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Senior Manager  Joined: 15 Sep 2011
Posts: 305
Location: United States
WE: Corporate Finance (Manufacturing)
Re: How many integers from 0 to 50, inclusive, have a remainder  [#permalink]

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Calculated the same. $$\frac{Last R1 - First R1}{3} +1 = \frac{(49 - 1)}{3} +1 = 16 + 1 = 17$$.

Testing the answer choices if another multiple is needed or if there are one too many. This proved that C was the correct answer.

A. 15 * 3 = 45. 50-45 = R5. Too low.
B. 16 * 3 = 48. 50-48 = R2.
C. 17 * 3 = 51. 50-51 = R1, which is in line with what the question asks.
D. 18 * 3 = 54. 54-50= R4. Too high
E. 19 * 3 = 57. 57-50= R7.Too high
Intern  Joined: 25 Jul 2016
Posts: 7
GMAT 1: 740 Q50 V40 How many integers from 0 to 50, inclusive, have a remainder  [#permalink]

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Economist wrote:
How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3 ?

A. 15
B. 16
C. 17
D. 18
E. 19

First, when a integer is divided by 3, it can have the remainder 0,1 or 2.

There are 50-0+1=51 integers between 0 and 50 inclusive.

Just a quick example:
integer 0: remainder 0
integer 1: remainder 1
integer 2: remainder 2
integer 3: remainder 0
integer 4: remainder 1 and so on. Practicaly, there are 51/3=17 remainders of 0, 17 remainders of 1 and 17 remainders of 2. Answer: C V
Status: Preparing for GMAT
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Re: How many integers from 0 to 50, inclusive, have a remainder  [#permalink]

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Number of integers from 0 to 50 with remainder of 1 when divided by 3 can be found out as follows:
Number of integers divisible by 3 from 0 to 50: 48 = 3 + (n-1) * 3 ; using the formula for nth term of series in arithmetic progression ,where nth term is 48, 1st term is 3 and difference between terms is 3.
Therefore n=16, since adding 1 to each of these numbers will give remainder 1, however, when 1 is divided by 3, it gives remainder of 1, hence we have to include 1, giving the total number of integers as 16+1 =17
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How many integers from 0 to 50, inclusive, have a remainder  [#permalink]

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How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3 ?

A. 15
B. 16
C. 17
D. 18
E. 19

let x=number of integers with a remainder of 1 when divided by 3
range is 1-49
1+3(x-1)=49
x=17
C.
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Re: How many integers from 0 to 50, inclusive, have a remainder  [#permalink]

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Economist wrote:
How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3 ?

A. 15
B. 16
C. 17
D. 18
E. 19

The first number that has a remainder of 1 when divided by 3 is 1, and the last number is 49.

Thus, the number of integers from 0 to 50 inclusive that have a remainder of 1 when divided by 3 is:

(49 - 1)/3 + 1 = 17

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Joined: 09 Mar 2016
Posts: 1229
How many integers from 0 to 50, inclusive, have a remainder  [#permalink]

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abhishekik wrote:
My ans is also C.17.

Explanation:

1 also gives 1 remainder when divided by 3, another number is 4, then 7 and so on.
Hence we have an arithmetic progression: 1, 4, 7, 10,..... 49, which are in the form 3n+1.
Now we have to find out number of terms.
tn=a+(n-1)d, where tn is the nth term of an AP, a is the first term and d is the common difference.
so, 49 = 1+(n-1)3
or, (n-1)3 = 48
or, n-1 = 16
or, n = 17

hello there from this $$49 = 1+(n-1)3$$ I get following:

$$49 = 1+3n-3$$

$$49=3n-2$$

$$3n=51$$

how did you get these value --- >

or, (n-1)3 = 48
or, n-1 = 16
or, n = 17
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How many integers from 0 to 50, inclusive, have a remainder  [#permalink]

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mejia401 wrote:
Calculated the same. $$\frac{Last R1 - First R1}{3} +1 = \frac{(49 - 1)}{3} +1 = 16 + 1 = 17$$.

Testing the answer choices if another multiple is needed or if there are one too many. This proved that C was the correct answer.

A. 15 * 3 = 45. 50-45 = R5. Too low.
B. 16 * 3 = 48. 50-48 = R2.
C. 17 * 3 = 51. 50-51 = R1, which is in line with what the question asks.
D. 18 * 3 = 54. 54-50= R4. Too high
E. 19 * 3 = 57. 57-50= R7.Too high

hello there how about this pushpitkc many thanks! Isnt it formula of number of multiples that you used .... but if it is formula of number of multiples in X range than isnt it used when we need to find number of terms divided by 3 and not when we need to find how many integers leave remainder 1 as in my post (#7) https://gmatclub.com/forum/arithmetic-p ... l#p2035478

Originally posted by dave13 on 01 Apr 2018, 06:13.
Last edited by dave13 on 01 Apr 2018, 07:36, edited 1 time in total.
Senior PS Moderator V
Joined: 26 Feb 2016
Posts: 3302
Location: India
GPA: 3.12
Re: How many integers from 0 to 50, inclusive, have a remainder  [#permalink]

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1
dave13 wrote:
abhishekik wrote:
My ans is also C.17.

Explanation:

1 also gives 1 remainder when divided by 3, another number is 4, then 7 and so on.
Hence we have an arithmetic progression: 1, 4, 7, 10,..... 49, which are in the form 3n+1.
Now we have to find out number of terms.
tn=a+(n-1)d, where tn is the nth term of an AP, a is the first term and d is the common difference.
so, 49 = 1+(n-1)3
or, (n-1)3 = 48
or, n-1 = 16
or, n = 17

hello there from this $$49 = 1+(n-1)3$$ I get following:

$$49 = 1+3n-3$$

$$49=3n-2$$

$$3n=51$$

how did you get these value --- >

or, (n-1)3 = 48
or, n-1 = 16
or, n = 17

Hi dave13

We have $$49 = 1+(n-1)3$$

This can be re-written as $$(n-1)3 = 49 - 1 = 48$$

Dividing by 3 on both sides, we get $$n-1 = \frac{48}{3} = 16$$

Therefore, we can arrive at n = 17

Hope this helps you!
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Senior PS Moderator V
Joined: 26 Feb 2016
Posts: 3302
Location: India
GPA: 3.12
Re: How many integers from 0 to 50, inclusive, have a remainder  [#permalink]

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1
dave13 wrote:
mejia401 wrote:
Calculated the same. $$\frac{Last R1 - First R1}{3} +1 = \frac{(49 - 1)}{3} +1 = 16 + 1 = 17$$.

Testing the answer choices if another multiple is needed or if there are one too many. This proved that C was the correct answer.

A. 15 * 3 = 45. 50-45 = R5. Too low.
B. 16 * 3 = 48. 50-48 = R2.
C. 17 * 3 = 51. 50-51 = R1, which is in line with what the question asks.
D. 18 * 3 = 54. 54-50= R4. Too high
E. 19 * 3 = 57. 57-50= R7.Too high

hello there how about this pushpitkc many thanks! Isnt it formula of number of multiples that you used .... but if it is formula of number of multiples in X range than isnt it used when we need to find number of terms divided by 3 and not when we need to find how many integers leave remainder 1 as in my post (#7) https://gmatclub.com/forum/arithmetic-p ... l#p2035478

Hi dave13

Yes, the formula only calculates the number of terms divided by 3. But if you observe
every(a number divisible by 3 + 1) will give a remainder of 1 when divided by 3

Here, between 0 and 50(both inclusive), there are
17 numbers which are divisible by 3 - 0(0*3) to 48(16*3)
17 numbers which leave the remainder 1(0*3 + 1) - 49(16*3 + 1)

Hope this helps clear your confusion!
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Intern  B
Joined: 02 Oct 2016
Posts: 23
Re: How many integers from 0 to 50, inclusive, have a remainder  [#permalink]

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1
The range of numbers that leave a remainder of 1:
1-49

Therefore total number of integers:
(49-1)/3 + 1 = 17

The last +1 is because the question is inclusive.

Hence C
Intern  B
Joined: 02 Jun 2018
Posts: 3
Re: How many integers from 0 to 50, inclusive, have a remainder  [#permalink]

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50/3 = 16 (int) ----- 16*3= 48. ∴ all these 16 numbers will have a (n+1) value. E.g. 48+1= 49. And include ‘1’ = 0+1. Making the total into 17.
Manager  S
Joined: 26 Jan 2015
Posts: 93
Re: How many integers from 0 to 50, inclusive, have a remainder  [#permalink]

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From 0 to 50, both included, we have 51 numbers.
The cyclicity of 3 is 3. That means, 3 leaves only 3 reminders, 0,1,2 and the cycle repeats. You can check as below.
Reminder of 0 when divided by 3 = 0
Reminder of 1 when divided by 3 = 1
Reminder of 2 when divided by 3 = 2
Reminder of 3 when divided by 3 = 0
Reminder of 4 when divided by 3 = 1
.
.
.
So, out of 51 numbers, there are only 3 reminders possible when divided by 3. So we can divide 51 numbers into 3 sets.
1.Ones that give 0 as reminder
2.Ones that give 1 as reminder
3.Ones that give 2 as reminder

So 51/3=17.
So the final answer is C.17
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Joined: 30 Sep 2017
Posts: 460
GMAT 1: 720 Q49 V40 GPA: 3.8
Re: How many integers from 0 to 50, inclusive, have a remainder  [#permalink]

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The range of numbers that leave a remainder of 1 when divided by 3:
1,4,...,49

Therefore total number of integers:
Un = U1 + (n-1)*b
49 = 1 + (n-1)*3
n=17

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Joined: 18 Dec 2017
Posts: 850
Location: United States (KS)
Re: How many integers from 0 to 50, inclusive, have a remainder  [#permalink]

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Economist wrote:
How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3 ?

A. 15
B. 16
C. 17
D. 18
E. 19

Just list the numbers. (If you have solved enough remainder problems you know that the least possible number is remainder itself)

So list from 1,4,7,...........,40,43,47,50.

Count them.

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