Difference between revisions of "1990 AHSME Problems/Problem 19"
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== Solution 2 == | == Solution 2 == | ||
− | Rearranging the expression <math>N^2 + 7, we get < | + | Rearranging the expression <math>N^2 + 7,</math> we get <math>N^2 + 7 = (N+4)(N-4) + 23.</math> |
− | Thus, < | + | Thus, <math> \frac{(N+4)(N-4) + 23}{N+4} = N - 4 + \frac{23}{N+4}.</math> |
− | Thus, < | + | Thus, <math>N+4 \equiv 0 \mod 23.</math> |
− | Hence, < | + | Hence, <math>N= 23x -4,</math> for some arbitrary number <math>x.</math> |
− | Solving for < | + | Solving for <math>x</math> in <math>23x-4<1990,</math> we find the answer is <math> x= 86,</math> or <math>\fbox{B}.</math> |
~coolmath2017 | ~coolmath2017 |
Revision as of 22:29, 21 June 2020
Contents
Problem
For how many integers between and is the improper fraction in lowest terms?
Solution 1
What we want to know is for how many is We start by setting for some arbitrary . This shows that evenly divides . Next we want to see under which conditions also divides . We know from the previous statement that and thus Next we simply add to get However, we also want which leads to from the previous statement. Since from that statement divides evenly, must be of the form , for some arbitrary integer . After this, we can set and Finally, we must find the largest such that This is a simple linear inequality for which the answer is , or .
Solution 2
Rearranging the expression we get
Thus,
Thus,
Hence, for some arbitrary number
Solving for in we find the answer is or
~coolmath2017
See also
1990 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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 | ||
All AHSME Problems and Solutions |
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