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Difference between revisions of "2007 iTest Problems/Problem 56"

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== Problem ==
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''The following problem is from the Ultimate Question of the [[2007 iTest]], where solving this problem required the answer of a previous problem.  When the problem is rewritten, the T-value is substituted.''
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Let <math>T=\text{TNFTPP}</math>. In the binary expansion of <math>\dfrac{2^{2007}-1}{2^T-1}</math>, how many of the first <math>10,000</math> digits to the right of the radix point are <math>0</math>'s?
 
  
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== Solution ==
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==Problem==
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In the binary expansion of <math>\dfrac{2^{2007}-1}{2^{225}-1}</math>, how many of the first <math>10,000</math> digits to the right of the radix point are <math>0</math>'s?
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==Solution==
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We can approach this problem by using long division in base 2 because long division takes advantage of regrouping.  The number <math>2^{2007} - 1</math> has <math>2007</math> ones while the number <math>2^{225} - 1</math> has <math>225</math> ones.
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              <u>          </u>
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    [225 ones])[2007 ones]
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Note that <math>2^{2007} - 1</math> can be rewritten as <math>(2^{225} - 1)(2^{1782} + 2^{1557} + 2^{1332} + \cdots 2^{207}) + 2^{207} - 1,</math> so the remainder when <math>2^{2007} - 1</math> is divided by <math>2^{225} - 1</math> is the same when <math>2^{207} - 1</math> is divided by <math>2^{225} - 1.</math>
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              <u>        </u>
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    [225 1’s])[207 1’s]
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There are not enough digits for <math>2^{207} - 1</math> to divide evenly into <math>2^{225} - 1,</math> so we need to bring down more zeroes.  Since <math>2^{225} - 2^{18} < 2^{225} - 1 < 2^{226} - 2^{19}</math>, we need to bring down 19 zeroes, resulting in 18 zeroes to the right of the [[radix point]].
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              <u>      0.[18 0’s]</u>
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    [225 1’s])[207 1’s][18 0’s]0
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Now we can subtract <math>2^{225} - 1</math> in the long division.
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              <u>      0.[18 0’s]1</u>
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    [225 1’s])[207 1’s][18 0’s]0
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            -<u>[206 1’s][18 1’s]1</u>
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              [206 1’s][18 0’s]1
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We can bring down more zeroes and repeat the iteration.
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              <u>      0.[18 0’s]111</u>
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    [225 1’s])[207 1’s][18 0’s]0
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            -<u>[206 1’s][18 1’s]1</u>
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              [206 1’s][18 0’s]10
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            -<u>[205 1’s][18 1’s]11</u>
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              [205 1’s][18 0’s]110
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            -<u>[204 1’s][18 1’s]111</u>
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              [204 1’s][18 0’s]111
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Notice that no zeroes are placed to the right of the last values after each iteration.  Also, there is a pattern after doing the subtraction in each iteration, where there are <math>n</math> ones followed by <math>18</math> zeroes and <math>207-n</math> ones.
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<br>
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To confirm this, we note that in each iteration, we multiply by <math>2</math> and subtract <math>2^{2007} - 1.</math>  Doing this results in
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  [n  1’s][18 0’s][207-n 1’s]0
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  <u>-[n-1 1’s][18 1’s][207-n 1’s]1</u>
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  [n-1 1’s][18 0’s][207-(n-1) 1’s]
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Thus, we see that the pattern holds.  In fact, tthe pattern repeats <math>206</math> times, and after that, the number from the subtraction just has <math>207</math> consecutive ones.  That means the digits repeats every <math>225</math> digits, and <math>18</math> of these digits are zeroes.
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<br>
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That means of the <math>9900</math> digits past the radix point, <math>792</math> of them are zeroes.  After <math>9918</math> digits, there are <math>810</math> zeroes, and since the <math>82</math> digits after are ones, we confirm that there are <math>\boxed{810}</math> zeroes of the first <math>10,000</math> digits past the radix point.
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{{alternate solutions}}
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==See Also==
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{{iTest box|year=2007|num-b=55|num-a=57}}
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[[Category:Olympiad Number Theory Problems]]

Latest revision as of 15:44, 19 August 2018

The following problem is from the Ultimate Question of the 2007 iTest, where solving this problem required the answer of a previous problem. When the problem is rewritten, the T-value is substituted.

Problem

In the binary expansion of $\dfrac{2^{2007}-1}{2^{225}-1}$, how many of the first $10,000$ digits to the right of the radix point are $0$'s?

Solution

We can approach this problem by using long division in base 2 because long division takes advantage of regrouping. The number $2^{2007} - 1$ has $2007$ ones while the number $2^{225} - 1$ has $225$ ones. 

                        
   [225 ones])[2007 ones]

Note that $2^{2007} - 1$ can be rewritten as $(2^{225} - 1)(2^{1782} + 2^{1557} + 2^{1332} + \cdots 2^{207}) + 2^{207} - 1,$ so the remainder when $2^{2007} - 1$ is divided by $2^{225} - 1$ is the same when $2^{207} - 1$ is divided by $2^{225} - 1.$ 

                      
   [225 1’s])[207 1’s]

There are not enough digits for $2^{207} - 1$ to divide evenly into $2^{225} - 1,$ so we need to bring down more zeroes. Since $2^{225} - 2^{18} < 2^{225} - 1 < 2^{226} - 2^{19}$, we need to bring down 19 zeroes, resulting in 18 zeroes to the right of the radix point. 

                    0.[18 0’s]
   [225 1’s])[207 1’s][18 0’s]0

Now we can subtract $2^{225} - 1$ in the long division. 

                    0.[18 0’s]1
   [225 1’s])[207 1’s][18 0’s]0
            -[206 1’s][18 1’s]1
             [206 1’s][18 0’s]1

We can bring down more zeroes and repeat the iteration. 

                    0.[18 0’s]111
   [225 1’s])[207 1’s][18 0’s]0
            -[206 1’s][18 1’s]1
             [206 1’s][18 0’s]10
            -[205 1’s][18 1’s]11
             [205 1’s][18 0’s]110
            -[204 1’s][18 1’s]111
             [204 1’s][18 0’s]111

Notice that no zeroes are placed to the right of the last values after each iteration. Also, there is a pattern after doing the subtraction in each iteration, where there are $n$ ones followed by $18$ zeroes and $207-n$ ones.


To confirm this, we note that in each iteration, we multiply by $2$ and subtract $2^{2007} - 1.$ Doing this results in 

  [n   1’s][18 0’s][207-n 1’s]0
 -[n-1 1’s][18 1’s][207-n 1’s]1
  [n-1 1’s][18 0’s][207-(n-1) 1’s]

Thus, we see that the pattern holds. In fact, tthe pattern repeats $206$ times, and after that, the number from the subtraction just has $207$ consecutive ones. That means the digits repeats every $225$ digits, and $18$ of these digits are zeroes.


That means of the $9900$ digits past the radix point, $792$ of them are zeroes. After $9918$ digits, there are $810$ zeroes, and since the $82$ digits after are ones, we confirm that there are $\boxed{810}$ zeroes of the first $10,000$ digits past the radix point.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See Also

2007 iTest (Problems, Answer Key)
Preceded by:
Problem 55 		Followed by:
Problem 57
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • TB1 • TB2 • TB3 • TB4
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