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Difference between revisions of "1992 IMO Problems/Problem 1"

(Created page with "Find all integers a, b, c satisfying 1 < a < b < c such that (a - 1)(b -1)(c - 1) is a divisor of abc - 1.")
 
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Find all integers a, b, c satisfying 1 < a < b < c such that (a - 1)(b -1)(c - 1) is a divisor of abc - 1.
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==Problem==
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Find all integers <math>a</math>, <math>b</math>, <math>c</math> satisfying <math>1 < a < b < c</math> such that <math>(a - 1)(b -1)(c - 1)</math> is a divisor of <math>abc - 1</math>.
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== Solution ==
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<math>1<\frac{abc-1}{(a-1)(b-1)(c-1)}<\frac{abc}{(a-1)(b-1)(c-1)}</math>
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<math>1<\frac{abc-1}{(a-1)(b-1)(c-1)}<\left(\frac{a}{a-1}\right) \left(\frac{b}{b-1}\right) \left(\frac{c}{c-1}\right)</math>
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With <math>1<a<b<c</math> it implies that <math>a \ge 2</math>, <math>b \ge 3</math>, <math>c \ge 4</math>
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Therefore, <math>\frac{a}{a-1}=1+\frac{1}{a-1}</math>
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which for <math>a</math> gives: <math>\frac{a}{a-1} \le 1+\frac{1}{2-1}</math>, which gives :<math>\frac{a}{a-1} \le 2</math>
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for <math>b</math> gives: <math>\frac{b}{b-1} \le 1+\frac{1}{3-1}</math>, which gives :<math>\frac{b}{b-1} \le \frac{3}{2}</math>
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for <math>c</math> gives: <math>\frac{c}{c-1} \le 1+\frac{1}{4-1}</math>, which gives :<math>\frac{c}{c-1} \le \frac{4}{3}</math>
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Substituting those inequalities into the original inequality gives:
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<math>1<\frac{abc-1}{(a-1)(b-1)(c-1)}<\left( 2 \right) \left(\frac{3}{2}\right) \left(\frac{4}{3}\right)</math>
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<math>1<\frac{abc-1}{(a-1)(b-1)(c-1)}<4</math>
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Since <math>\frac{abc-1}{(a-1)(b-1)(c-1)}</math> needs to be integer,
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then <math>\frac{abc-1}{(a-1)(b-1)(c-1)}=2</math> or <math>\frac{abc-1}{(a-1)(b-1)(c-1)}=3</math>
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Case 1: <math>\frac{abc-1}{(a-1)(b-1)(c-1)}=2</math>
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<math>abc-1=2(a-1)(b-1)(c-1)=2abc-2(ab+bc+ac)+2(a+b+c)-2</math>
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Case 1, subcase <math>a=2</math>:
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<math>2bc-1=2bc-2(b+c)+2</math> gives: <math>2(b+c)=3</math> which has no solution because <math>2(b+c)</math> is even.
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Case 1, subcase <math>a=3</math>:
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<math>3bc-1=4bc-4(b+c)+4</math>
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<math>bc-4b-4c+5=0</math>
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<math>(b-4)(c-4)=11</math>
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<math>b-4=1</math> and <math>c-4=11</math> provides solution <math>(a,b,c)=(3,5,15)</math>
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Case 2: <math>\frac{abc-1}{(a-1)(b-1)(c-1)}=3</math>
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<math>abc-1=3(a-1)(b-1)(c-1)=3abc-3(ab+bc+ac)+3(a+b+c)-3</math>
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Case 2, subcase <math>a=2</math>:
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<math>2bc-1=3bc-3(b+c)+3</math>
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<math>bc-3b-3c+4=0</math>
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<math>(b-3)(c-3)=5</math>
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<math>b-3=1</math> and <math>c-3=5</math> provides solution <math>(a,b,c)=(2,4,8)</math>
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Case 2, subcase <math>a=3</math>:
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<math>3bc-1=6bc-6(b+c)+6</math>
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Since <math>(3bc-1</math>) mod <math>3 = -1</math> and <math>(6bc-6(b+c)+6)</math> mod <math>3 = 0</math>, then there is no solution for this subcase.
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Now we verify our two solutions:
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when <math>(a,b,c)=(2,4,8)</math>
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<math>abc-1=(2)(4)(8)-1=63</math> and <math>(a-1)(b-1)(c-1)=(1)(3)(7)=21</math>
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Since <math>21</math> is a factor of <math>63</math>, this solutions is correct.
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when <math>(a,b,c)=(3,5,15)</math>
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<math>abc-1=(3)(5)(15)-1=224</math> and <math>(a-1)(b-1)(c-1)=(2)(4)(14)=112</math>
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Since <math>112</math> is a factor of <math>224</math>, this solutions is also correct.
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The solutions are: <math>(a,b,c)=(2,4,8)</math> and <math>(a,b,c)=(3,5,15)</math>
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 +
~ Tomas Diaz. orders@tomasdiaz.com
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{{alternate solutions}}
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==See Also==
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{{IMO box|year=1992|before=First Question|num-a=2}}
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[[Category:Olympiad Geometry Problems]]
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[[Category:3D Geometry Problems]]

