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The remainder when positive integer N is divided by 2 is 1, when divid

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

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New post 24 Apr 2018, 23:30
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A
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C
D
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Re: The remainder when positive integer N is divided by 2 is 1, when divid  [#permalink]

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New post 24 Apr 2018, 23:49
9
12

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
    o N = 3b + 2 = 3b’ – 1
    o N = 4c + 3 = 4c’ – 1
    o N = 5d + 4 = 5d’ – 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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New post 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 are-19,29,39,49,59,69,79,89,99
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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Re: The remainder when positive integer N is divided by 2 is 1, when divid  [#permalink]

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New post 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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The remainder when positive integer N is divided by 2 is 1, when divid  [#permalink]

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New post 13 May 2018, 13:02
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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=(n-1)/2-(n-2)/3➡(n+1)/6
thus, n+1 is multiple of 6
let y=difference between last two quotients:
y=(n-3)/4-(n-4)/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
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Re: The remainder when positive integer N is divided by 2 is 1, when divid  [#permalink]

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New post 14 May 2018, 02:28
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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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New post 14 May 2018, 20:10
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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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New post 15 May 2018, 16:14
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2
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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Re: The remainder when positive integer N is divided by 2 is 1, when divid  [#permalink]

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New post 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
    o N = 3b + 2 = 3b’ – 1
    o N = 4c + 3 = 4c’ – 1
    o N = 5d + 4 = 5d’ – 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


Hello,
Would you please tell me where I can find this rule: N= lcm Q + Constant R?

Thank you very much

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

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New post 20 Sep 2018, 09:57
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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.
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The remainder when positive integer N is divided by 2 is 1, when divid  [#permalink]

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

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New post 20 Sep 2018, 19:19
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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 dave13

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

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New post 08 Dec 2018, 02:21
JeffTargetTestPrep wrote:

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.
Answer: C


Hi....
can you please expand on the bolded part and explain on how you recognized this?
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The remainder when positive integer N is divided by 2 is 1, when divid  [#permalink]

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New post 04 Jan 2019, 02:04
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1
Mansoor50 wrote:
JeffTargetTestPrep wrote:

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.
Answer: C


Hi....
can you please expand on the bolded part and explain on how you recognized this?


Hello Mansoor50,

I'll try to explain the bold part by touching upon the basics of remainders and remainder theorem.

The remainder when any positive integer is divided by a divisor has to be non-negative and less than the divisor. This is the basic rule for remainders.
For e.g Remainders of:
1) 3/3 = 0
2) 4/3 = 1
3) 5/3 = 2
...we cannot have a remainder of 3, as remainder has to be less than divisor as discussed.

Now what does it mean to have a negative remainder?
Take a look at the below examples, when we apply the remainder theorem formula of Dividend = (Divisor * Quotient) + Remainder
1) N=3a - 1 (....means that when N is divided by 3, it leaves a remainder of -1).
However, we know that remainder cannot be negative. So we rewrite -1 as the difference of 2 numbers, so that upon expansion, we get a positive remainder
i.e N=3a-[3-2]......further expansion, we get....N=3a-3+2 .....and still further N=3(a-1)+2.....now notice that 3(a-1) is divisible by 3. And 2 divided by 3 gives a remainder of 2.

If you understand the above, then you'll notice that negative remainders can be converted to "equivalent" positive remainders by subtracting the negative remainder from the divisor.
As we saw, when divisor is 3 and remainder is -1, the equivalent positive remainder is (3-1) = 2.
And if remainder is -2, then the corresponding remainder will be (3-2) = 1

Converse is also true.... i.e when remainder is 2 and divisor is 3...the corresponding negative remainder is 2-3, i.e -1.
You can check this with any combinations...say remainder is 5 when you divide a number by 8...therefore it's corresponding negative remainder will always by (5-8) = -3
Or when remainder is 7 when you divide a number by 8, it's corresponding negative remainder will always be (7-8) = -1
The -1 remainder concept helps solve several GMAT remainder problems like this one, because simply put, whenever the remainder is exactly 1 less than the divisor, it means remainder is -1.

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

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New post 04 Jan 2019, 05:28
Darshi04 wrote:
Mansoor50 wrote:
JeffTargetTestPrep wrote:

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.
Answer: C


Hi....
can you please expand on the bolded part and explain on how you recognized this?


Hello Mansoor50,

I'll try to explain the bold part by touching upon the basics of remainders and remainder theorem.

The remainder when any positive integer is divided by a divisor has to be non-negative and less than the divisor. This is the basic rule for remainders.
For e.g Remainders of:
1) 3/3 = 0
2) 4/3 = 1
3) 5/3 = 2
...we cannot have a remainder of 3, as remainder has to be less than divisor as discussed.

Now what does it mean to have a negative remainder?
Take a look at the below examples, when we apply the remainder theorem formula of Dividend = (Divisor * Quotient) + Remainder
1) N=3a - 1 (....means that when N is divided by 3, it leaves a remainder of -1).
However, we know that remainder cannot be negative. So we rewrite -1 as the difference of 2 numbers, so that upon expansion, we get a positive remainder
i.e N=3a-[3-2]......further expansion, we get....N=3a-3+2 .....and still further N=3(a-1)+2.....now notice that 3(a-1) is divisible by 3. And 2 divided by 3 gives a remainder of 2.

If you understand the above, then you'll notice that negative remainders can be converted to "equivalent" positive remainders by subtracting the negative remainder from the divisor.
As we saw, when divisor is 3 and remainder is -1, the equivalent positive remainder is (3-1) = 2.
And if remainder is -2, then the corresponding remainder will be (3-2) = 1

Converse is also true.... i.e when remainder is 2 and divisor is 3...the corresponding negative remainder is 2-3, i.e -1.
You can check this with any combinations...say remainder is 5 when you divide a number by 8...therefore it's corresponding negative remainder will always by (5-8) = -3
Or when remainder is 7 when you divide a number by 8, it's corresponding negative remainder will always be (7-8) = -1
The -1 remainder concept helps solve several GMAT remainder problems like this one, because simply put, whenever the remainder is exactly 1 less than the divisor, it means remainder is -1.

Hope it is clear now.


thank you!!!
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Re: The remainder when positive integer N is divided by 2 is 1, when divid   [#permalink] 04 Jan 2019, 05:28
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