Difference between revisions of "2006 AIME II Problems/Problem 7"
Lookcloser (talk | contribs) (→Solution) |
Lookcloser (talk | contribs) (→Solution 3) |
||
Line 23: | Line 23: | ||
=== Solution 3 === | === 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. | + | 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 1 < a < | + | When <math>1</math> < <math>a</math> < <math>1004 there are </math>9<math> X </math>9<math> = </math>814 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 100 < a < 200 there are 9 X 8 =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. | + | 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>. |
− | This can clearly be extended to 100k < a < 100(k+1) where k is an integer and 0 < k < 9. | + | 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>. |
− | Thus for 100 < a < 500 there are 72 X 4 = 288 possible values of a. | + | Thus for <math>100</math> < a < <math>500</math> there are <math>72</math> X 4 = <math>288</math> possible values of <math>a</math>. |
− | Thus when 1 < a < 500 there are 288 + 81 = 369 possible values of a and b. | + | 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>. |
− | Doubling this yields | + | Doubling this yields <math>369</math> X <math>2</math> = <math>738</math>. |
== See also == | == See also == |
Revision as of 21:16, 7 March 2011
Problem
Find the number of ordered pairs of positive integers such that and neither nor has a zero digit.
Solution
Solution 1
There are numbers up to 1000 that have 0 as their units digit. All of the other excluded possibilities are when or have a 0 in the tens digit, and since the equation is symmetric, we will just count when has a 0 in the tens digit and multiply by 2 (notice that the only time both and 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 numbers in every hundred numbers that have a tens digit of 0 (this is true from 100 to 900), totaling such numbers; considering also and we have . Therefore, there are such ordered pairs.
Solution 2
Let and be 3 digit numbers:
cde +fgh ---- 1000
and must add up to , and must add up to , and and must add up to . Since none of the digits can be 0, there are possibilites if both numbers are three digits.
There are two other scenarios. and 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 possibilities (the two accounting for whether or has three digits) and for the second case there are possibilities. Thus, thus total possibilities for is .
Solution 3
We first must notice that we can find all the possible values of between and and then double that result.
When < < 99814 possible solution for a so that neither a nor b has a zero in it, counting through , through ... through . When < < there are X = possible solution for a so that neither a nor b has a zero in it, counting through , through ... through . This can clearly be extended to < < where is an integer and < < . Thus for < a < there are X 4 = possible values of .
Thus when < < there are + = possible values of and .
Doubling this yields X = .
See also
2006 AIME II (Problems • Answer Key • Resources) | ||
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 |