Reverse addition
$begingroup$
I have a number 46 which is the result of addition of 12+17+17.
Is there a way, given the result 46 and 12 , 17 as the numbers used to get the result, to find out in what combination 12 and 17 was used to get the result 46
arithmetic
asked 3 hours ago
user10183910user10183910
111
New contributor
user10183910 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I have a number 46 which is the result of addition of 12+17+17.
Is there a way, given the result 46 and 12 , 17 as the numbers used to get the result, to find out in what combination 12 and 17 was used to get the result 46
arithmetic
asked 3 hours ago
user10183910user10183910
111
New contributor
user10183910 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
You can relatively easy determine all solutions over the integers. If there is only one solution over the positive integers, you can recover the representation.
$endgroup$
– Peter
3 hours ago
$begingroup$
Not sure this is clear. For instance, If you are told you get $5$ using $2's$ and $1's$ you could have $5=2+1+1+1$ or $5=2+2+1$ , for example.
$endgroup$
– lulu
3 hours ago
$begingroup$
suppose the number is 169 and 12s and 17s are the numbers used.. The answer I need is its the addition of 5times17 and 7times 12.. Is there a formula to deduce this
$endgroup$
– user10183910
3 hours ago
add a comment |
$begingroup$
I have a number 46 which is the result of addition of 12+17+17.
Is there a way, given the result 46 and 12 , 17 as the numbers used to get the result, to find out in what combination 12 and 17 was used to get the result 46
arithmetic
asked 3 hours ago
user10183910user10183910
111
New contributor
user10183910 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I have a number 46 which is the result of addition of 12+17+17.
Is there a way, given the result 46 and 12 , 17 as the numbers used to get the result, to find out in what combination 12 and 17 was used to get the result 46
arithmetic
arithmetic
asked 3 hours ago
user10183910user10183910
111
New contributor
user10183910 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
user10183910user10183910
111
New contributor
user10183910 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
user10183910user10183910
111
New contributor
user10183910 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
user10183910user10183910
111
asked 3 hours ago
user10183910user10183910
111
111
New contributor
user10183910 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user10183910 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
user10183910 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
You can relatively easy determine all solutions over the integers. If there is only one solution over the positive integers, you can recover the representation.
$endgroup$
– Peter
3 hours ago
$begingroup$
Not sure this is clear. For instance, If you are told you get $5$ using $2's$ and $1's$ you could have $5=2+1+1+1$ or $5=2+2+1$ , for example.
$endgroup$
– lulu
3 hours ago
$begingroup$
suppose the number is 169 and 12s and 17s are the numbers used.. The answer I need is its the addition of 5times17 and 7times 12.. Is there a formula to deduce this
$endgroup$
– user10183910
3 hours ago
add a comment |
$begingroup$
You can relatively easy determine all solutions over the integers. If there is only one solution over the positive integers, you can recover the representation.
$endgroup$
– Peter
3 hours ago
$begingroup$
Not sure this is clear. For instance, If you are told you get $5$ using $2's$ and $1's$ you could have $5=2+1+1+1$ or $5=2+2+1$ , for example.
$endgroup$
– lulu
3 hours ago
$begingroup$
suppose the number is 169 and 12s and 17s are the numbers used.. The answer I need is its the addition of 5times17 and 7times 12.. Is there a formula to deduce this
$endgroup$
– user10183910
3 hours ago
$begingroup$
You can relatively easy determine all solutions over the integers. If there is only one solution over the positive integers, you can recover the representation.
$endgroup$
– Peter
3 hours ago
$begingroup$
You can relatively easy determine all solutions over the integers. If there is only one solution over the positive integers, you can recover the representation.
$endgroup$
– Peter
3 hours ago
$begingroup$
Not sure this is clear. For instance, If you are told you get $5$ using $2's$ and $1's$ you could have $5=2+1+1+1$ or $5=2+2+1$ , for example.
