Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1999 AHSME Problems/Problem 25"

m (→‎Solution 4)
m (→‎Solution 3 (The Easiest and Most Intuitive): corrected an error)
 
(One intermediate revision 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>a_3=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>
+
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 rest of the right hand side will not add up to be <math>5</math>, <math>a_2 = 1</math>
+
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>

Latest revision as of 00:43, 4 October 2023

Problem

There are unique integers $a_{2},a_{3},a_{4},a_{5},a_{6},a_{7}$ such that

\[\frac {5}{7} = \frac {a_{2}}{2!} + \frac {a_{3}}{3!} + \frac {a_{4}}{4!} + \frac {a_{5}}{5!} + \frac {a_{6}}{6!} + \frac {a_{7}}{7!}\]

where $0\leq a_{i} < i$ for $i = 2,3,\ldots,7$. Find $a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}$.

$\textrm{(A)} \ 8 \qquad \textrm{(B)} \ 9 \qquad \textrm{(C)} \ 10 \qquad \textrm{(D)} \ 11 \qquad \textrm{(E)} \ 12$

Solution 1(Modular Functions)

Multiply out the $7!$ to get

\[5 \cdot 6! = (3 \cdot 4 \cdots 7)a_2 + (4 \cdots 7)a_3 + (5 \cdot 6 \cdot 7)a_4 + 42a_5 + 7a_6 + a_7 .\]

By Wilson's Theorem (or by straightforward division), $a_7 + 7(a_6 + 6a_5 + \cdots) \equiv 5 \cdot 6! \equiv -5 \equiv 2 \pmod{7}$, so $a_7 = 2$. Then we move $a_7$ to the left and divide through by $7$ to obtain

\[\frac{5 \cdot 6!-2}{7} = 514 = 360a_2 + 120a_3 + 30a_4 + 6a_5 + a_6.\]

We then repeat this procedure $\pmod{6}$, from which it follows that $a_6 \equiv 514 \equiv 4 \pmod{6}$, and so forth. Continuing, we find the unique solution to be $(a_2, a_3, a_4, a_5, a_6, a_7) = (1,1,1,0,4,2)$ (uniqueness is assured by the Division Theorem). The answer is $9 \Longrightarrow \boxed{\mathrm{(B)}}$.

Solution 2(Basic Algebra and Bashing)

We start by multiplying both sides by $7!$, and we get: \[3600=2520a_2+840a_3+210a_4+42a_5+7a_6+a_7\] After doing some guess and check, we find that the answer is $\boxed{\textbf{(B)~9}}$.

~aopspandy

Solution 3 (The Easiest and Most Intuitive)

Let's clear up the fractions: \[\frac{5}{7}=\frac{2520a_2+840a_3+210a_4+42a_5+7a_6+a_7}{7!}\] \[3600=2520a_2+840a_3+210a_4+42a_5+7a_6+a_7\] Notice that if we divide everything by $7$ then we would have:\[514+\frac{2}{7}=360a_2+120a_3+30a_4+6a_5+a_6+\frac{1}{7}a_7\] Since $0 \le a_7<7$ and $a_7$ must be an integer, then we have $\frac{2}{7}=\frac{1}{7}a_7$, so $a_7=2$.

Similarly, if we divide everything by $6$, then we would have: \[85+\frac{4}{6}=60a_2+20a_3+5a_4+a_5+\frac{1}{6}a_6\] Again, since $0 \le a_6<6$ and $a_6$ must be an integer, we have $\frac{4}{6}=\frac{1}{6}a_6$, so $a_6=4$.

The pattern repeats itself, so in the end we have $a_2=1$, $a_3=1$, $a_4=1$, $a_5=0$, $a_6=4$, $a_7=2$. So $a_2+a_3+a_4+a_5+a_6+a_7=\boxed{\textbf{(B)~9}}$

~BurpSuite, with a help from ostriches88

Solution 4

By multiplying both sides by $7$ we get

\[5 = \frac72 a_2 + \frac76 a_3+ \frac{7}{24} a_4 + \frac{7}{120} a_5 + \frac{7}{720} a_6 + \frac{a_7}{720}\]

since $0 \le a_2 < 2$, if $a_2 = 0$ the rest of the right hand side will not add up to be $5$, so $a_2 = 1$

\[\frac32 = \frac76 a_3+ \frac{7}{24} a_4 + \frac{7}{120} a_5 + \frac{7}{720} a_6 + \frac{a_7}{720}\]

If $a_3 = 2$, $\frac76 a_3 = \frac73 > \frac32$, so $a_3 = 1$

\[\frac13 = \frac{7}{24} a_4 + \frac{7}{120} a_5 + \frac{7}{720} a_6 + \frac{a_7}{720}\]

If $a_4 = 2$, $\frac{7}{24} a_4 = \frac{7}{12} > \frac13$, so $a_4 = 1$

\[\frac{1}{24} = \frac{7}{120} a_5 + \frac{7}{720} a_6 + \frac{a_7}{720}\]

If $a_5 = 1$, $\frac{7}{120} a_5 = \frac{7}{120} > \frac{1}{24}$, so $a_5 = 0$

\[\frac{1}{24} = \frac{7}{720} a_6 + \frac{a_7}{720}\]

$30 = 7 a_6 + a_7$. Since $a_7 < 7$, $a_7 = 2$ and $a_6=4$

Therefore, $a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7} = 1 + 1 + 1 + 0 + 4 + 2= \boxed{\textbf{(B) } 9}$.

~isabelchen

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1999_AHSME_Problems/Problem_25&oldid=199113"