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Re: Remainder Problem [#permalink]
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
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Re: How many integers from 0-50 have a remainder of 3 when divid [#permalink]
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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

Ans : B
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Re: How many integers from 0-50 have a remainder of 3 when divid [#permalink]
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

....
-9-18-27-36-45-
- = number of integers have a remainder of 3
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How many integers from 0-50 have a remainder of 3 when divid [#permalink]
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


3+9x<51
9x<48
x<5 1/3
5+1(for 3/9 remainder)=6
B.
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Re: How many integers from 0-50 have a remainder of 3 when divid [#permalink]
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Re: How many integers from 0-50 have a remainder of 3 when divid [#permalink]
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