective divisor (for example, 3 is the largest possible remainder when a number is divided by 4). Therefore, if we can the find a number that has remainder of 0 when it’s divided by 2, 3, 4 and 5, and then subtract 1 from it, we will have a suitable value for N. Since we want the least possible value of N, we want to find the smallest positive integer that is divisible by 2, 3, 4 and 5. This number is, of course, the least common multiple of the 4 numbers and it is: LCM(2, 3, 4, 5) = LCM(2, 3, 2^2, 5) = 2^2 x 3 x 5 = 60. Since 60 is the LCM of (2, 3, 4, 5), 60  1 = 59 is the least possible value of N. Let’s verify it: 59/2 = 29 R 1; 59/3 = 19 R 2; 59/4 = 14 R 3 and 59/5 = 11 R 4. Thus the sum of the digits of N = 5 + 9 = 14. Answer: C pushpitkc does the below in red mean that since the DIFFERENCE between each divisor and each remainder is 1, we need to subtract 1 from LCM of divisor ? can you please elaborate on this ? thanks! Notice that each remainder is the largest possible remainder of their respective divisor (for example, 3 is the largest possible remainder when a number is divided by 4). Therefore, if we can the find a number that has remainder of 0 when it’s divided by 2, 3, 4 and 5, and then subtract 1 from it, we will have a suitable value for N
_________________
In English I speak with a dictionary, and with people I am shy.



BSchool Forum Moderator
Joined: 26 Feb 2016
Posts: 3141
Location: India
GPA: 3.12

Re: The remainder when positive integer N is divided by 2 is 1, when divid
[#permalink]
Show Tags
20 Sep 2018, 19:19
dave13 wrote: Bunuel wrote: The remainder when positive integer N is divided by 2 is 1, when divided by 3 is 2, when divided by 4 is 3, and when divided by 5 is 4. What is the sum of the digits of the least possible value of N?
(A) 11 (B) 13 (C) 14 (D) 16 (E) 17 pushpitkc hello there on what rule is this statement based ? can you pls explain ? Recall that if the remainder is 4 when a number is divided by 5, the units digit must be either 4 or 9. Hey dave13Let's consider some examples of numbers which give us a remainder of 4 when divided by 5 One of the options is when the number ends with 4 itself  4,14,24..... Another option is when the number ends with 9  9,19,29..... (each of these options, when divided by 5, will yield a remainder of 4) Unfortunately, there is no RULE for this  you will need to use the PATTERN method in order to understand which of the numbers give us a remainder of 4. Hope that helps you!
_________________
You've got what it takes, but it will take everything you've got




Re: The remainder when positive integer N is divided by 2 is 1, when divid &nbs
[#permalink]
20 Sep 2018, 19:19






