document.write( "Question 662231: How many numbers divisible by 5 and lying 5000 and 6000(both inclusive) can be formed from the digits 5,6,7,8 and 9? \n" ); document.write( "

Algebra.Com's Answer #412129 by KMST(5328) $\"\"$ $\"About$

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I understand the problem to mean that
\n" ); document.write( "we can only use the digits 5, 6, 7, 8, and 9.
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\n" ); document.write( "To be between 5000 and 6000, the number will have to start with 5,
\n" ); document.write( "because the only number 5000 and 6000 (both inclusive)
\n" ); document.write( "that does not start with 5 is 6000, which requires three zeros.
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\n" ); document.write( "Numbers divisible by 5 must end in 5 or 10.
\n" ); document.write( "If we can only use the digits 5, 6, 7, 8, and 9,
\n" ); document.write( "a number divisible by 5 will have to end in 5.
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\n" ); document.write( "That only allows us to choose the two middle digits.
\n" ); document.write( "Nothing is said about not repeating digits,
\n" ); document.write( "so there are 5 independent choices for each of the middle digits,
\n" ); document.write( "which makes $\"5%2A5=25\"$ choices overall. \n" ); document.write( "

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