10 Digit Number PuzzleDate: 09/14/2004 at 17:01:24 From: Kranti Subject: Number Puzzle Write a 10 digit number where the first digit tells you the number of zeros in the 10 digit number, the second digit is the number of ones in the number, the third digit is the number of twos in the number and so on until the 10th digit is the number of nines in the number. For example, 9000000001 is not true because there are eight zeros but the first digit is 9, not 8. Can you help? Date: 09/14/2004 at 17:14:17 From: Doctor Vogler Subject: Re: Number Puzzle Hi Kranti, Thanks for writing to Dr. Math. I've seen various problems like this one. Let's try thinking about it this way. Your number is abcdefghij where there are a zeros, b ones, c twos, and so on up to j nines. Consider what happens when you add all of the digits together. You get a+b+c+d+e+f+g+h+i+j but since there are a zeros, b ones, c twos, etc, then this should be the same thing as 0a+1b+2c+3d+4e+5f+6g+7h+8i+9j, which gives us an equation: b+2c+3d+4e+5f+6g+7h+8i+9j = a+b+c+d+e+f+g+h+i+j, which we can simplify as c+2d+3e+4f+5g+6h+7i+8j = a by subtracting the rest of the right hand side from both sides of the equation. Now a can't be bigger than 9, which means that g, h, i, j can't be bigger than 1; in fact, their sum can't be bigger than 1, which means that either all of them are zero or one of them is 1 and the rest are all zero. And f can't be bigger than 2. In fact, f+g+h+i+j can't be bigger than 2. And e+f+g+h+i+j can't be bigger than 3. And d+e+f+g+h+i+j can't be bigger than 4. Now try a few cases. What happens if j = 1 ? That means that either a=8 and c to i are all zero, or a=9, c=1, and d to i are all zero. Do either of those work for any number b? If not, then you have to conclude that j=0. Now what happens if i = 1 ? Keep doing this, and you will soon find that you can check all of the possibilities fairly quickly, and so you can find all possible solutions. If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ |
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