$endgroup$
– lulu
3 hours ago
$begingroup$
Not sure this is clear. For instance, If you are told you get $5$ using $2's$ and $1's$ you could have $5=2+1+1+1$ or $5=2+2+1$ , for example.
$endgroup$
– lulu
3 hours ago
$begingroup$
suppose the number is 169 and 12s and 17s are the numbers used.. The answer I need is its the addition of 5times17 and 7times 12.. Is there a formula to deduce this
$endgroup$
– user10183910
3 hours ago
$begingroup$
suppose the number is 169 and 12s and 17s are the numbers used.. The answer I need is its the addition of 5times17 and 7times 12.. Is there a formula to deduce this
$endgroup$
– user10183910
3 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You could try using modulo algebra:
$$n=alpha 12 + beta 17implies [n]_{17}=[12]_{17}[alpha]_{17} implies [alpha]_{17}=[n]_{17}([12]_{17})^{-1}=[n]_{17}[5]_{17}$$
You get $([12]_{17})^{-1}=[5]_{17}$ with the extended euclidean algorithm
This only works well because 17 is prime (or more specifically gcd(12,17)=1). So then you know $alpha=5n + mcdot 17$.
That is already an improvement over brute force trying all whole numbers for alpha. Since 5 is a whole number (and this generalizes) $5n>n>alpha$ so you know that m is negative and basically walk down the negative integers.
You can try playing with that approach a bit.
So pseudo algorithm:
extended euclidean algoritm yields: x,y, gcd(12,17) with 12x+12y=gcd(12,17)
if(gcd(12,17)=1)
loop m=0,1,2...
try alpha=x*n-m*17 (calculate beta - is int?)
answered 3 hours ago
Felix B.Felix B.
769214
$endgroup$
$begingroup$
Does this help, though? I can take the original, not reduce it mod $17$, and get $12 alpha=n-17beta$. With $n$ known, this is still $alpha$ as a function of another unknown $beta$. You have $alpha$ as a function of an unknown $m$, which is still in terms of one free variable.
$endgroup$
– Randall
2 hours ago
$begingroup$
Yes but how would you find $alpha$ and $beta$ that satisfy this equation? by starting with $alpha=1$ and incrementing and checking whether the solution for beta is an integer. With my formula you can increment over the alpha with larger steps (and also check whether beta is an integer). Or am I missing something?
$endgroup$
– Felix B.
2 hours ago
$begingroup$
I'm not sure that you're missing anything. It was an honest question.
$endgroup$
– Randall
2 hours ago
$begingroup$
And sometimes I do transformations and end up with something that just rephrases the original. Could have been my mistake too, so it is good if someone makes a sanity check.
$endgroup$
– Felix B.
2 hours ago
add a comment |
$begingroup$
If "the number is 169 and 12s and 17s are the numbers used." then you have the Diophantine equation 12x+ 17y= 169. 12 divides into 17 once with remainder 5: 17- 12= 5. 5 divides into 12 twice with remainder 2: 12- 2(5)= 2. Finally, 2 divides into 5 once with remainder 1: 5- 2(2)= 1. Replace that (2) by 12- 2(5): 5- 2(12- 2(5))= 5(5)- 2(12)= 1. Replace that (5) by 17- 12: 5(17- 12)- 2(12)= 5(17)- 7(12)= 1. Multiply by 169: 845(17)- 1183(12)= 169.
So one solution is x= -1183, y= 845. But adding any multiple of 17 to x and subtracting that same multiple of 12 from y gives another solution: 12(-1183+ 17k)+ 17(845- 12k)= -14196+ 12(17)k+ 14365- 17(12)k= 169. In particular, taking k= 70 gives positive values for both x and y: x= -1183+ 17(70)= 7 and y= 845- 12(70)= 5. 169= 7(12)+ 5(17).
answered 2 hours ago
user247327user247327
10.5k1515
$endgroup$
$begingroup$
got confused. i didnt get the replacing part
$endgroup$
– user10183910
2 hours ago
$begingroup$
why have to considered k=70?
