document.write( "Question 1174984: Find a three-digit number such that the hundreds’ digit minus the tens’ digit is 1, the tens’ digit
\n" ); document.write( "minus the units’ digit is 1, and the sum of the digits is 15.
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Algebra.Com's Answer #800493 by math_tutor2020(3817)\"\" \"About 
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\n" ); document.write( "h = hundreds digit
\n" ); document.write( "t = tens digit
\n" ); document.write( "u = units digit\r
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\n" ); document.write( "\n" ); document.write( "A three digit number like 789 means that h = 7, t = 8, u = 9
\n" ); document.write( "We can also construct that number like so
\n" ); document.write( "789 = 700+80+9
\n" ); document.write( "789 = 7*100+8*10+9*1\r
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\n" ); document.write( "\n" ); document.write( "Going back to the general format, the three digit number \"htu\" can be written as \"100h%2B10t%2Bu\"\r
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\n" ); document.write( "\n" ); document.write( "As another example, let's say
\n" ); document.write( "h = 5
\n" ); document.write( "t = 4
\n" ); document.write( "u = 2
\n" ); document.write( "\"htu+=+100h%2B10t%2Bu\"
\n" ); document.write( "\"542+=+100%2A5%2B10%2A4%2B2\"
\n" ); document.write( "\"542+=+500%2B40%2B2\"
\n" ); document.write( "\"542+=+542\"\r
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\n" ); document.write( "\n" ); document.write( "In this very narrow context, the notation \"htu\" does not mean \"h times t times u\", even though with many algebra problems, we're conditioned to see two letters together and think \"that's implied multiplication\". \r
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\n" ); document.write( "\n" ); document.write( "With \"htu\", I'm simply using three blank placeholders as the three digits.\r
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\n" ); document.write( "\n" ); document.write( "Anyways, we're told that \"hundreds’ digit minus the tens’ digit is 1\" which means we can form the equation \"h-t+=+1\". Solving for h gets us \"h+=+1%2Bt\"\r
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\n" ); document.write( "\n" ); document.write( "We're also told that \"the tens’ digit minus the units’ digit is 1\" which forms \"t-u=1\" and that solves to \"t+=+1%2Bu\"\r
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\n" ); document.write( "\n" ); document.write( "The two equations \"h+=+1%2Bt\" and \"t+=+1%2Bu\" can be combined like so
\n" ); document.write( "\"h+=+1%2Bt\"
\n" ); document.write( "\"h+=+1%2B%281%2Bu%29\" Replace t with 1+u
\n" ); document.write( "\"h+=+2%2Bu\" \r
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\n" ); document.write( "\n" ); document.write( "So whatever the hundreds digit (h) is, we add on 2 to get the units digit (u).\r
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\n" ); document.write( "\n" ); document.write( "The last piece of useful information is that \"the sum of the digits is 15\". So the three digits add to 15.\r
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\n" ); document.write( "\n" ); document.write( "\"h%2Bt%2Bu+=+15\"\r
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\n" ); document.write( "\n" ); document.write( "\"h%2B%281%2Bu%29%2Bu+=+15\" Plug in \"t+=+1%2Bu\"\r
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\n" ); document.write( "\n" ); document.write( "\"h%2B1%2B2u+=+15\"\r
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\n" ); document.write( "\n" ); document.write( "\"2%2Bu%2B1%2B2u+=+15\" Plug in \"h+=+2%2Bu\". Let's solve for the variable 'u'.\r
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\n" ); document.write( "\n" ); document.write( "\"3u%2B3+=+15\"\r
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\n" ); document.write( "\n" ); document.write( "\"3u%2B3-3+=+15-3\" Subtract 3 from both sides\r
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\n" ); document.write( "\n" ); document.write( "\"3u+=+12\"\r
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\n" ); document.write( "\n" ); document.write( "\"3u%2F3+=+12%2F3\" Divide both sides by 3\r
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\n" ); document.write( "\n" ); document.write( "\"u+=+4\" This is the units digit\r
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\n" ); document.write( "\n" ); document.write( "Once we know that, we can find the values of t and h
\n" ); document.write( "\"t+=+1%2Bu+=+1%2B4+=+5\"
\n" ); document.write( "\"h+=+2%2Bu+=+2%2B4+=+6\"\r
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\n" ); document.write( "\n" ); document.write( "We have
\n" ); document.write( "h = 6
\n" ); document.write( "t = 5
\n" ); document.write( "u = 4\r
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\n" ); document.write( "\n" ); document.write( "To check, we see that
\n" ); document.write( "h+t+u = 6+5+4 = 11+4 = 15
\n" ); document.write( "so that confirms the third equation and statement.\r
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\n" ); document.write( "\n" ); document.write( "Also, we can see that the difference from the hundreds and tens digit (6 and 5) is 1 unit, so is the difference from the tens to the units digit. So the answer is fully confirmed.\r
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\n" ); document.write( "\n" ); document.write( "Side note: you could use a trial and error process to guess three digit numbers, and see if you can land on the answer. But I think the algebraic method is most efficient.\r
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\n" ); document.write( "\n" ); document.write( "Answer: 654
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