Difference between revisions of "1999 AHSME Problems/Problem 25"
Isabelchen (talk | contribs) m (→Solution 4) |
Isabelchen (talk | contribs) m (→Solution 3 (The Easiest and Most Intuitive): corrected an error) |
||
(3 intermediate revisions by the same user not shown) | |||
Line 35: | Line 35: | ||
Again, since <math>0 \le a_6<6</math> and <math>a_6</math> must be an integer, we have <math>\frac{4}{6}=\frac{1}{6}a_6</math>, so <math>a_6=4</math>. | Again, since <math>0 \le a_6<6</math> and <math>a_6</math> must be an integer, we have <math>\frac{4}{6}=\frac{1}{6}a_6</math>, so <math>a_6=4</math>. | ||
− | The pattern repeats itself, so in the end we have <math>a_2=1</math>, <math>a_3=1</math>, <math> | + | The pattern repeats itself, so in the end we have <math>a_2=1</math>, <math>a_3=1</math>, <math>a_4=1</math>, <math>a_5=0</math>, <math>a_6=4</math>, <math>a_7=2</math>. So <math>a_2+a_3+a_4+a_5+a_6+a_7=\boxed{\textbf{(B)~9}}</math> |
~BurpSuite, with a help from ostriches88 | ~BurpSuite, with a help from ostriches88 | ||
Line 45: | Line 45: | ||
<cmath>5 = \frac72 a_2 + \frac76 a_3+ \frac{7}{24} a_4 + \frac{7}{120} a_5 + \frac{7}{720} a_6 + \frac{a_7}{720}</cmath> | <cmath>5 = \frac72 a_2 + \frac76 a_3+ \frac{7}{24} a_4 + \frac{7}{120} a_5 + \frac{7}{720} a_6 + \frac{a_7}{720}</cmath> | ||
− | since <math>0 \le a_2 < 2</math>, if <math>a_2 = 0</math> the | + | since <math>0 \le a_2 < 2</math>, if <math>a_2 = 0</math> the rest of the right hand side will not add up to be <math>5</math>, so <math>a_2 = 1</math> |
<cmath>\frac32 = \frac76 a_3+ \frac{7}{24} a_4 + \frac{7}{120} a_5 + \frac{7}{720} a_6 + \frac{a_7}{720}</cmath> | <cmath>\frac32 = \frac76 a_3+ \frac{7}{24} a_4 + \frac{7}{120} a_5 + \frac{7}{720} a_6 + \frac{a_7}{720}</cmath> | ||
Line 63: | Line 63: | ||
<math>30 = 7 a_6 + a_7</math>. Since <math>a_7 < 7</math>, <math>a_7 = 2</math> and <math>a_6=4</math> | <math>30 = 7 a_6 + a_7</math>. Since <math>a_7 < 7</math>, <math>a_7 = 2</math> and <math>a_6=4</math> | ||
− | Therefore, <math>a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7} = 1 + 1 + 1 + 0 + 2 | + | Therefore, <math>a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7} = 1 + 1 + 1 + 0 + 4 + 2= \boxed{\textbf{(B) } 9}</math>. |
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] |
Latest revision as of 01:43, 4 October 2023
Contents
Problem
There are unique integers such that
where for
. Find
.
Solution 1(Modular Functions)
Multiply out the to get
By Wilson's Theorem (or by straightforward division), , so
. Then we move
to the left and divide through by
to obtain
We then repeat this procedure , from which it follows that
, and so forth. Continuing, we find the unique solution to be
(uniqueness is assured by the Division Theorem). The answer is
.
Solution 2(Basic Algebra and Bashing)
We start by multiplying both sides by , and we get:
After doing some guess and check, we find that the answer is
.
~aopspandy
Solution 3 (The Easiest and Most Intuitive)
Let's clear up the fractions:
Notice that if we divide everything by
then we would have:
Since
and
must be an integer, then we have
, so
.
Similarly, if we divide everything by , then we would have:
Again, since
and
must be an integer, we have
, so
.
The pattern repeats itself, so in the end we have ,
,
,
,
,
. So
~BurpSuite, with a help from ostriches88
Solution 4
By multiplying both sides by we get
since , if
the rest of the right hand side will not add up to be
, so
If ,
, so
If ,
, so
If ,
, so
. Since
,
and
Therefore, .
See also
1999 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Problem 26 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.