Mixing
$2017-(a+b+c+d)=n$ where $n=abcd_{(10)}$ (Local Math Olympiad Qualification Exam Question)
$begingroup$
The problem text states: "'Person A' was born in 1962, in 1980 he turned 18 which is the sum of the numbers in his birth year (1+9+6+2). 'Person B' has a younger brother 'Person C', in 2017 the age of each one of them was equal to the sum of the digits in their birth year, when was 'Person B' born?".
How can I approach this problem? I enumerated the cases using a script and found out that the possible numbers n that satisfy the requirements are 2012 and 1994.
contest-math
asked Jan 11 at 21:40
Elhamer YacineElhamer Yacine
102
$endgroup$
add a comment |
$begingroup$
The problem text states: "'Person A' was born in 1962, in 1980 he turned 18 which is the sum of the numbers in his birth year (1+9+6+2). 'Person B' has a younger brother 'Person C', in 2017 the age of each one of them was equal to the sum of the digits in their birth year, when was 'Person B' born?".
How can I approach this problem? I enumerated the cases using a script and found out that the possible numbers n that satisfy the requirements are 2012 and 1994.
contest-math
asked Jan 11 at 21:40
Elhamer YacineElhamer Yacine
102
$endgroup$
$begingroup$
It seems you already have your answer. You could have done the enumeration manually, it would not have taken long.
$endgroup$
– Daniel Mathias
Jan 11 at 21:45
$begingroup$
@DanielMathias Yeah, but I assume there's a more solid solution/method/proof/explanation for it
$endgroup$
– Elhamer Yacine
Jan 11 at 21:48
$begingroup$
It is possible that you were born in October of 2007 and in Februrary of 2017 you would be 9 years old.
$endgroup$
– Doug M
Jan 11 at 21:49
$begingroup$
@DougM lol, that's a clever way to think of it, but I don't think that's what they meant. In that case I think you're kinda going to start counting from 2008 to 2017 or something along those lines
$endgroup$
– Elhamer Yacine
Jan 11 at 21:54
add a comment |
$begingroup$
The problem text states: "'Person A' was born in 1962, in 1980 he turned 18 which is the sum of the numbers in his birth year (1+9+6+2). 'Person B' has a younger brother 'Person C', in 2017 the age of each one of them was equal to the sum of the digits in their birth year, when was 'Person B' born?".
How can I approach this problem? I enumerated the cases using a script and found out that the possible numbers n that satisfy the requirements are 2012 and 1994.
contest-math
asked Jan 11 at 21:40
Elhamer YacineElhamer Yacine
102
$endgroup$
The problem text states: "'Person A' was born in 1962, in 1980 he turned 18 which is the sum of the numbers in his birth year (1+9+6+2). 'Person B' has a younger brother 'Person C', in 2017 the age of each one of them was equal to the sum of the digits in their birth year, when was 'Person B' born?".
How can I approach this problem? I enumerated the cases using a script and found out that the possible numbers n that satisfy the requirements are 2012 and 1994.
contest-math
contest-math
asked Jan 11 at 21:40
Elhamer YacineElhamer Yacine
102
asked Jan 11 at 21:40
Elhamer YacineElhamer Yacine
102
asked Jan 11 at 21:40
Elhamer YacineElhamer Yacine
102
asked Jan 11 at 21:40
Elhamer YacineElhamer Yacine
102
asked Jan 11 at 21:40
Elhamer YacineElhamer Yacine
102
102
$begingroup$
It seems you already have your answer. You could have done the enumeration manually, it would not have taken long.
$endgroup$
– Daniel Mathias
Jan 11 at 21:45
$begingroup$
@DanielMathias Yeah, but I assume there's a more solid solution/method/proof/explanation for it
$endgroup$
– Elhamer Yacine
Jan 11 at 21:48
$begingroup$
It is possible that you were born in October of 2007 and in Februrary of 2017 you would be 9 years old.
$endgroup$
– Doug M
Jan 11 at 21:49
$begingroup$
@DougM lol, that's a clever way to think of it, but I don't think that's what they meant. In that case I think you're kinda going to start counting from 2008 to 2017 or something along those lines
$endgroup$
– Elhamer Yacine
Jan 11 at 21:54
add a comment |
$begingroup$
It seems you already have your answer. You could have done the enumeration manually, it would not have taken long.
$endgroup$
– Daniel Mathias
Jan 11 at 21:45
$begingroup$
@DanielMathias Yeah, but I assume there's a more solid solution/method/proof/explanation for it
$endgroup$
– Elhamer Yacine
Jan 11 at 21:48
$begingroup$
It is possible that you were born in October of 2007 and in Februrary of 2017 you would be 9 years old.
$endgroup$
– Doug M
Jan 11 at 21:49
$begingroup$
@DougM lol, that's a clever way to think of it, but I don't think that's what they meant. In that case I think you're kinda going to start counting from 2008 to 2017 or something along those lines
$endgroup$
– Elhamer Yacine
Jan 11 at 21:54
$begingroup$
It seems you already have your answer. You could have done the enumeration manually, it would not have taken long.
$endgroup$
– Daniel Mathias
Jan 11 at 21:45
$begingroup$
It seems you already have your answer. You could have done the enumeration manually, it would not have taken long.
$endgroup$
– Daniel Mathias
Jan 11 at 21:45
$begingroup$
@DanielMathias Yeah, but I assume there's a more solid solution/method/proof/explanation for it
$endgroup$
– Elhamer Yacine
Jan 11 at 21:48
$begingroup$
@DanielMathias Yeah, but I assume there's a more solid solution/method/proof/explanation for it
$endgroup$
– Elhamer Yacine
Jan 11 at 21:48
$begingroup$
It is possible that you were born in October of 2007 and in Februrary of 2017 you would be 9 years old.
$endgroup$
– Doug M
Jan 11 at 21:49
$begingroup$
It is possible that you were born in October of 2007 and in Februrary of 2017 you would be 9 years old.
$endgroup$
– Doug M
Jan 11 at 21:49
$begingroup$
@DougM lol, that's a clever way to think of it, but I don't think that's what they meant. In that case I think you're kinda going to start counting from 2008 to 2017 or something along those lines
$endgroup$
– Elhamer Yacine
Jan 11 at 21:54
$begingroup$
@DougM lol, that's a clever way to think of it, but I don't think that's what they meant. In that case I think you're kinda going to start counting from 2008 to 2017 or something along those lines
$endgroup$
– Elhamer Yacine
Jan 11 at 21:54
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The possible cases for $a$ are $a=1$ and $a=2$.
Case $a=2$.
We have $2017-2-b-c-d=2000+100b+10c+d$, that is,
$$
15=101b+11c+2d
$$
Therefore $b=0$. Note that $c$ must be odd, so $c=1$ and $d=2$: the year is $2012$.
Case $a=1$
We have $2017-1-b-c-d=1000+100b+10c+d$, that is,
$$
1016=101b+11c+2d
$$
Note that $11c+2dle 99+18=117$, so we need $101bge899$, that is, $bge9$. Hence $b=9$ and $11c+2d=107$. Since $2dle18$, we have $11cge89$, so $cge9$, hence $c=9$ and $d=4$. The year is 1994.
Comment
The younger brother is needed, or person B could have been born either in 1994 or 2012.
answered Jan 11 at 22:05
egregegreg
181k1485203
$endgroup$
add a comment |
$begingroup$
If you want do to this without a script.
Born in 2010 - 2017.
$7-d = 3+d\
d = 5$
Born in 2000 - 2009.
$17 - d = 2+d\
2d = 15\
d = 7frac 12$
If you are looking for integer solutions, none exist. But, as you don't turn your age until your birthday comes, born in $2007$ can work.
Born in 1990 - 1999.
$27-d = 19+d\
d = 4$
Born in 1980 - 1989.
$37-d = 18+d\
d = 9frac 12$
$1989$ is works depending on your rules.
Your age must be less than $1+9+9+9 = 28$
We could perhaps generalize it a little more and say
$17 - 10c - d = 2 + c + d\
9c + 2d = 15$
or
$117 - 10c - d = 10 + c + d\
9c + 2d = 107$
If we want $c, d$ to be in ${0,1,2,3,4,5,6,7,8,9}, c $ must be odd and sufficiently small in one equation and large in the other.
answered Jan 11 at 22:22
Doug MDoug M
44.9k31854
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The possible cases for $a$ are $a=1$ and $a=2$.
Case $a=2$.
We have $2017-2-b-c-d=2000+100b+10c+d$, that is,
$$
15=101b+11c+2d
$$
Therefore $b=0$. Note that $c$ must be odd, so $c=1$ and $d=2$: the year is $2012$.
Case $a=1$
We have $2017-1-b-c-d=1000+100b+10c+d$, that is,
$$
1016=101b+11c+2d
$$
Note that $11c+2dle 99+18=117$, so we need $101bge899$, that is, $bge9$. Hence $b=9$ and $11c+2d=107$. Since $2dle18$, we have $11cge89$, so $cge9$, hence $c=9$ and $d=4$. The year is 1994.
Comment
The younger brother is needed, or person B could have been born either in 1994 or 2012.
answered Jan 11 at 22:05
egregegreg
181k1485203
$endgroup$
add a comment |
$begingroup$
The possible cases for $a$ are $a=1$ and $a=2$.
Case $a=2$.
We have $2017-2-b-c-d=2000+100b+10c+d$, that is,
$$
15=101b+11c+2d
$$
Therefore $b=0$. Note that $c$ must be odd, so $c=1$ and $d=2$: the year is $2012$.
Case $a=1$
We have $2017-1-b-c-d=1000+100b+10c+d$, that is,
$$
1016=101b+11c+2d
$$
Note that $11c+2dle 99+18=117$, so we need $101bge899$, that is, $bge9$. Hence $b=9$ and $11c+2d=107$. Since $2dle18$, we have $11cge89$, so $cge9$, hence $c=9$ and $d=4$. The year is 1994.
Comment
The younger brother is needed, or person B could have been born either in 1994 or 2012.
answered Jan 11 at 22:05
egregegreg
181k1485203
$endgroup$
add a comment |
$begingroup$
The possible cases for $a$ are $a=1$ and $a=2$.
Case $a=2$.
We have $2017-2-b-c-d=2000+100b+10c+d$, that is,
$$
15=101b+11c+2d
$$
Therefore $b=0$. Note that $c$ must be odd, so $c=1$ and $d=2$: the year is $2012$.
Case $a=1$
We have $2017-1-b-c-d=1000+100b+10c+d$, that is,
$$
1016=101b+11c+2d
$$
Note that $11c+2dle 99+18=117$, so we need $101bge899$, that is, $bge9$. Hence $b=9$ and $11c+2d=107$. Since $2dle18$, we have $11cge89$, so $cge9$, hence $c=9$ and $d=4$. The year is 1994.
Comment
The younger brother is needed, or person B could have been born either in 1994 or 2012.
answered Jan 11 at 22:05
egregegreg
181k1485203
$endgroup$
The possible cases for $a$ are $a=1$ and $a=2$.
Case $a=2$.
We have $2017-2-b-c-d=2000+100b+10c+d$, that is,
$$
15=101b+11c+2d
$$
Therefore $b=0$. Note that $c$ must be odd, so $c=1$ and $d=2$: the year is $2012$.
Case $a=1$
We have $2017-1-b-c-d=1000+100b+10c+d$, that is,
$$
1016=101b+11c+2d
$$
Note that $11c+2dle 99+18=117$, so we need $101bge899$, that is, $bge9$. Hence $b=9$ and $11c+2d=107$. Since $2dle18$, we have $11cge89$, so $cge9$, hence $c=9$ and $d=4$. The year is 1994.
Comment
The younger brother is needed, or person B could have been born either in 1994 or 2012.
answered Jan 11 at 22:05
egregegreg
181k1485203
answered Jan 11 at 22:05
egregegreg
181k1485203
answered Jan 11 at 22:05
egregegreg
181k1485203
answered Jan 11 at 22:05
egregegreg
181k1485203
181k1485203
add a comment |
add a comment |
$begingroup$
If you want do to this without a script.
Born in 2010 - 2017.
$7-d = 3+d\
d = 5$
Born in 2000 - 2009.
$17 - d = 2+d\
2d = 15\
d = 7frac 12$
If you are looking for integer solutions, none exist. But, as you don't turn your age until your birthday comes, born in $2007$ can work.
Born in 1990 - 1999.
$27-d = 19+d\
d = 4$
Born in 1980 - 1989.
$37-d = 18+d\
d = 9frac 12$
$1989$ is works depending on your rules.
