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Re: Remainder Problem [#permalink]
17 Dec 2009, 02:13

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zaarathelab wrote:

How many integers from 0-50 have a remainder of 3 when divided by 9?

A. 5 B 6 C 7 D 8 E 9

A remainder of 3 when divided by 9 --> \(n=9q+3\), where q is an integer \(\geq{0}\).

\(9q+3<50\) --> \(q<5\frac{2}{9}\), hence q can take 6 values from 0 to \(5\).

Answer: B (6).

Or, one can just manually list the numbers of the form \(n=9q+3\), which are less than 50: 3, 12, 21, 30, 39, and 48. Total of 6 numbers. _________________

Re: Remainder Problem [#permalink]
06 Feb 2013, 03:01

Range of numbers is 50-0+1= 51. Possible remainders= 0,1,2,3,4,5,6,7,8 51/9 gives a quotient 5 and a remainder 6. This means that after completing 5 rounds from 0 to 8 ,it found numbers with remainders from 0 to 5 and this has 1 extra count for remainder 3.Hence it is 5+1=6

Re: How many integers from 0-50 have a remainder of 3 when divid [#permalink]
08 Aug 2013, 23:12

The first number when divided by 9 leaves a remainder of 3 is 3. The last number below 50 when divided by 9 leaves a remainder of 3 is 48 so (48 - 3)/9 + 1 = 6

Re: How many integers from 0-50 have a remainder of 3 when divid [#permalink]
03 Apr 2015, 03:38

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