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# 51) How many integers from 0 to 50, inclusive, have a

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Senior Manager
Joined: 02 May 2004
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51) How many integers from 0 to 50, inclusive, have a [#permalink]  29 Jul 2007, 20:22
51) 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
Senior Manager
Joined: 02 May 2004
Posts: 313
Followers: 1

Kudos [?]: 6 [0], given: 0

How do you get 17? I was looking for an explanation lol
Senior Manager
Joined: 04 Jun 2007
Posts: 348
Followers: 1

Kudos [?]: 15 [0], given: 0

Re: Inclusive Integers/Divisibility [#permalink]  29 Jul 2007, 21:31
jamesrwrightiii wrote:
51) 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

I can explain this using arithmetic progression. Hope it is useful.
Between 0 and 50 (both inclusive) the least and greatest number satisfying the given condition are 1 and 49. The common difference of the sequence is 3.

(last term) = (first term) + (n-1)(common difference), where is the number of terms.
so, 49 = 1 + (n-1)3
or, 48 = 3n - 3
or, 3n = 51
or, n = 17

Since the range is pretty small you can even try to write down all the terms (1,4,7,10,....,46,49) and count them !
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Re: Inclusive Integers/Divisibility [#permalink]  30 Jul 2007, 03:35
jamesrwrightiii wrote:
51) 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

I'm getting B, 16.

Any number that is 1 great than a multiple of 3 will leave a remainder of 1.
Since there are 16 multiples of 3 which are less than 50, there are 16 numbers that leave a remainder of 1 when divided by 3 (the first one being 4 and the last one being 49).

Am I missing something?
Senior Manager
Joined: 04 Jun 2007
Posts: 348
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Kudos [?]: 15 [0], given: 0

Re: Inclusive Integers/Divisibility [#permalink]  30 Jul 2007, 03:55
GK_Gmat wrote:
jamesrwrightiii wrote:
51) 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

I'm getting B, 16.

Any number that is 1 great than a multiple of 3 will leave a remainder of 1.
Since there are 16 multiples of 3 which are less than 50, there are 16 numbers that leave a remainder of 1 when divided by 3 (the first one being 4 and the last one being 49).

Am I missing something?

I think you are missing 1.
1 divided by 3 leaves a remainder of 1.
Re: Inclusive Integers/Divisibility   [#permalink] 30 Jul 2007, 03:55
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# 51) How many integers from 0 to 50, inclusive, have a

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