Your age must be less than $1+9+9+9 = 28$
We could perhaps generalize it a little more and say
$17 - 10c - d = 2 + c + d\
9c + 2d = 15$
or
$117 - 10c - d = 10 + c + d\
9c + 2d = 107$
If we want $c, d$ to be in ${0,1,2,3,4,5,6,7,8,9}, c $ must be odd and sufficiently small in one equation and large in the other.
answered Jan 11 at 22:22
Doug MDoug M
44.9k31854
$endgroup$
add a comment |
$begingroup$
If you want do to this without a script.
Born in 2010 - 2017.
$7-d = 3+d\
d = 5$
Born in 2000 - 2009.
$17 - d = 2+d\
2d = 15\
d = 7frac 12$
If you are looking for integer solutions, none exist. But, as you don't turn your age until your birthday comes, born in $2007$ can work.
Born in 1990 - 1999.
$27-d = 19+d\
d = 4$
Born in 1980 - 1989.
$37-d = 18+d\
d = 9frac 12$
$1989$ is works depending on your rules.
Your age must be less than $1+9+9+9 = 28$
We could perhaps generalize it a little more and say
$17 - 10c - d = 2 + c + d\
9c + 2d = 15$
or
$117 - 10c - d = 10 + c + d\
9c + 2d = 107$
If we want $c, d$ to be in ${0,1,2,3,4,5,6,7,8,9}, c $ must be odd and sufficiently small in one equation and large in the other.
answered Jan 11 at 22:22
Doug MDoug M
44.9k31854
$endgroup$
add a comment |
$begingroup$
If you want do to this without a script.
Born in 2010 - 2017.
$7-d = 3+d\
d = 5$
Born in 2000 - 2009.
$17 - d = 2+d\
2d = 15\
d = 7frac 12$
If you are looking for integer solutions, none exist. But, as you don't turn your age until your birthday comes, born in $2007$ can work.
Born in 1990 - 1999.
$27-d = 19+d\
d = 4$
Born in 1980 - 1989.
$37-d = 18+d\
d = 9frac 12$
$1989$ is works depending on your rules.
Your age must be less than $1+9+9+9 = 28$
We could perhaps generalize it a little more and say
$17 - 10c - d = 2 + c + d\
9c + 2d = 15$
or
$117 - 10c - d = 10 + c + d\
9c + 2d = 107$
If we want $c, d$ to be in ${0,1,2,3,4,5,6,7,8,9}, c $ must be odd and sufficiently small in one equation and large in the other.
answered Jan 11 at 22:22
Doug MDoug M
44.9k31854
$endgroup$
If you want do to this without a script.
Born in 2010 - 2017.
$7-d = 3+d\
d = 5$
Born in 2000 - 2009.
$17 - d = 2+d\
2d = 15\
d = 7frac 12$
If you are looking for integer solutions, none exist. But, as you don't turn your age until your birthday comes, born in $2007$ can work.
Born in 1990 - 1999.
$27-d = 19+d\
d = 4$
Born in 1980 - 1989.
$37-d = 18+d\
d = 9frac 12$
$1989$ is works depending on your rules.
Your age must be less than $1+9+9+9 = 28$
We could perhaps generalize it a little more and say
$17 - 10c - d = 2 + c + d\
9c + 2d = 15$
or
$117 - 10c - d = 10 + c + d\
9c + 2d = 107$
If we want $c, d$ to be in ${0,1,2,3,4,5,6,7,8,9}, c $ must be odd and sufficiently small in one equation and large in the other.
answered Jan 11 at 22:22
Doug MDoug M
44.9k31854
answered Jan 11 at 22:22
Doug MDoug M
44.9k31854
answered Jan 11 at 22:22
Doug MDoug M
44.9k31854
answered Jan 11 at 22:22
Doug MDoug M
44.9k31854
44.9k31854
add a comment |
add a comment |
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$begingroup$
It seems you already have your answer. You could have done the enumeration manually, it would not have taken long.
$endgroup$
– Daniel Mathias
Jan 11 at 21:45
$begingroup$
@DanielMathias Yeah, but I assume there's a more solid solution/method/proof/explanation for it
$endgroup$
– Elhamer Yacine
Jan 11 at 21:48
$begingroup$
It is possible that you were born in October of 2007 and in Februrary of 2017 you would be 9 years old.
$endgroup$
– Doug M
Jan 11 at 21:49
$begingroup$
@DougM lol, that's a clever way to think of it, but I don't think that's what they meant. In that case I think you're kinda going to start counting from 2008 to 2017 or something along those lines
$endgroup$
– Elhamer Yacine
Jan 11 at 21:54