Given a three digit number $n$, let $f(n)$ be the sum of digits of $n$, their products in pairs, and the...












3












$begingroup$


This is my first time posting so do correct me if I am doing anything wrong.



Please help me with this math problem from the British Maths Olympiad (1994 British Maths Olympiad1 Q1 Number Theory).




Starting with any three digit number $n$ (such as $n = 625$) we obtain a new number $f(n)$ which is equal to the sum of the three digits of $n$, their three products in pairs, and the product of all three digits.
Find all three digit numbers such that $frac{n}{f(n)}=1$.




The only solution I found is $199$, can someone verify it please?






elementary-number-theory contest-math







share|cite|improve this question











$endgroup$












  • $begingroup$
    in the definition of $f$, are you concatenating the three results? Can you give an example of a pair $(n,f(n))$?
    $endgroup$
    – mathworker21
    Nov 19 '18 at 5:44










  • $begingroup$
    So $f(199) = 19, 9918, 81$?
    $endgroup$
    – steven gregory
    Nov 19 '18 at 5:45






  • 1




    $begingroup$
    @mathworker21`@stevengregory For example if $n=625$, $f(n)=6+2+5+6*2+6*5+2*5+6*2*5$
    $endgroup$
    – 3684
    Nov 19 '18 at 5:51






  • 2




    $begingroup$
    to do this problem, write $n = 100a+10b+c$ and just write everything out and solve the equation you get
    $endgroup$
    – mathworker21
    Nov 19 '18 at 5:56






  • 1




    $begingroup$
    @mathworker21, I have tried that but I didn't get too far, can you try it if you have time? I also thought about factorising a+b+c+ab+ac+bc+abc as (a+1)(b+1)(c+1)-1 but still wasn't able to get far. For number theory Olympiad problems are there systematic methods or does it require a different method every time.
    $endgroup$
    – 3684
    Nov 19 '18 at 6:00
















3












$begingroup$


This is my first time posting so do correct me if I am doing anything wrong.



Please help me with this math problem from the British Maths Olympiad (1994 British Maths Olympiad1 Q1 Number Theory).




Starting with any three digit number $n$ (such as $n = 625$) we obtain a new number $f(n)$ which is equal to the sum of the three digits of $n$, their three products in pairs, and the product of all three digits.
Find all three digit numbers such that $frac{n}{f(n)}=1$.




The only solution I found is $199$, can someone verify it please?






elementary-number-theory contest-math







share|cite|improve this question











$endgroup$












  • $begingroup$
    in the definition of $f$, are you concatenating the three results? Can you give an example of a pair $(n,f(n))$?
    $endgroup$
    – mathworker21
    Nov 19 '18 at 5:44










  • $begingroup$
    So $f(199) = 19, 9918, 81$?
    $endgroup$
    – steven gregory
    Nov 19 '18 at 5:45






  • 1




    $begingroup$
    @mathworker21`@stevengregory For example if $n=625$, $f(n)=6+2+5+6*2+6*5+2*5+6*2*5$
    $endgroup$
    – 3684
    Nov 19 '18 at 5:51






  • 2




    $begingroup$
    to do this problem, write $n = 100a+10b+c$ and just write everything out and solve the equation you get
    $endgroup$
    – mathworker21
    Nov 19 '18 at 5:56






  • 1




    $begingroup$
    @mathworker21, I have tried that but I didn't get too far, can you try it if you have time? I also thought about factorising a+b+c+ab+ac+bc+abc as (a+1)(b+1)(c+1)-1 but still wasn't able to get far. For number theory Olympiad problems are there systematic methods or does it require a different method every time.
    $endgroup$
    – 3684
    Nov 19 '18 at 6:00














3












3








3


1



$begingroup$


This is my first time posting so do correct me if I am doing anything wrong.



Please help me with this math problem from the British Maths Olympiad (1994 British Maths Olympiad1 Q1 Number Theory).




Starting with any three digit number $n$ (such as $n = 625$) we obtain a new number $f(n)$ which is equal to the sum of the three digits of $n$, their three products in pairs, and the product of all three digits.
Find all three digit numbers such that $frac{n}{f(n)}=1$.




The only solution I found is $199$, can someone verify it please?






elementary-number-theory contest-math







share|cite|improve this question











$endgroup$




This is my first time posting so do correct me if I am doing anything wrong.



Please help me with this math problem from the British Maths Olympiad (1994 British Maths Olympiad1 Q1 Number Theory).




Starting with any three digit number $n$ (such as $n = 625$) we obtain a new number $f(n)$ which is equal to the sum of the three digits of $n$, their three products in pairs, and the product of all three digits.
Find all three digit numbers such that $frac{n}{f(n)}=1$.




The only solution I found is $199$, can someone verify it please?






elementary-number-theory contest-math




elementary-number-theory contest-math






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 19 '18 at 12:31









amWhy

1




1










asked Nov 19 '18 at 5:36









36843684

1297




1297












  • $begingroup$
    in the definition of $f$, are you concatenating the three results? Can you give an example of a pair $(n,f(n))$?
    $endgroup$
    – mathworker21
    Nov 19 '18 at 5:44










  • $begingroup$
    So $f(199) = 19, 9918, 81$?
    $endgroup$
    – steven gregory
    Nov 19 '18 at 5:45






  • 1




    $begingroup$
    @mathworker21`@stevengregory For example if $n=625$, $f(n)=6+2+5+6*2+6*5+2*5+6*2*5$
    $endgroup$
    – 3684
    Nov 19 '18 at 5:51






  • 2




    $begingroup$
    to do this problem, write $n = 100a+10b+c$ and just write everything out and solve the equation you get
    $endgroup$
    – mathworker21
    Nov 19 '18 at 5:56






  • 1




    $begingroup$
    @mathworker21, I have tried that but I didn't get too far, can you try it if you have time? I also thought about factorising a+b+c+ab+ac+bc+abc as (a+1)(b+1)(c+1)-1 but still wasn't able to get far. For number theory Olympiad problems are there systematic methods or does it require a different method every time.
    $endgroup$
    – 3684
    Nov 19 '18 at 6:00


















  • $begingroup$
    in the definition of $f$, are you concatenating the three results? Can you give an example of a pair $(n,f(n))$?
    $endgroup$
    – mathworker21
    Nov 19 '18 at 5:44










  • $begingroup$
    So $f(199) = 19, 9918, 81$?
    $endgroup$
    – steven gregory
    Nov 19 '18 at 5:45






  • 1




    $begingroup$
    @mathworker21`@stevengregory For example if $n=625$, $f(n)=6+2+5+6*2+6*5+2*5+6*2*5$
    $endgroup$
    – 3684
    Nov 19 '18 at 5:51






  • 2




    $begingroup$
    to do this problem, write $n = 100a+10b+c$ and just write everything out and solve the equation you get
    $endgroup$
    – mathworker21
    Nov 19 '18 at 5:56






  • 1




    $begingroup$
    @mathworker21, I have tried that but I didn't get too far, can you try it if you have time? I also thought about factorising a+b+c+ab+ac+bc+abc as (a+1)(b+1)(c+1)-1 but still wasn't able to get far. For number theory Olympiad problems are there systematic methods or does it require a different method every time.
    $endgroup$
    – 3684
    Nov 19 '18 at 6:00
















$begingroup$
in the definition of $f$, are you concatenating the three results? Can you give an example of a pair $(n,f(n))$?
$endgroup$
– mathworker21
Nov 19 '18 at 5:44




$begingroup$
in the definition of $f$, are you concatenating the three results? Can you give an example of a pair $(n,f(n))$?
$endgroup$
– mathworker21
Nov 19 '18 at 5:44












$begingroup$
So $f(199) = 19, 9918, 81$?
$endgroup$
– steven gregory
Nov 19 '18 at 5:45




$begingroup$
So $f(199) = 19, 9918, 81$?
$endgroup$
– steven gregory
Nov 19 '18 at 5:45




1




1




$begingroup$
@mathworker21`@stevengregory For example if $n=625$, $f(n)=6+2+5+6*2+6*5+2*5+6*2*5$
$endgroup$
– 3684
Nov 19 '18 at 5:51




$begingroup$
@mathworker21`@stevengregory For example if $n=625$, $f(n)=6+2+5+6*2+6*5+2*5+6*2*5$
$endgroup$
– 3684
Nov 19 '18 at 5:51




2




2




$begingroup$
to do this problem, write $n = 100a+10b+c$ and just write everything out and solve the equation you get
$endgroup$
– mathworker21
Nov 19 '18 at 5:56




$begingroup$
to do this problem, write $n = 100a+10b+c$ and just write everything out and solve the equation you get
$endgroup$
– mathworker21
Nov 19 '18 at 5:56




1




1




$begingroup$
@mathworker21, I have tried that but I didn't get too far, can you try it if you have time? I also thought about factorising a+b+c+ab+ac+bc+abc as (a+1)(b+1)(c+1)-1 but still wasn't able to get far. For number theory Olympiad problems are there systematic methods or does it require a different method every time.
$endgroup$
– 3684
Nov 19 '18 at 6:00




$begingroup$
@mathworker21, I have tried that but I didn't get too far, can you try it if you have time? I also thought about factorising a+b+c+ab+ac+bc+abc as (a+1)(b+1)(c+1)-1 but still wasn't able to get far. For number theory Olympiad problems are there systematic methods or does it require a different method every time.
$endgroup$
– 3684
Nov 19 '18 at 6:00










2 Answers
2






active

oldest

votes


















9












$begingroup$

Let $n=100a+10b+c,$ where $a> 0$ and $b,cgeq 0$. We are trying to solve $$100a+10b+c=a+b+c+ab+ac+bc+abc \ implies 99a+9b=abc+ab+ac+bc \ implies a(99-b-c-bc)=b(c-9) \$$$c-9leq 0$, but $b+c+bcleq 99$. So the above equation holds iff $b=c=9$, which means $a$ can take any value.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    May I ask how you got to the solution so quick, do you just see the solution?
    $endgroup$
    – 3684
    Nov 19 '18 at 6:13










  • $begingroup$
    @3684 this problem is fairly 'routine'. Plus, I might've seen this before/its associated solution, but I wouldn't remember if I did
    $endgroup$
    – user574848
    Nov 19 '18 at 6:15










  • $begingroup$
    How many main 'routine' methods to solve number theory problems would you say there are? At higher levels do you think each question requires some different insight.
    $endgroup$
    – 3684
    Nov 19 '18 at 6:19



















8












$begingroup$

Here's part $(b)$ because I'm assuming you don't need help with part $(a)$:




We want to compute all possible integers $n$ such that $frac{n}{f(n)} = 1$. Since we know that, by assumption, $n$ is a three-digit number, we can write



$$n = 100a + 10b + c,$$



where $a, b, c$ are integers. If this is the case, in terms of our newly defined variables $a$, $b$, and $c$, we can express $f(n)$ as follows:



$$f(n) = abc + ab + bc + ac + a + b + c.$$



Now, in order to have $frac{n}{f(n)} = 1,$ we must have $n = f(n)$. This happens when



$$99a + 9b = abc + ab + bc + ac$$



$$Longleftrightarrow (9-c)b = a(bc + b + c - 99) $$



Also, we must have $b, c leq 9,$ which implies $bc + b + c - 99 leq 0$. However, since $a neq 0$ (if $a = 0$, we would be able to form a two or one-digit number instead of a three-digit one!), we conclude $b = c = 9$. Therefore, our solution set is given by



$$boxed{{199, 299, 399, 499, 599, 699, 799, 899, 999}}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for your answer, sorry I could only accept one answer. May I ask how you see the solution so quick?
    $endgroup$
    – 3684
    Nov 19 '18 at 6:15











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









9












$begingroup$

Let $n=100a+10b+c,$ where $a> 0$ and $b,cgeq 0$. We are trying to solve $$100a+10b+c=a+b+c+ab+ac+bc+abc \ implies 99a+9b=abc+ab+ac+bc \ implies a(99-b-c-bc)=b(c-9) \$$$c-9leq 0$, but $b+c+bcleq 99$. So the above equation holds iff $b=c=9$, which means $a$ can take any value.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    May I ask how you got to the solution so quick, do you just see the solution?
    $endgroup$
    – 3684
    Nov 19 '18 at 6:13










  • $begingroup$
    @3684 this problem is fairly 'routine'. Plus, I might've seen this before/its associated solution, but I wouldn't remember if I did
    $endgroup$
    – user574848
    Nov 19 '18 at 6:15










  • $begingroup$
    How many main 'routine' methods to solve number theory problems would you say there are? At higher levels do you think each question requires some different insight.
    $endgroup$
    – 3684
    Nov 19 '18 at 6:19
















9












$begingroup$

Let $n=100a+10b+c,$ where $a> 0$ and $b,cgeq 0$. We are trying to solve $$100a+10b+c=a+b+c+ab+ac+bc+abc \ implies 99a+9b=abc+ab+ac+bc \ implies a(99-b-c-bc)=b(c-9) \$$$c-9leq 0$, but $b+c+bcleq 99$. So the above equation holds iff $b=c=9$, which means $a$ can take any value.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    May I ask how you got to the solution so quick, do you just see the solution?
    $endgroup$
    – 3684
    Nov 19 '18 at 6:13










  • $begingroup$
    @3684 this problem is fairly 'routine'. Plus, I might've seen this before/its associated solution, but I wouldn't remember if I did
    $endgroup$
    – user574848
    Nov 19 '18 at 6:15










  • $begingroup$
    How many main 'routine' methods to solve number theory problems would you say there are? At higher levels do you think each question requires some different insight.
    $endgroup$
    – 3684
    Nov 19 '18 at 6:19














9












9








9





$begingroup$

Let $n=100a+10b+c,$ where $a> 0$ and $b,cgeq 0$. We are trying to solve $$100a+10b+c=a+b+c+ab+ac+bc+abc \ implies 99a+9b=abc+ab+ac+bc \ implies a(99-b-c-bc)=b(c-9) \$$$c-9leq 0$, but $b+c+bcleq 99$. So the above equation holds iff $b=c=9$, which means $a$ can take any value.






share|cite|improve this answer









$endgroup$



Let $n=100a+10b+c,$ where $a> 0$ and $b,cgeq 0$. We are trying to solve $$100a+10b+c=a+b+c+ab+ac+bc+abc \ implies 99a+9b=abc+ab+ac+bc \ implies a(99-b-c-bc)=b(c-9) \$$$c-9leq 0$, but $b+c+bcleq 99$. So the above equation holds iff $b=c=9$, which means $a$ can take any value.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 19 '18 at 6:07









user574848user574848

40417




40417












  • $begingroup$
    May I ask how you got to the solution so quick, do you just see the solution?
    $endgroup$
    – 3684
    Nov 19 '18 at 6:13










  • $begingroup$
    @3684 this problem is fairly 'routine'. Plus, I might've seen this before/its associated solution, but I wouldn't remember if I did
    $endgroup$
    – user574848
    Nov 19 '18 at 6:15










  • $begingroup$
    How many main 'routine' methods to solve number theory problems would you say there are? At higher levels do you think each question requires some different insight.
    $endgroup$
    – 3684
    Nov 19 '18 at 6:19


















  • $begingroup$
    May I ask how you got to the solution so quick, do you just see the solution?
    $endgroup$
    – 3684
    Nov 19 '18 at 6:13










  • $begingroup$
    @3684 this problem is fairly 'routine'. Plus, I might've seen this before/its associated solution, but I wouldn't remember if I did
    $endgroup$
    – user574848
    Nov 19 '18 at 6:15










  • $begingroup$
    How many main 'routine' methods to solve number theory problems would you say there are? At higher levels do you think each question requires some different insight.
    $endgroup$
    – 3684
    Nov 19 '18 at 6:19
















$begingroup$
May I ask how you got to the solution so quick, do you just see the solution?
$endgroup$
– 3684
Nov 19 '18 at 6:13




$begingroup$
May I ask how you got to the solution so quick, do you just see the solution?
$endgroup$
– 3684
Nov 19 '18 at 6:13












$begingroup$
@3684 this problem is fairly 'routine'. Plus, I might've seen this before/its associated solution, but I wouldn't remember if I did
$endgroup$
– user574848
Nov 19 '18 at 6:15




$begingroup$
@3684 this problem is fairly 'routine'. Plus, I might've seen this before/its associated solution, but I wouldn't remember if I did
$endgroup$
– user574848
Nov 19 '18 at 6:15












$begingroup$
How many main 'routine' methods to solve number theory problems would you say there are? At higher levels do you think each question requires some different insight.
$endgroup$
– 3684
Nov 19 '18 at 6:19




$begingroup$
How many main 'routine' methods to solve number theory problems would you say there are? At higher levels do you think each question requires some different insight.
$endgroup$
– 3684
Nov 19 '18 at 6:19











8












$begingroup$

Here's part $(b)$ because I'm assuming you don't need help with part $(a)$:




We want to compute all possible integers $n$ such that $frac{n}{f(n)} = 1$. Since we know that, by assumption, $n$ is a three-digit number, we can write



$$n = 100a + 10b + c,$$



where $a, b, c$ are integers. If this is the case, in terms of our newly defined variables $a$, $b$, and $c$, we can express $f(n)$ as follows:



$$f(n) = abc + ab + bc + ac + a + b + c.$$



Now, in order to have $frac{n}{f(n)} = 1,$ we must have $n = f(n)$. This happens when



$$99a + 9b = abc + ab + bc + ac$$



$$Longleftrightarrow (9-c)b = a(bc + b + c - 99) $$



Also, we must have $b, c leq 9,$ which implies $bc + b + c - 99 leq 0$. However, since $a neq 0$ (if $a = 0$, we would be able to form a two or one-digit number instead of a three-digit one!), we conclude $b = c = 9$. Therefore, our solution set is given by



$$boxed{{199, 299, 399, 499, 599, 699, 799, 899, 999}}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for your answer, sorry I could only accept one answer. May I ask how you see the solution so quick?
    $endgroup$
    – 3684
    Nov 19 '18 at 6:15
















8












$begingroup$

Here's part $(b)$ because I'm assuming you don't need help with part $(a)$:




We want to compute all possible integers $n$ such that $frac{n}{f(n)} = 1$. Since we know that, by assumption, $n$ is a three-digit number, we can write



$$n = 100a + 10b + c,$$



where $a, b, c$ are integers. If this is the case, in terms of our newly defined variables $a$, $b$, and $c$, we can express $f(n)$ as follows:



$$f(n) = abc + ab + bc + ac + a + b + c.$$



Now, in order to have $frac{n}{f(n)} = 1,$ we must have $n = f(n)$. This happens when



$$99a + 9b = abc + ab + bc + ac$$



$$Longleftrightarrow (9-c)b = a(bc + b + c - 99) $$



Also, we must have $b, c leq 9,$ which implies $bc + b + c - 99 leq 0$. However, since $a neq 0$ (if $a = 0$, we would be able to form a two or one-digit number instead of a three-digit one!), we conclude $b = c = 9$. Therefore, our solution set is given by



$$boxed{{199, 299, 399, 499, 599, 699, 799, 899, 999}}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for your answer, sorry I could only accept one answer. May I ask how you see the solution so quick?
    $endgroup$
    – 3684
    Nov 19 '18 at 6:15














8












8








8





$begingroup$

Here's part $(b)$ because I'm assuming you don't need help with part $(a)$:




We want to compute all possible integers $n$ such that $frac{n}{f(n)} = 1$. Since we know that, by assumption, $n$ is a three-digit number, we can write



$$n = 100a + 10b + c,$$



where $a, b, c$ are integers. If this is the case, in terms of our newly defined variables $a$, $b$, and $c$, we can express $f(n)$ as follows:



$$f(n) = abc + ab + bc + ac + a + b + c.$$



Now, in order to have $frac{n}{f(n)} = 1,$ we must have $n = f(n)$. This happens when



$$99a + 9b = abc + ab + bc + ac$$



$$Longleftrightarrow (9-c)b = a(bc + b + c - 99) $$



Also, we must have $b, c leq 9,$ which implies $bc + b + c - 99 leq 0$. However, since $a neq 0$ (if $a = 0$, we would be able to form a two or one-digit number instead of a three-digit one!), we conclude $b = c = 9$. Therefore, our solution set is given by



$$boxed{{199, 299, 399, 499, 599, 699, 799, 899, 999}}$$






share|cite|improve this answer









$endgroup$



Here's part $(b)$ because I'm assuming you don't need help with part $(a)$:




We want to compute all possible integers $n$ such that $frac{n}{f(n)} = 1$. Since we know that, by assumption, $n$ is a three-digit number, we can write



$$n = 100a + 10b + c,$$



where $a, b, c$ are integers. If this is the case, in terms of our newly defined variables $a$, $b$, and $c$, we can express $f(n)$ as follows:



$$f(n) = abc + ab + bc + ac + a + b + c.$$



Now, in order to have $frac{n}{f(n)} = 1,$ we must have $n = f(n)$. This happens when



$$99a + 9b = abc + ab + bc + ac$$



$$Longleftrightarrow (9-c)b = a(bc + b + c - 99) $$



Also, we must have $b, c leq 9,$ which implies $bc + b + c - 99 leq 0$. However, since $a neq 0$ (if $a = 0$, we would be able to form a two or one-digit number instead of a three-digit one!), we conclude $b = c = 9$. Therefore, our solution set is given by



$$boxed{{199, 299, 399, 499, 599, 699, 799, 899, 999}}$$







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answered Nov 19 '18 at 6:08









Ekesh KumarEkesh Kumar

1,01228




1,01228












  • $begingroup$
    Thank you for your answer, sorry I could only accept one answer. May I ask how you see the solution so quick?
    $endgroup$
    – 3684
    Nov 19 '18 at 6:15


















  • $begingroup$
    Thank you for your answer, sorry I could only accept one answer. May I ask how you see the solution so quick?
    $endgroup$
    – 3684
    Nov 19 '18 at 6:15
















$begingroup$
Thank you for your answer, sorry I could only accept one answer. May I ask how you see the solution so quick?
$endgroup$
– 3684
Nov 19 '18 at 6:15




$begingroup$
Thank you for your answer, sorry I could only accept one answer. May I ask how you see the solution so quick?
$endgroup$
– 3684
Nov 19 '18 at 6:15


















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