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Question from the 2011 IMC (international mathematics competition) key stage III paper, about the evaluation of a+b+c

$a+b+c+ab+bc+ca+abc=1000$ find the minimum value of $a+b+c$LCM and GCD equationQuadratic equation, math olympiad questionHow many positive integers less than $2011$ cannot be expressed in the form $4a + 5b$, where $a$ and $b$ are positive integers?Help me finding $a+b+c$ in the given questionGood points in a triangleFind a six-digit perfect square of a particular form – BMO 1993 P1Let $p$, $q$ be prime numbers such that $n^{3pq} – n$ is a multiple of $3pq$ for all positive integers $n$. Find the least possible value of $p + q$.About International Mathematics Competition (IMC)Minimize LCM / GCDQuestion from the 2011 IMC (International Mathematics Competition) Key Stage III paper, about the evaluation of a quadratic equation

2

$begingroup$

Let $a,b,c$ be positive integers such that
$$ab+bc+ca+2(a+b+c)=8045qquad abc-a-b-c=-2$$
Find the value of $a+b+c$.

I originally saying that $a+b+c=abc+2$.

However, it was to no avail, as I concluded nowhere.

elementary-number-theory contest-math systems-of-equations problem-solving

share|cite|improve this question

edited 42 mins ago

Bill Dubuque

211k29193646

asked 2 hours ago

kenithkenith

354

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of $a+b+c+ab+bc+ca+abc=1000$ find the minimum value of $a+b+c$
  $endgroup$
  – Bill Dubuque
  44 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is not a duplicate, because, a+b+c can only have one value
  $endgroup$
  – kenith
  36 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The same methods there work here. If you know how to solve one than you can solve the other. It's not good to clutter the site with infinitely many minor tweaks on these problems.
  $endgroup$
  – Bill Dubuque
  33 mins ago
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

Let $a,b,c$ be positive integers such that
$$ab+bc+ca+2(a+b+c)=8045qquad abc-a-b-c=-2$$
Find the value of $a+b+c$.

I originally saying that $a+b+c=abc+2$.

However, it was to no avail, as I concluded nowhere.

elementary-number-theory contest-math systems-of-equations problem-solving

share|cite|improve this question

edited 42 mins ago

Bill Dubuque

211k29193646

asked 2 hours ago

kenithkenith

354

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of $a+b+c+ab+bc+ca+abc=1000$ find the minimum value of $a+b+c$
  $endgroup$
  – Bill Dubuque
  44 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is not a duplicate, because, a+b+c can only have one value
  $endgroup$
  – kenith
  36 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The same methods there work here. If you know how to solve one than you can solve the other. It's not good to clutter the site with infinitely many minor tweaks on these problems.
  $endgroup$
  – Bill Dubuque
  33 mins ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Let $a,b,c$ be positive integers such that
$$ab+bc+ca+2(a+b+c)=8045qquad abc-a-b-c=-2$$
Find the value of $a+b+c$.

I originally saying that $a+b+c=abc+2$.

However, it was to no avail, as I concluded nowhere.

elementary-number-theory contest-math systems-of-equations problem-solving

share|cite|improve this question

edited 42 mins ago

Bill Dubuque

211k29193646

asked 2 hours ago

kenithkenith

354

$endgroup$

share|cite|improve this question

edited 42 mins ago

Bill Dubuque

211k29193646

asked 2 hours ago

kenithkenith

354

share|cite|improve this question

edited 42 mins ago

Bill Dubuque

211k29193646

asked 2 hours ago

kenithkenith

354

edited 42 mins ago

Bill Dubuque

211k29193646

edited 42 mins ago

Bill Dubuque

211k29193646

asked 2 hours ago

kenithkenith

354

asked 2 hours ago

kenithkenith

354

3 Answers
3

active

oldest

votes

5

$begingroup$

it may be useful to recall that:
$$
(1+a)(1+b)(1+c) = 1 + (a+b+c) + (ab + bc + ca) + abc
$$

share|cite|improve this answer

answered 1 hour ago

David HoldenDavid Holden

14.9k21224

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, but how will I work out the value of a+b+c
  $endgroup$
  – kenith
  59 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  ???????????????
  $endgroup$
  – kenith
  59 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @kenith: can you factor $8044$?
  $endgroup$
  – robjohn♦
  53 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes I can factor it
  $endgroup$
  – kenith
  50 mins ago
  

  
  
  
  

add a comment | 

0

$begingroup$

You have that
$$begin{cases} ab+bc+ca+2(a+b+c)=8045\ abc-a-b-c=-2 end{cases}$$ Therefore
$$ab+bc+ca+2(a+b+c)+abc-a-b-c=abc+ab+bc+ca+a+b+c=8045-2=8043$$
As @David Holden observed

$$abc+ab+bc+ca+a+b+c+1=(1+a)(1+b)(1+c)=8044$$

Observe that the divisors of $8044$ are ${1, 2, 4, 2011, 8044}$. Since $a,b,cin mathbb Z_{gt0}$, none of the factors $(1+a), (1+b)$ or $(1+c)$ will be $1$, so forget about the divisors $1text{ and } 8044$. Now you just have to combine the factors and work out all solutions. Notice that they're going to be symmetric.

