Find $5$ numbers where the sum of all pairs gives the results $110, 112, 113, 114, 115, 116, 117, 118, 120,...
$begingroup$
Find $5$ numbers where the sum of the pairs gives the results $110, 112, 113, 114, 115, 116, 117, 118, 120, 121$
I have been trying on this question for some time and it seems easy at first glance, but later on, I was unable to start on the question. I tried turning these into equations with $a_1, a_2, a_3, a_4, text{ and }a_5$ but couldn't as it is not determined which $2$ numbers give which sum. I was, however, able to achieve that $a_1+a_2+a_3+a_4+a_5=289$ (which is obvious). Now is there anyway I could continue without guessing. Thank you.
elementary-number-theory
asked Dec 8 '18 at 11:40
user587054user587054
47511
$endgroup$
add a comment |
$begingroup$
Find $5$ numbers where the sum of the pairs gives the results $110, 112, 113, 114, 115, 116, 117, 118, 120, 121$
I have been trying on this question for some time and it seems easy at first glance, but later on, I was unable to start on the question. I tried turning these into equations with $a_1, a_2, a_3, a_4, text{ and }a_5$ but couldn't as it is not determined which $2$ numbers give which sum. I was, however, able to achieve that $a_1+a_2+a_3+a_4+a_5=289$ (which is obvious). Now is there anyway I could continue without guessing. Thank you.
elementary-number-theory
asked Dec 8 '18 at 11:40
user587054user587054
47511
$endgroup$
add a comment |
$begingroup$
Find $5$ numbers where the sum of the pairs gives the results $110, 112, 113, 114, 115, 116, 117, 118, 120, 121$
I have been trying on this question for some time and it seems easy at first glance, but later on, I was unable to start on the question. I tried turning these into equations with $a_1, a_2, a_3, a_4, text{ and }a_5$ but couldn't as it is not determined which $2$ numbers give which sum. I was, however, able to achieve that $a_1+a_2+a_3+a_4+a_5=289$ (which is obvious). Now is there anyway I could continue without guessing. Thank you.
elementary-number-theory
asked Dec 8 '18 at 11:40
user587054user587054
47511
$endgroup$
Find $5$ numbers where the sum of the pairs gives the results $110, 112, 113, 114, 115, 116, 117, 118, 120, 121$
I have been trying on this question for some time and it seems easy at first glance, but later on, I was unable to start on the question. I tried turning these into equations with $a_1, a_2, a_3, a_4, text{ and }a_5$ but couldn't as it is not determined which $2$ numbers give which sum. I was, however, able to achieve that $a_1+a_2+a_3+a_4+a_5=289$ (which is obvious). Now is there anyway I could continue without guessing. Thank you.
elementary-number-theory
elementary-number-theory
asked Dec 8 '18 at 11:40
user587054user587054
47511
asked Dec 8 '18 at 11:40
user587054user587054
47511
asked Dec 8 '18 at 11:40
user587054user587054
47511
asked Dec 8 '18 at 11:40
user587054user587054
47511
asked Dec 8 '18 at 11:40
user587054user587054
47511
47511
add a comment |
add a comment |
1 Answer
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$begingroup$
As there are no duplicates among the pair sums, the five numbers are distinct. So wlog. $a_1<a_2<a_3<a_4<a_5$. Then $110=a_1+a_2$, $112=a_1+a_3$, which makes $a_3=a_2+2$. Similarly, we find $a_4=a_3+1$ from the last two entries $120,121$. So far we have found that the numbers are
$$a_1, a_2, a_2+2, a_2+3, a_5.$$
This also tells us that $a_1+a_4=113$ and that $a_2+a_5=118$.
