document.write( "Question 1194539: Hi
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document.write( "In a 3 digit number the units digit is 3 more than the tens digit and the sum of the digits is 11. If the units and hundreds digits are interchanged the number increased by 99.
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document.write( "What is original number.
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document.write( "Thanks \n" );
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Algebra.Com's Answer #826764 by math_tutor2020(3817)![]() ![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "Answer: 425 is the original number\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "See below for the work shown and explanation.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "==================================================================================================================\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "x = hundreds digit \n" ); document.write( "y = tens digit \n" ); document.write( "z = units or ones digit \n" ); document.write( "These variables represent nonnegative single-digit whole numbers from the set {0,1,2,3,4,5,6,7,8,9} \n" ); document.write( "Though I should be more careful to point out that x = 0 is not possible since something like 012 = 12 is a two digit number. \n" ); document.write( "Luckily, zero is allowed for the other digits.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Any three-digit number is of the form 100x+10y+z \n" ); document.write( "For example, if we had the following three values: \n" ); document.write( "x = 1 \n" ); document.write( "y = 2 \n" ); document.write( "z = 3 \n" ); document.write( "then, \n" ); document.write( "100x+10y+z = 100*1+10*2+3 = 123\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "We're told that \"the units digit is 3 more than the tens digit\", which means symbolically we would say z = y+3 \n" ); document.write( "Whatever y is, add on 3 to get z.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "We also know that \"the sum of the digits is 11\", so, \n" ); document.write( "x+y+z = 11 \n" ); document.write( "Let's replace z with y+3 and simplify \n" ); document.write( "x+y+z = 11 \n" ); document.write( "x+y+y+3 = 11 \n" ); document.write( "x+2y+3 = 11 \n" ); document.write( "x+2y = 11-3 \n" ); document.write( "x+2y = 8 \n" ); document.write( "Then we can solve for x \n" ); document.write( "x+2y = 8 \n" ); document.write( "x = 8-2y \n" ); document.write( "which will be useful later in a substitution step.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Let A = 100x+10y+z be the original three-digit number, and B = 100z+10y+x be the reversed version of the original number (eg: A = 123 and B = 321) \n" ); document.write( "As you can see, I swapped the hundreds digit and units digit to reverse the number.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "We're told that \"if the units and hundreds digits are interchanged the number increased by 99.\" \n" ); document.write( "meaning, \n" ); document.write( "new = old+99 \n" ); document.write( "B = A+99 \n" ); document.write( "100z+10y+x = 100x+10y+z+99\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "From here, we will plug in these items \n" ); document.write( "x = 8-2y \n" ); document.write( "z = y+3 \n" ); document.write( "so that everything is now in terms of y. This will allow us to solve for y.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "100z+10y+x = 100x+10y+z+99 \n" ); document.write( "100(y+3)+10y+8-2y = 100(8-2y)+10y+y+3+99 \n" ); document.write( "100y+300+10y+8-2y = 800-200y+10y+y+3+99 \n" ); document.write( "108y+308 = 902-189y \n" ); document.write( "108y+189y = 902-308 \n" ); document.write( "297y = 594 \n" ); document.write( "y = 594/297 \n" ); document.write( "y = 2\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Then use this to find x and z \n" ); document.write( "x = 8-2y = 8-2*2 = 4 \n" ); document.write( "z = y+3 = 2+3 = 5\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The original number is: \n" ); document.write( "A = 100x+10y+z \n" ); document.write( "A = 100*4+10*2+5 \n" ); document.write( "A = 400+20+5 \n" ); document.write( "A = 425 \n" ); document.write( "The reversed number is \n" ); document.write( "B = 100z+10y+x \n" ); document.write( "B = 100*5+10*2+4 \n" ); document.write( "B = 500+20+4 \n" ); document.write( "B = 524\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Notice how B-A = 524-425 = 99 \n" ); document.write( "which helps show that B = A+99 is correct. \n" ); document.write( "Or you could verify like so: \n" ); document.write( "B = A+99 \n" ); document.write( "524 = 425+99 \n" ); document.write( "524 = 524\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Also, \n" ); document.write( "x+y+z = 4+2+5 = 6+5 = 11 \n" ); document.write( "confirms the digits add up to 11\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Lastly, the units digit (5) is exactly three more than the tens digit (2).\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Therefore, we have fully confirmed the answer of 425 \n" ); document.write( "\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |