document.write( "Question 987601: Mr. X had a habit of spending money according to dates. i. e. If date is 19 he was spending Rs.19 and if date is 15, he was spending Rs. 15. \r
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document.write( "One night he calculated total spending of 5 consecutive days - Monday to Friday and he found that he spent Rs. 61 in 5 days.\r
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document.write( "So, identify the dates.??? \n" );
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Algebra.Com's Answer #608314 by ikleyn(52810)![]() ![]() You can put this solution on YOUR website! ' \n" ); document.write( "Answer. There is only one solution:\r \n" ); document.write( "\n" ); document.write( " 27 and 28 of February and 1, 2, and 3 of March.\r \n" ); document.write( "\n" ); document.write( " (27 + 28 + 1 + 2 + 3 = 61). \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Solution\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "If these days would be inside one month, then the dates are (x-2), (x-1), x, (x+1) and (x+2), where x is the date of the middle of 5 days. \r \n" ); document.write( "\n" ); document.write( "Then the sum must be multiple of 5, since \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "(x-2) + (x-1) + x + (x+1) + (x+2) = 5x.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "But the integer 61 is not multiple of 5. Contradiction. \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Hence, the dates are partly the end of some month and the beginning of the next month. \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Then, it is easy to check the the dates 27, 28, 1, 2 and 3 satisfy the condition 27 + 28 + 1 + 2 + 3 = 61. \r \n" ); document.write( "\n" ); document.write( "Next, it is easy to see that there is no other solution.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |