document.write( "Question 1169499: how many three digit numbers can be formed from the set 0-9 if repetition are allowed and the number must be a).even b).divisible by 5 c).divisible by 10
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Algebra.Com's Answer #794440 by ikleyn(52866) $\"\"$ $\"About$

You can put this solution on YOUR website!
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\n" ); document.write( "\n" ); document.write( " In such problems, the term \"three digit number\" means, by default, the number having NON-ZERO leading digit.\r
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document.write( "(a)  the last, ones digit, is any of 5 options 0, 2, 4, 6, 8.\r\n" );
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document.write( "     the tens digit is  any from 0 to 9 : 10 options;\r\n" );
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document.write( "     the leading digit: any from 1 to 9 :  9 options.\r\n" );
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document.write( "     Total amount of such numbers is  9*10*5 = 450.\r\n" );
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document.write( "     The same answer you can obtain by different way\r\n" );
document.write( "         noticing that there are 900 three-digit numbers from 100 to 999,\r\n" );
document.write( "         and exactly half of them are even numbers.\r\n" );
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document.write( "(b)  If a three-digit number is divisible by 5, its ones digit must be 0 or 5.\r\n" );
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document.write( "     So, the amount of such three-difit numbers is 9*10*2 = 180.\r\n" );
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document.write( "     Again, it is exactly    part of the total 900 three-digit numbers.\r\n" );
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document.write( "(c)  Let me be short in this last case:\r\n" );
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document.write( "     The amount of the three-digit numbers divisible by 10 is exactly    part \r\n" );
document.write( "     of the total 900 three-digit numbers, i.e. 90, in total.\r\n" );
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