Latest revision as of 23:40, 16 November 2023

Problem

Find all integers $a$, $b$, $c$ satisfying $1 < a < b < c$ such that $(a - 1)(b -1)(c - 1)$ is a divisor of $abc - 1$.

Solution

$1<\frac{abc-1}{(a-1)(b-1)(c-1)}<\frac{abc}{(a-1)(b-1)(c-1)}$

$1<\frac{abc-1}{(a-1)(b-1)(c-1)}<\left(\frac{a}{a-1}\right) \left(\frac{b}{b-1}\right) \left(\frac{c}{c-1}\right)$

With $1<a<b<c$ it implies that $a \ge 2$, $b \ge 3$, $c \ge 4$

Therefore, $\frac{a}{a-1}=1+\frac{1}{a-1}$

which for $a$ gives: $\frac{a}{a-1} \le 1+\frac{1}{2-1}$, which gives :$\frac{a}{a-1} \le 2$

for $b$ gives: $\frac{b}{b-1} \le 1+\frac{1}{3-1}$, which gives :$\frac{b}{b-1} \le \frac{3}{2}$

for $c$ gives: $\frac{c}{c-1} \le 1+\frac{1}{4-1}$, which gives :$\frac{c}{c-1} \le \frac{4}{3}$

Substituting those inequalities into the original inequality gives:

$1<\frac{abc-1}{(a-1)(b-1)(c-1)}<\left( 2 \right) \left(\frac{3}{2}\right) \left(\frac{4}{3}\right)$

$1<\frac{abc-1}{(a-1)(b-1)(c-1)}<4$

Since $\frac{abc-1}{(a-1)(b-1)(c-1)}$ needs to be integer,

then $\frac{abc-1}{(a-1)(b-1)(c-1)}=2$ or $\frac{abc-1}{(a-1)(b-1)(c-1)}=3$

Case 1: $\frac{abc-1}{(a-1)(b-1)(c-1)}=2$

$abc-1=2(a-1)(b-1)(c-1)=2abc-2(ab+bc+ac)+2(a+b+c)-2$

Case 1, subcase $a=2$:

$2bc-1=2bc-2(b+c)+2$ gives: $2(b+c)=3$ which has no solution because $2(b+c)$ is even.

Case 1, subcase $a=3$:

$3bc-1=4bc-4(b+c)+4$

$bc-4b-4c+5=0$

$(b-4)(c-4)=11$

$b-4=1$ and $c-4=11$ provides solution $(a,b,c)=(3,5,15)$

Case 2: $\frac{abc-1}{(a-1)(b-1)(c-1)}=3$

$abc-1=3(a-1)(b-1)(c-1)=3abc-3(ab+bc+ac)+3(a+b+c)-3$

Case 2, subcase $a=2$:

$2bc-1=3bc-3(b+c)+3$

$bc-3b-3c+4=0$

$(b-3)(c-3)=5$

$b-3=1$ and $c-3=5$ provides solution $(a,b,c)=(2,4,8)$

Case 2, subcase $a=3$:

$3bc-1=6bc-6(b+c)+6$

Since $(3bc-1$) mod $3 = -1$ and $(6bc-6(b+c)+6)$ mod $3 = 0$, then there is no solution for this subcase.

Now we verify our two solutions:

when $(a,b,c)=(2,4,8)$

$abc-1=(2)(4)(8)-1=63$ and $(a-1)(b-1)(c-1)=(1)(3)(7)=21$

Since $21$ is a factor of $63$, this solutions is correct.

when $(a,b,c)=(3,5,15)$

$abc-1=(3)(5)(15)-1=224$ and $(a-1)(b-1)(c-1)=(2)(4)(14)=112$

Since $112$ is a factor of $224$, this solutions is also correct.

The solutions are: $(a,b,c)=(2,4,8)$ and $(a,b,c)=(3,5,15)$

~ Tomas Diaz. orders@tomasdiaz.com

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See Also

1992 IMO (Problems) • Resources
Preceded by
First Question 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 2
All IMO Problems and Solutions
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