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pretttyune
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(P/S) Remainder 1 26 Nov 2007, 05:16
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I got below question on a web, and translated it into English. It could sound strange, but please help me to solve it.

Q) When divide the integers of which are between 0 and 50 by 3, how many has 1 as its remainder?

Sorry for poor translation.



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(P/S) Remainder 1 26 Nov 2007, 06:07
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17.

[0..50] ==> [1..49] (1 and 49 is edge integer with remainder=1

N=(49-1)/3+1=16+1=17
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Deckard
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(P/S) Remainder 1 26 Nov 2007, 12:39
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walker
17.

[0..50] ==> [1..49] (1 and 49 is edge integer with remainder=1

N=(49-1)/3+1=16+1=17
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Walker, may i ask you why have you added the 1?
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(P/S) Remainder 1 26 Nov 2007, 13:22
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Deckard
Walker, may i ask you why have you added the 1?
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X1--X--X--X--X--Xn+1 we have n segments but n+1 numbers

For example,

4,5,6,7,8,9,10,11,12,13
segments: (13-4)/3=3
numbers: 3+1=4
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GMATBLACKBELT
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(P/S) Remainder 1 26 Nov 2007, 18:07
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walker
17.

[0..50] ==> [1..49] (1 and 49 is edge integer with remainder=1

N=(49-1)/3+1=16+1=17
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Nice, I had 16. 49-4 =45/3 (3 b/c each "jump" is 3) --> 15+1=16

But I forgot about the 1/3. doh!
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(P/S) Remainder 1 26 Nov 2007, 18:20
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walker
17.

[0..50] ==> [1..49] (1 and 49 is edge integer with remainder=1

N=(49-1)/3+1=16+1=17
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good explanation! i forgot about 1..



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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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