Difference between revisions of "2006 AIME II Problems/Problem 7"

m (Solution 3)
Line 25: Line 25:
 
We first must notice that we can find all the possible values of <math>a</math> between <math>1</math> and <math>500</math> and then double that result.  
 
We first must notice that we can find all the possible values of <math>a</math> between <math>1</math> and <math>500</math> and then double that result.  
  
When <math>1</math> < <math>a</math> < <math>100</math> there are <math>9</math> X <math>9</math> = <math>81</math> possible solution for a so that neither a nor b has a zero in it, counting <math>1</math> through <math>9</math>, <math>11</math> through <math>19</math> ... <math>81</math> through <math>89</math>.
+
When <math>1 < a < 100</math> there are <math>9\times9 = 81</math> possible solution for <math>a</math> so that neither <math>a</math> nor <math>b</math> has a zero in it, counting <math>1</math> through <math>9</math>, <math>11</math> through <math>19</math>, ..., <math>81</math> through <math>89</math>.
When <math>100</math> < <math>a</math> < <math>200</math> there are <math>9</math> X <math>8</math> =<math>72</math> possible solution for a so that neither a nor b has a zero in it, counting  <math>111</math> through <math>119</math>, <math>121</math> through <math>129</math> ... <math>181</math> through <math>189</math>.
+
When <math>100 < a < 200</math> there are <math>9\times8 =72</math> possible solution for a so that neither a nor b has a zero in it, counting  <math>111</math> through <math>119</math>, <math>121</math> through <math>129</math>, ..., <math>181</math> through <math>189</math>.
This can clearly be extended to <math>100k</math> < <math>a</math> < <math>100(k+1)</math> where <math>k</math> is an integer and <math>0</math> < <math>k</math> < <math>9</math>.  
+
This can clearly be extended to <math>100k < a < 100(k+1)</math> where <math>k</math> is an integer and <math>0 <k < 9</math>.  
Thus for <math>100</math> < a < <math>500</math> there are <math>72</math> X <math>4</math> = <math>288</math> possible values of <math>a</math>.
+
Thus for <math>100 < a < 500</math> there are <math>72\times4</math> = <math>288</math> possible values of <math>a</math>.
  
Thus when <math>1</math> < <math>a</math> < <math>500</math> there are <math>288</math> + <math>81</math> = <math>369</math> possible values of <math>a</math> and <math>b</math>.  
+
Thus when <math>1 < a < 500</math> there are <math>288 + 81 =369</math> possible values of <math>a</math> and <math>b</math>.  
  
Doubling this yields <math>369</math> X <math>2</math> = <math>738</math>.
+
Doubling this yields <math>369\times2= 738</math>.
  
 
== See also ==
 
== See also ==

Revision as of 20:03, 13 March 2015

Problem

Find the number of ordered pairs of positive integers $(a,b)$ such that $a+b=1000$ and neither $a$ nor $b$ has a zero digit.

Solution

Solution 1

There are $\left\lfloor\frac{999}{10}\right\rfloor = 99$ numbers up to 1000 that have 0 as their units digit. All of the other excluded possibilities are when $a$ or $b$ have a 0 in the tens digit, and since the equation is symmetric, we will just count when $a$ has a 0 in the tens digit and multiply by 2 (notice that the only time both $a$ and $b$ can have a 0 in the tens digit is when they are divisible by 100, which falls into the above category, so we do not have to worry about overcounting).

Excluding the numbers divisible by 100, which were counted already, there are $9$ numbers in every hundred numbers that have a tens digit of 0 (this is true from 100 to 900), totaling $9 \cdot 9 = 81$ such numbers; considering $b$ also and we have $81 \cdot 2 = 162$. Therefore, there are $999 - (99 + 162) = \boxed{738}$ such ordered pairs.

Solution 2

Let $a = \overline{cde}$ and $b = \overline{fgh}$ be 3 digit numbers:

 cde
+fgh
----
1000

$e$ and $h$ must add up to $10$, $d$ and $g$ must add up to $9$, and $c$ and $f$ must add up to $9$. Since none of the digits can be 0, there are $9 \times 8 \times 8=576$ possibilites if both numbers are three digits.

There are two other scenarios. $a$ and $b$ can be a three digit number and a two digit number, or a three digit number and a one digit number. For the first scenario, there are $9 \times 8 \times 2=144$ possibilities (the two accounting for whether $a$ or $b$ has three digits) and for the second case there are $9 \times 2=18$ possibilities. Thus, thus total possibilities for $(a,b)$ is $576+144+18=738$.


Solution 3

We first must notice that we can find all the possible values of $a$ between $1$ and $500$ and then double that result.

When $1 < a < 100$ there are $9\times9 = 81$ possible solution for $a$ so that neither $a$ nor $b$ has a zero in it, counting $1$ through $9$, $11$ through $19$, ..., $81$ through $89$. When $100 < a < 200$ there are $9\times8 =72$ possible solution for a so that neither a nor b has a zero in it, counting $111$ through $119$, $121$ through $129$, ..., $181$ through $189$. This can clearly be extended to $100k < a < 100(k+1)$ where $k$ is an integer and $0 <k < 9$. Thus for $100 < a < 500$ there are $72\times4$ = $288$ possible values of $a$.

Thus when $1 < a < 500$ there are $288 + 81 =369$ possible values of $a$ and $b$.

Doubling this yields $369\times2= 738$.

See also

2006 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS