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# How many integers from 0 to 50, inclusive, have a remainder

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Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink]

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01 Jul 2015, 16:55
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
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Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink]

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14 Sep 2016, 10:11
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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14 Sep 2016, 11:03
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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26 Sep 2017, 16:24
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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How many integers from 0 to 50, inclusive, have a remainder [#permalink]

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01 Apr 2018, 06:04
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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Updated on: 01 Apr 2018, 07:36
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

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.
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Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink]

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01 Apr 2018, 06:25
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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Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink]

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01 Apr 2018, 10:51
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

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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Joined: 02 Oct 2016
Posts: 25
Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink]

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08 Apr 2018, 06:58
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
Re: How many integers from 0 to 50, inclusive, have a remainder   [#permalink] 08 Apr 2018, 06:58

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