Now that the three pair sums $a_2+a_3=2a_2+2$, $a_2+a_4=2a_2+3$, $a_3+a_4=2a_2+5$ must occur among $114,115,116,117$. The only possible match is that $a_2+a_3=114$, i.e., $a_2=56$. The rest follows easily from here.
answered Dec 8 '18 at 11:56
Hagen von EitzenHagen von Eitzen
278k23269501
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
As there are no duplicates among the pair sums, the five numbers are distinct. So wlog. $a_1<a_2<a_3<a_4<a_5$. Then $110=a_1+a_2$, $112=a_1+a_3$, which makes $a_3=a_2+2$. Similarly, we find $a_4=a_3+1$ from the last two entries $120,121$. So far we have found that the numbers are
$$a_1, a_2, a_2+2, a_2+3, a_5.$$
This also tells us that $a_1+a_4=113$ and that $a_2+a_5=118$.
Now that the three pair sums $a_2+a_3=2a_2+2$, $a_2+a_4=2a_2+3$, $a_3+a_4=2a_2+5$ must occur among $114,115,116,117$. The only possible match is that $a_2+a_3=114$, i.e., $a_2=56$. The rest follows easily from here.
answered Dec 8 '18 at 11:56
Hagen von EitzenHagen von Eitzen
278k23269501
$endgroup$
add a comment |
$begingroup$
As there are no duplicates among the pair sums, the five numbers are distinct. So wlog. $a_1<a_2<a_3<a_4<a_5$. Then $110=a_1+a_2$, $112=a_1+a_3$, which makes $a_3=a_2+2$. Similarly, we find $a_4=a_3+1$ from the last two entries $120,121$. So far we have found that the numbers are
$$a_1, a_2, a_2+2, a_2+3, a_5.$$
This also tells us that $a_1+a_4=113$ and that $a_2+a_5=118$.
Now that the three pair sums $a_2+a_3=2a_2+2$, $a_2+a_4=2a_2+3$, $a_3+a_4=2a_2+5$ must occur among $114,115,116,117$. The only possible match is that $a_2+a_3=114$, i.e., $a_2=56$. The rest follows easily from here.
answered Dec 8 '18 at 11:56
Hagen von EitzenHagen von Eitzen
278k23269501
$endgroup$
add a comment |
$begingroup$
As there are no duplicates among the pair sums, the five numbers are distinct. So wlog. $a_1<a_2<a_3<a_4<a_5$. Then $110=a_1+a_2$, $112=a_1+a_3$, which makes $a_3=a_2+2$. Similarly, we find $a_4=a_3+1$ from the last two entries $120,121$. So far we have found that the numbers are
$$a_1, a_2, a_2+2, a_2+3, a_5.$$
This also tells us that $a_1+a_4=113$ and that $a_2+a_5=118$.
Now that the three pair sums $a_2+a_3=2a_2+2$, $a_2+a_4=2a_2+3$, $a_3+a_4=2a_2+5$ must occur among $114,115,116,117$. The only possible match is that $a_2+a_3=114$, i.e., $a_2=56$. The rest follows easily from here.
answered Dec 8 '18 at 11:56
Hagen von EitzenHagen von Eitzen
278k23269501
$endgroup$
As there are no duplicates among the pair sums, the five numbers are distinct. So wlog. $a_1<a_2<a_3<a_4<a_5$. Then $110=a_1+a_2$, $112=a_1+a_3$, which makes $a_3=a_2+2$. Similarly, we find $a_4=a_3+1$ from the last two entries $120,121$. So far we have found that the numbers are
$$a_1, a_2, a_2+2, a_2+3, a_5.$$
This also tells us that $a_1+a_4=113$ and that $a_2+a_5=118$.
Now that the three pair sums $a_2+a_3=2a_2+2$, $a_2+a_4=2a_2+3$, $a_3+a_4=2a_2+5$ must occur among $114,115,116,117$. The only possible match is that $a_2+a_3=114$, i.e., $a_2=56$. The rest follows easily from here.
answered Dec 8 '18 at 11:56
Hagen von EitzenHagen von Eitzen
278k23269501
answered Dec 8 '18 at 11:56
Hagen von EitzenHagen von Eitzen
278k23269501
answered Dec 8 '18 at 11:56
Hagen von EitzenHagen von Eitzen
278k23269501
answered Dec 8 '18 at 11:56
Hagen von EitzenHagen von Eitzen
278k23269501
278k23269501
add a comment |
add a comment |
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