Added later:
You can only express $8044$ as a product of three positive integers greater than $1$ as $$8044=2011·2·2$$ See why? Hence

$$a+b+c=2010+1+1=2012$$

share|cite|improve this answer

edited 26 mins ago

answered 44 mins ago

Dr. MathvaDr. Mathva

1,880322

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes but using you're method, how am I going to work out the value of a+b+c
  $endgroup$
  – kenith
  37 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You forgot the +1. RHS should be 8044, no?
  $endgroup$
  – eric_kernfeld
  28 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @kenith Done!...
  $endgroup$
  – Dr. Mathva
  22 mins ago
  

  
  
  
  

add a comment | 

0

$begingroup$

If you don't want to pull a clever factorization out of nowhere, another strategy is to notice that a, b, and c have to be very small in order to satisfy the second equation. In fact, even 2, 2, and 1 is too big, because 4 - 5 is bigger than -2. Thus, two of the three must be 1. This leaves the third to be whatever you want, as a*1*1 - x - 1 - 1 = -2 for all a. Simplify the first equation using b=c=1 and you'll get the answer pretty plainly.

This strategy also doesn't require you to manually verify that 2011 is prime.

share|cite|improve this answer

answered 21 mins ago

eric_kernfelderic_kernfeld

1011

$endgroup$

add a comment | 

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5

$begingroup$

it may be useful to recall that:
$$
(1+a)(1+b)(1+c) = 1 + (a+b+c) + (ab + bc + ca) + abc
$$

share|cite|improve this answer

answered 1 hour ago

David HoldenDavid Holden

14.9k21224

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, but how will I work out the value of a+b+c
  $endgroup$
  – kenith
  59 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  ???????????????
  $endgroup$
  – kenith
  59 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @kenith: can you factor $8044$?
  $endgroup$
  – robjohn♦
  53 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes I can factor it
  $endgroup$
  – kenith
  50 mins ago
  

  
  
  
  

add a comment | 

5

$begingroup$

it may be useful to recall that:
$$
(1+a)(1+b)(1+c) = 1 + (a+b+c) + (ab + bc + ca) + abc
$$

share|cite|improve this answer

answered 1 hour ago

David HoldenDavid Holden

14.9k21224

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, but how will I work out the value of a+b+c
  $endgroup$
  – kenith
  59 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  ???????????????
  $endgroup$
  – kenith
  59 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @kenith: can you factor $8044$?
  $endgroup$
  – robjohn♦
  53 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes I can factor it
  $endgroup$
  – kenith
  50 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

it may be useful to recall that:
$$
(1+a)(1+b)(1+c) = 1 + (a+b+c) + (ab + bc + ca) + abc
$$

share|cite|improve this answer

answered 1 hour ago

David HoldenDavid Holden

14.9k21224

$endgroup$

share|cite|improve this answer

answered 1 hour ago

David HoldenDavid Holden

14.9k21224

answered 1 hour ago

David HoldenDavid Holden

14.9k21224

answered 1 hour ago

David HoldenDavid Holden

14.9k21224

0

$begingroup$

You have that
$$begin{cases} ab+bc+ca+2(a+b+c)=8045\ abc-a-b-c=-2 end{cases}$$ Therefore
$$ab+bc+ca+2(a+b+c)+abc-a-b-c=abc+ab+bc+ca+a+b+c=8045-2=8043$$
As @David Holden observed

$$abc+ab+bc+ca+a+b+c+1=(1+a)(1+b)(1+c)=8044$$

Observe that the divisors of $8044$ are ${1, 2, 4, 2011, 8044}$. Since $a,b,cin mathbb Z_{gt0}$, none of the factors $(1+a), (1+b)$ or $(1+c)$ will be $1$, so forget about the divisors $1text{ and } 8044$. Now you just have to combine the factors and work out all solutions. Notice that they're going to be symmetric.

Added later:
You can only express $8044$ as a product of three positive integers greater than $1$ as $$8044=2011·2·2$$ See why? Hence

$$a+b+c=2010+1+1=2012$$

share|cite|improve this answer

edited 26 mins ago

answered 44 mins ago

Dr. MathvaDr. Mathva

1,880322

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes but using you're method, how am I going to work out the value of a+b+c
  $endgroup$
  – kenith
  37 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You forgot the +1. RHS should be 8044, no?
  $endgroup$
  – eric_kernfeld
  28 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @kenith Done!...
  $endgroup$
  – Dr. Mathva
  22 mins ago
  

  
  
  
  

add a comment | 

0

$begingroup$

You have that
$$begin{cases} ab+bc+ca+2(a+b+c)=8045\ abc-a-b-c=-2 end{cases}$$ Therefore
$$ab+bc+ca+2(a+b+c)+abc-a-b-c=abc+ab+bc+ca+a+b+c=8045-2=8043$$
As @David Holden observed

$$abc+ab+bc+ca+a+b+c+1=(1+a)(1+b)(1+c)=8044$$

Observe that the divisors of $8044$ are ${1, 2, 4, 2011, 8044}$. Since $a,b,cin mathbb Z_{gt0}$, none of the factors $(1+a), (1+b)$ or $(1+c)$ will be $1$, so forget about the divisors $1text{ and } 8044$. Now you just have to combine the factors and work out all solutions. Notice that they're going to be symmetric.

