Print out all odd number from 1 to 100 using a for loop!

Published July 19, 2007 C++ , Reference

The very first thought likely to hit will be that of taking mod 2(%2) of numbers from 1 to 100 in a loop and checking whether it yield an remainder or not. But there is even better a way to deal with it.

                 for( unsigned int i = 1; i < = 100; i++ )
                        if( i & 0×00000001 )
                                 cout << i<<” “;

The catch is, only odd number’s binary representation ends with ‘1’.Check the below table

Dec	Hex	Bin
000	00	00000000
001	01	00000001
002	02	00000010
003	03	00000011
004	04	00000100
005	05	00000101
006	06	00000110
007	07	00000111
008	08	00001000
009	09	00001001
010	0A	00001010
011	0B	00001011
012	0C	00001100
013	0D	00001101
014	0E	00001110
015	0F	00001111

6 Responses to “Print out all odd number from 1 to 100 using a for loop!”

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1 Shanil July 20, 2007 at 9:13 am

Gud technique!!!
2 mohanraj September 24, 2007 at 12:47 pm

do you have someother methods
3 Vineeth December 31, 2007 at 12:52 pm

Good
4 Hash January 21, 2008 at 10:33 pm

What was meant by the statement :
“The catch is, only prime number’s binary representation ends with ‘1’”

9 is not a prime. However it also ends in ‘1′
“009 09 00001001″

Or did you mean something else?
5 Gov January 22, 2008 at 9:29 am

It was a typo,I have corrected it
6 Chris Buckley June 30, 2008 at 4:01 pm

How about

for (unsigned int i = 1; i <= 100; i += 2)
cout << i << ” “;