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The remainder when positive integer N is divided by 2 is 1, when divid [#permalink]
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24 Apr 2018, 23:30
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Re: The remainder when positive integer N is divided by 2 is 1, when divid [#permalink]
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24 Apr 2018, 23:49
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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
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Re: The remainder when positive integer N is divided by 2 is 1, when divid [#permalink]
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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
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Re: The remainder when positive integer N is divided by 2 is 1, when divid [#permalink]
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13 May 2018, 03:30
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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
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The remainder when positive integer N is divided by 2 is 1, when divid [#permalink]
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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



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Re: The remainder when positive integer N is divided by 2 is 1, when divid [#permalink]
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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.
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Re: The remainder when positive integer N is divided by 2 is 1, when divid [#permalink]
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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
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Re: The remainder when positive integer N is divided by 2 is 1, when divid [#permalink]
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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
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