Added later:
You can only express $8044$ as a product of three positive integers greater than $1$ as $$8044=2011·2·2$$ See why? Hence

$$a+b+c=2010+1+1=2012$$

share|cite|improve this answer

edited 26 mins ago

answered 44 mins ago

Dr. MathvaDr. Mathva

1,880322

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes but using you're method, how am I going to work out the value of a+b+c
  $endgroup$
  – kenith
  37 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You forgot the +1. RHS should be 8044, no?
  $endgroup$
  – eric_kernfeld
  28 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @kenith Done!...
  $endgroup$
  – Dr. Mathva
  22 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

You have that
$$begin{cases} ab+bc+ca+2(a+b+c)=8045\ abc-a-b-c=-2 end{cases}$$ Therefore
$$ab+bc+ca+2(a+b+c)+abc-a-b-c=abc+ab+bc+ca+a+b+c=8045-2=8043$$
As @David Holden observed

$$abc+ab+bc+ca+a+b+c+1=(1+a)(1+b)(1+c)=8044$$

Observe that the divisors of $8044$ are ${1, 2, 4, 2011, 8044}$. Since $a,b,cin mathbb Z_{gt0}$, none of the factors $(1+a), (1+b)$ or $(1+c)$ will be $1$, so forget about the divisors $1text{ and } 8044$. Now you just have to combine the factors and work out all solutions. Notice that they're going to be symmetric.

Added later:
You can only express $8044$ as a product of three positive integers greater than $1$ as $$8044=2011·2·2$$ See why? Hence

$$a+b+c=2010+1+1=2012$$

share|cite|improve this answer

edited 26 mins ago

answered 44 mins ago

Dr. MathvaDr. Mathva

1,880322

$endgroup$

share|cite|improve this answer

edited 26 mins ago

answered 44 mins ago

Dr. MathvaDr. Mathva

1,880322

answered 44 mins ago

Dr. MathvaDr. Mathva

1,880322

answered 44 mins ago

Dr. MathvaDr. Mathva

1,880322

0

$begingroup$

If you don't want to pull a clever factorization out of nowhere, another strategy is to notice that a, b, and c have to be very small in order to satisfy the second equation. In fact, even 2, 2, and 1 is too big, because 4 - 5 is bigger than -2. Thus, two of the three must be 1. This leaves the third to be whatever you want, as a*1*1 - x - 1 - 1 = -2 for all a. Simplify the first equation using b=c=1 and you'll get the answer pretty plainly.

This strategy also doesn't require you to manually verify that 2011 is prime.

share|cite|improve this answer

answered 21 mins ago

eric_kernfelderic_kernfeld

1011

$endgroup$

add a comment | 

0

$begingroup$

If you don't want to pull a clever factorization out of nowhere, another strategy is to notice that a, b, and c have to be very small in order to satisfy the second equation. In fact, even 2, 2, and 1 is too big, because 4 - 5 is bigger than -2. Thus, two of the three must be 1. This leaves the third to be whatever you want, as a*1*1 - x - 1 - 1 = -2 for all a. Simplify the first equation using b=c=1 and you'll get the answer pretty plainly.

This strategy also doesn't require you to manually verify that 2011 is prime.

share|cite|improve this answer

answered 21 mins ago

eric_kernfelderic_kernfeld

1011

$endgroup$

add a comment | 

$begingroup$

If you don't want to pull a clever factorization out of nowhere, another strategy is to notice that a, b, and c have to be very small in order to satisfy the second equation. In fact, even 2, 2, and 1 is too big, because 4 - 5 is bigger than -2. Thus, two of the three must be 1. This leaves the third to be whatever you want, as a*1*1 - x - 1 - 1 = -2 for all a. Simplify the first equation using b=c=1 and you'll get the answer pretty plainly.

This strategy also doesn't require you to manually verify that 2011 is prime.

share|cite|improve this answer

answered 21 mins ago

eric_kernfelderic_kernfeld

1011

$endgroup$

share|cite|improve this answer

answered 21 mins ago

eric_kernfelderic_kernfeld

1011

answered 21 mins ago

eric_kernfelderic_kernfeld

1011

answered 21 mins ago

eric_kernfelderic_kernfeld

1011