Interger

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Interger

by cartera » Sun Nov 23, 2008 7:52 am
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


OA
C [spoiler][(49 - 1)/3]+1=17[/spoiler]
Source: — Problem Solving |

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by orel » Sun Nov 23, 2008 8:07 am
isn't it 16?

0<=3n+1<=50
n<=16

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by 4meonly » Sun Nov 23, 2008 8:07 am
the least integer that will satisfy given options is 1
the greates - 49
49-1 /3 = 16
do not forget to add 1
16+1 = 17

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by orel » Sun Nov 23, 2008 8:40 am
4meonly wrote:the least integer that will satisfy given options is 1
the greates - 49
49-1 /3 = 16
do not forget to add 1
16+1 = 17
hi!

how the remainder can be 1 when 1 is devided by 3? so, the least integer satisfying the option is 4.
but even if we count one by one, the integers satisfying the option are:

4, 7,10,13,16,19,22,25,28,31,34,37,40,43,46,49
so, 16
am i missing smth???
please, explain!!!

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by 4meonly » Sun Nov 23, 2008 11:41 am
Yes, u still miss 1
How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3 ?

n=3q+1
where n - your integer
3 - factor
Q- quotent
1 - remainder
you should find the number of n that are 0<n<50

4=3*1 +1
7=3*2 +1
10 = 3*3 +1

and 1 = 3*0 +1

OK?

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by logitech » Sun Nov 23, 2008 2:05 pm
if you divide X with Y

and

X < Y

The remainder is X itself.

Ex:

3 is divided by 4

the remainder is 3

1 is divided by 3

the remainder is 1
LGTCH
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by orel » Mon Nov 24, 2008 12:25 am
now it makes sense to me!

thanks guys!

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by GMATCHPOINT » Mon Nov 24, 2008 1:21 pm
Logitech,
If one day you decide to share your flashcards, I would love to hv a copy. hehehe.