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Director  Joined: 01 Apr 2008
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How many integers from 0 to 50, inclusive, have a remainder  [#permalink]

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Question Stats: 60% (01:21) correct 40% (01:29) wrong based on 1115 sessions

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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

Originally posted by Economist on 24 Mar 2009, 23:51.
Last edited by Bunuel on 06 Nov 2012, 03:52, edited 2 times in total.
Renamed the topic and edited the question.
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How many integers from 0 to 50, inclusive, have a remainder  [#permalink]

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20
19
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

Algebraic way:

Integer have a remainder of 1 when divided by 3 --> $$n=3p+1$$, where $$p$$ is an integer $$\geq{0}$$, so $$n$$ can take the following values: 1, 4, 7, ...

$$n=3p+1\leq{50}$$ --> $$3p\leq{49}$$ --> $$p\leq{16\frac{1}{3}}$$ --> so $$p$$, can take 17 values from 0 to 16, inclusive.

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Re: PS: 0 to 50 inclusive, remainder  [#permalink]

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13
15
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
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Manager  Joined: 09 Nov 2008
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Re: PS: 0 to 50 inclusive, remainder  [#permalink]

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5
3
C:17

I brute forced this one.
All the multiples of 3 + 1 will have a remainder of 1:
4,7,10,13,16,19,22,25,28,31,34,37,40,43,46,49 - 16 numbers total
But then I thought, no way it's this easy and thought about 1. 1/3 would also have a remainder of 1, making the answer 17.
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Re: PS: 0 to 50 inclusive, remainder  [#permalink]

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3
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
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Does it mean that if zero is included in any similar problem like this then we should consider it, no matter what the divisor is???
Ex- if question say any no. between 0 and 50, inclusive, divisible by 3
Answer will still be 17 Is it right _________________
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hardnstrong wrote:
Does it mean that if zero is included in any similar problem like this then we should consider it, no matter what the divisor is???
Ex- if question say any no. between 0 and 50, inclusive, divisible by 3
Answer will still be 17 Is it right 0 is a multiple of every integer (except zero itself), so there are $$\frac{48-0}{3}+1=17$$ numbers divisible by 3 in the range 0-50 inclusive (check this: totally-basic-94862.html?highlight=range).

But in original question 0 is not considered as one of the numbers: the lowest value of n is 1 (for p=0) and the highest value of n is 49 (for p=16), so total of 17 such numbers.
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bibha wrote:
How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3?
A.14 B.15. C.16 D.17 E.18 ((Last - First)/ n) +1 -> (49-1)/3 +1 = 17

We use 49 because that is the last that will produce a remainder of 1 when divided by 3 and 1/3 has a remainder of 1.
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GMAT 1: 710 Q49 V35 Re: PS: 0 to 50 inclusive, remainder  [#permalink]

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My answer is C.

There are 48/3=16 numbers that are divisible by 3 (0 is excluded). If we add 1 to each and every of these number, we still have 16 numbers that have a remainder of 1 when divided by 3 (say 4, 7, 10,..., 49). But we have not counted 0 yet, 0+1 equals 1, 1 has a remainder of 1 when divided by 3.

Hence C.
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Re: PS: 0 to 50 inclusive, remainder  [#permalink]

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I think we can find this very quickly by using this:
approximately 1/3 of the numbers from 0-50 are div by 3 so, (50/3) +1 = 16 + 1 = 17
approximately 1/2 of the numbers are div by 2 => (50/2 ) + 1 = 26
approximately 1/4 of the numbers are div by 4 => (50/4) + 1 =13
approximately 1/5 of the numbers are div by 5 => (50/5) + 1 =11
approximately 1/6 of the numbers are div by 6 => (50/6) + 1 = 9
I saw this on Bunuel's post somewhere and I thought to myself, how did he know? so I tried to calculate it and voila!, it is true...didn't try with range starting with non-zero numbers though
don't forget to add one..before you are done!
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Re: PS: 0 to 50 inclusive, remainder  [#permalink]

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when 1 is devided by 3 remainder is 1, and when 3 is devided remainder 0.
16*3=48. so upto 48 there will be 16 number which will give remainder 1 after 48, there is only 1 number up to 50 which will give 1 remainder so the answer 17
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Re: PS: 0 to 50 inclusive, remainder  [#permalink]

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Re: PS: 0 to 50 inclusive, remainder  [#permalink]

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First, lets look at the range. With 1 remainder, 4-49.

