Difference between revisions of "1993 UNCO Math Contest II Problems/Problem 2"
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== Problem == | == Problem == | ||
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== Solution == | == Solution == | ||
− | Fist we add them together and divide. We get the repeating decimal .134316134316. We can see that it repeats every six decimal places, so we need to find the remainder when 623 is divided by six. When we divide, we see that the remainder | + | == Solution 1 === |
+ | Fist we add them together and divide. We get the repeating decimal .134316134316. We can see that it repeats every six decimal places, so we need to find the remainder when 623 is divided by six. When we divide, we see that the remainder is five, so we go in five decimal places and find 1, and we are done. | ||
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+ | == Solution 2 == | ||
+ | The sum of <math>1/9</math> and <math>2/99</math> is <math>13/99</math>, and the sum of of <math>13/99</math> and <math>3/999</math> (aka <math>1/333</math>) is <math>164/1221</math>. Doing long division, we get the decimal <math>0.\overline{134316}</math>, so the <math>623^{rd}</math> place, we only need the <math>623\pod6=5^{th}</math> digit, which is <math>\boxed{1}</math> | ||
== See also == | == See also == |
Latest revision as of 23:24, 19 January 2023
Problem
Determine the digit in the place after the decimal point in the repeating decimal for:
Solution
Solution 1 =
Fist we add them together and divide. We get the repeating decimal .134316134316. We can see that it repeats every six decimal places, so we need to find the remainder when 623 is divided by six. When we divide, we see that the remainder is five, so we go in five decimal places and find 1, and we are done.
Solution 2
The sum of and is , and the sum of of and (aka ) is . Doing long division, we get the decimal , so the place, we only need the digit, which is
See also
1993 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
All UNCO Math Contest Problems and Solutions |