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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] New post 06 Feb 2013, 02:24
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] New post 06 May 2013, 04:41
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] New post 24 May 2014, 05:54
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Re: How many integers from 0 to 50, inclusive, have a remainder   [#permalink] 24 May 2014, 05:54
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