So, number of elements = (49-3)/3 + 1 = 16

Now, lets not forget 1, since 1/3 -> 1 as remainder (Good one!)

So, 16 +1 = 17
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Re: PS: 0 to 50 inclusive, remainder  [#permalink]

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General Formula : n= px+q
p=3
q=1
n=3x+1
Substituting Values n=0,4,7...49 .
x = 0 to 16, inclusive
Hence Total nos : 17
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Re: How many integers from 0 to 50, inclusive, have a remainder  [#permalink]

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Bunuel wrote:

Algebraic way:

Integer have a remainder of 1 when divided by 3 --> $$n=3p+1$$, where $$p$$ is an integer $$\geq{0}$$, so $$n$$ can take the following values: 1, 4, 7, ...

$$n=3p+1\leq{50}$$ --> $$3p\leq{49}$$ --> $$p\leq{16\frac{1}{3}}$$ --> so $$p$$, can take 17 values from 0 to 16, inclusive.

hi, i was wondering how we got 17 from the calculation that $$p\leq{16\frac{1}{3}}$$ ?

my approach was:

50-0+1 = 51 integers total

0/3 has r = 0
1/3 has r = 1
2/3 has r = 2

3/3 has r = 0
... and so on

thus there will be 1 value for every three that will have a remainder of 1 when divided by 3 (cyclicity of 3?)

so 51/3 = 17

would this method work for similar questions?

i was wondering how we treat the $$p\leq{16\frac{1}{3}}$$ term when trying the algebraic method? i.e. how do we know to arrive at 17 ?

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

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essarr wrote:
Bunuel wrote:

Algebraic way:

Integer have a remainder of 1 when divided by 3 --> $$n=3p+1$$, where $$p$$ is an integer $$\geq{0}$$, so $$n$$ can take the following values: 1, 4, 7, ...

$$n=3p+1\leq{50}$$ --> $$3p\leq{49}$$ --> $$p\leq{16\frac{1}{3}}$$ --> so $$p$$, can take 17 values from 0 to 16, inclusive.

hi, i was wondering how we got 17 from the calculation that $$p\leq{16\frac{1}{3}}$$ ?

i was wondering how we treat the $$p\leq{16\frac{1}{3}}$$ term when trying the algebraic method? i.e. how do we know to arrive at 17 ?

thanks!

We have that $$n=3p+1$$, where $$p$$ is an integer $$\geq{0}$$. So, n can be 1 (for p=0), 4 (for p=1), 7, 10, ..., and 49 (for p=16) --> 17 values for p --> 17 values for n.

OR: $$p\leq{16\frac{1}{3}}$$ implies that p can take integer values from 0 to 16, inclusive, thus it can take total of 17 values.

Hope it's clear.
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Re: How many integers from 0 to 50, inclusive, have a remainder  [#permalink]

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The range of given numbers is 50-0+1=51
Any number when divided by 3 can give remainders from the list {0,1,2}. This patten will repeat for every 3 numbers. Hence divide the range by number of available remainders. i.e 51/3=17
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Re: PS: 0 to 50 inclusive, 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,

I like this solution using the arithmetic progression but in the expression tn=a+(n-1)d why n-1 and not n ?

As I know Un= U0 + n r where U0 is the first terme and r is the common difference.

Thx in advance for the lighting !
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Re: How many integers from 0 to 50, inclusive, have a remainder  [#permalink]

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I chose C

Since 3 has 16 multiples b/t 0-50 there must be 16 with a remainder of 1. But also $$1/3$$ is included in this list. so 16+1=17=C Re: How many integers from 0 to 50, inclusive, have a remainder   [#permalink] 01 Jul 2015, 16:06

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