$endgroup$
– user10183910
1 hour ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You could try using modulo algebra:
$$n=alpha 12 + beta 17implies [n]_{17}=[12]_{17}[alpha]_{17} implies [alpha]_{17}=[n]_{17}([12]_{17})^{-1}=[n]_{17}[5]_{17}$$
You get $([12]_{17})^{-1}=[5]_{17}$ with the extended euclidean algorithm
This only works well because 17 is prime (or more specifically gcd(12,17)=1). So then you know $alpha=5n + mcdot 17$.
That is already an improvement over brute force trying all whole numbers for alpha. Since 5 is a whole number (and this generalizes) $5n>n>alpha$ so you know that m is negative and basically walk down the negative integers.
You can try playing with that approach a bit.
So pseudo algorithm:
extended euclidean algoritm yields: x,y, gcd(12,17) with 12x+12y=gcd(12,17)
if(gcd(12,17)=1)
loop m=0,1,2...
try alpha=x*n-m*17 (calculate beta - is int?)
answered 3 hours ago
Felix B.Felix B.
769214
$endgroup$
$begingroup$
Does this help, though? I can take the original, not reduce it mod $17$, and get $12 alpha=n-17beta$. With $n$ known, this is still $alpha$ as a function of another unknown $beta$. You have $alpha$ as a function of an unknown $m$, which is still in terms of one free variable.
$endgroup$
– Randall
2 hours ago
$begingroup$
Yes but how would you find $alpha$ and $beta$ that satisfy this equation? by starting with $alpha=1$ and incrementing and checking whether the solution for beta is an integer. With my formula you can increment over the alpha with larger steps (and also check whether beta is an integer). Or am I missing something?
$endgroup$
– Felix B.
2 hours ago
$begingroup$
I'm not sure that you're missing anything. It was an honest question.
$endgroup$
– Randall
2 hours ago
$begingroup$
And sometimes I do transformations and end up with something that just rephrases the original. Could have been my mistake too, so it is good if someone makes a sanity check.
$endgroup$
– Felix B.
2 hours ago
add a comment |
$begingroup$
You could try using modulo algebra:
$$n=alpha 12 + beta 17implies [n]_{17}=[12]_{17}[alpha]_{17} implies [alpha]_{17}=[n]_{17}([12]_{17})^{-1}=[n]_{17}[5]_{17}$$
You get $([12]_{17})^{-1}=[5]_{17}$ with the extended euclidean algorithm
This only works well because 17 is prime (or more specifically gcd(12,17)=1). So then you know $alpha=5n + mcdot 17$.
That is already an improvement over brute force trying all whole numbers for alpha. Since 5 is a whole number (and this generalizes) $5n>n>alpha$ so you know that m is negative and basically walk down the negative integers.
You can try playing with that approach a bit.
So pseudo algorithm:
extended euclidean algoritm yields: x,y, gcd(12,17) with 12x+12y=gcd(12,17)
if(gcd(12,17)=1)
loop m=0,1,2...
try alpha=x*n-m*17 (calculate beta - is int?)
answered 3 hours ago
Felix B.Felix B.
769214
$endgroup$
$begingroup$
Does this help, though? I can take the original, not reduce it mod $17$, and get $12 alpha=n-17beta$. With $n$ known, this is still $alpha$ as a function of another unknown $beta$. You have $alpha$ as a function of an unknown $m$, which is still in terms of one free variable.
$endgroup$
– Randall
2 hours ago
$begingroup$
Yes but how would you find $alpha$ and $beta$ that satisfy this equation? by starting with $alpha=1$ and incrementing and checking whether the solution for beta is an integer. With my formula you can increment over the alpha with larger steps (and also check whether beta is an integer). Or am I missing something?
$endgroup$
– Felix B.
2 hours ago
$begingroup$
I'm not sure that you're missing anything. It was an honest question.
$endgroup$
– Randall
2 hours ago
$begingroup$
And sometimes I do transformations and end up with something that just rephrases the original. Could have been my mistake too, so it is good if someone makes a sanity check.
$endgroup$
– Felix B.
2 hours ago
add a comment |
$begingroup$
You could try using modulo algebra:
$$n=alpha 12 + beta 17implies [n]_{17}=[12]_{17}[alpha]_{17} implies [alpha]_{17}=[n]_{17}([12]_{17})^{-1}=[n]_{17}[5]_{17}$$
You get $([12]_{17})^{-1}=[5]_{17}$ with the extended euclidean algorithm
This only works well because 17 is prime (or more specifically gcd(12,17)=1). So then you know $alpha=5n + mcdot 17$.
That is already an improvement over brute force trying all whole numbers for alpha. Since 5 is a whole number (and this generalizes) $5n>n>alpha$ so you know that m is negative and basically walk down the negative integers.
You can try playing with that approach a bit.
So pseudo algorithm:
extended euclidean algoritm yields: x,y, gcd(12,17) with 12x+12y=gcd(12,17)
if(gcd(12,17)=1)
loop m=0,1,2...
try alpha=x*n-m*17 (calculate beta - is int?)
answered 3 hours ago
Felix B.Felix B.
769214
$endgroup$
You could try using modulo algebra:
$$n=alpha 12 + beta 17implies [n]_{17}=[12]_{17}[alpha]_{17} implies [alpha]_{17}=[n]_{17}([12]_{17})^{-1}=[n]_{17}[5]_{17}$$
You get $([12]_{17})^{-1}=[5]_{17}$ with the extended euclidean algorithm
This only works well because 17 is prime (or more specifically gcd(12,17)=1). So then you know $alpha=5n + mcdot 17$.
That is already an improvement over brute force trying all whole numbers for alpha. Since 5 is a whole number (and this generalizes) $5n>n>alpha$ so you know that m is negative and basically walk down the negative integers.
You can try playing with that approach a bit.
So pseudo algorithm:
extended euclidean algoritm yields: x,y, gcd(12,17) with 12x+12y=gcd(12,17)
if(gcd(12,17)=1)
loop m=0,1,2...
try alpha=x*n-m*17 (calculate beta - is int?)
answered 3 hours ago
Felix B.Felix B.
769214
edited 2 hours ago
answered 3 hours ago
Felix B.Felix B.
769214
answered 3 hours ago
Felix B.Felix B.
769214
answered 3 hours ago
Felix B.Felix B.
769214
769214
$begingroup$
Does this help, though? I can take the original, not reduce it mod $17$, and get $12 alpha=n-17beta$. With $n$ known, this is still $alpha$ as a function of another unknown $beta$. You have $alpha$ as a function of an unknown $m$, which is still in terms of one free variable.
$endgroup$
– Randall
2 hours ago
$begingroup$
Yes but how would you find $alpha$ and $beta$ that satisfy this equation? by starting with $alpha=1$ and incrementing and checking whether the solution for beta is an integer. With my formula you can increment over the alpha with larger steps (and also check whether beta is an integer). Or am I missing something?
$endgroup$
– Felix B.
2 hours ago
$begingroup$
I'm not sure that you're missing anything. It was an honest question.
$endgroup$
– Randall
2 hours ago
$begingroup$
And sometimes I do transformations and end up with something that just rephrases the original. Could have been my mistake too, so it is good if someone makes a sanity check.
$endgroup$
– Felix B.
2 hours ago
add a comment |
$begingroup$
Does this help, though? I can take the original, not reduce it mod $17$, and get $12 alpha=n-17beta$. With $n$ known, this is still $alpha$ as a function of another unknown $beta$. You have $alpha$ as a function of an unknown $m$, which is still in terms of one free variable.
$endgroup$
– Randall
2 hours ago
$begingroup$
Yes but how would you find $alpha$ and $beta$ that satisfy this equation? by starting with $alpha=1$ and incrementing and checking whether the solution for beta is an integer. With my formula you can increment over the alpha with larger steps (and also check whether beta is an integer). Or am I missing something?
$endgroup$
– Felix B.
2 hours ago
$begingroup$
I'm not sure that you're missing anything. It was an honest question.
$endgroup$
– Randall
2 hours ago
$begingroup$
And sometimes I do transformations and end up with something that just rephrases the original. Could have been my mistake too, so it is good if someone makes a sanity check.
$endgroup$
– Felix B.
2 hours ago
$begingroup$
Does this help, though? I can take the original, not reduce it mod $17$, and get $12 alpha=n-17beta$. With $n$ known, this is still $alpha$ as a function of another unknown $beta$. You have $alpha$ as a function of an unknown $m$, which is still in terms of one free variable.
$endgroup$
– Randall
2 hours ago
$begingroup$
Does this help, though? I can take the original, not reduce it mod $17$, and get $12 alpha=n-17beta$. With $n$ known, this is still $alpha$ as a function of another unknown $beta$. You have $alpha$ as a function of an unknown $m$, which is still in terms of one free variable.
$endgroup$
– Randall
2 hours ago
$begingroup$
Yes but how would you find $alpha$ and $beta$ that satisfy this equation? by starting with $alpha=1$ and incrementing and checking whether the solution for beta is an integer. With my formula you can increment over the alpha with larger steps (and also check whether beta is an integer). Or am I missing something?
$endgroup$
– Felix B.
2 hours ago
$begingroup$
Yes but how would you find $alpha$ and $beta$ that satisfy this equation? by starting with $alpha=1$ and incrementing and checking whether the solution for beta is an integer. With my formula you can increment over the alpha with larger steps (and also check whether beta is an integer). Or am I missing something?
$endgroup$
– Felix B.
2 hours ago
$begingroup$
I'm not sure that you're missing anything. It was an honest question.
$endgroup$
– Randall
2 hours ago
$begingroup$
I'm not sure that you're missing anything. It was an honest question.
$endgroup$
– Randall
2 hours ago
$begingroup$
And sometimes I do transformations and end up with something that just rephrases the original. Could have been my mistake too, so it is good if someone makes a sanity check.
$endgroup$
– Felix B.
2 hours ago
$begingroup$
And sometimes I do transformations and end up with something that just rephrases the original. Could have been my mistake too, so it is good if someone makes a sanity check.
$endgroup$
– Felix B.
2 hours ago
add a comment |
$begingroup$
If "the number is 169 and 12s and 17s are the numbers used." then you have the Diophantine equation 12x+ 17y= 169. 12 divides into 17 once with remainder 5: 17- 12= 5. 5 divides into 12 twice with remainder 2: 12- 2(5)= 2. Finally, 2 divides into 5 once with remainder 1: 5- 2(2)= 1. Replace that (2) by 12- 2(5): 5- 2(12- 2(5))= 5(5)- 2(12)= 1. Replace that (5) by 17- 12: 5(17- 12)- 2(12)= 5(17)- 7(12)= 1. Multiply by 169: 845(17)- 1183(12)= 169.
So one solution is x= -1183, y= 845. But adding any multiple of 17 to x and subtracting that same multiple of 12 from y gives another solution: 12(-1183+ 17k)+ 17(845- 12k)= -14196+ 12(17)k+ 14365- 17(12)k= 169. In particular, taking k= 70 gives positive values for both x and y: x= -1183+ 17(70)= 7 and y= 845- 12(70)= 5. 169= 7(12)+ 5(17).
answered 2 hours ago
user247327user247327
10.5k1515
$endgroup$
$begingroup$
got confused. i didnt get the replacing part
$endgroup$
– user10183910
2 hours ago
$begingroup$
why have to considered k=70?
$endgroup$
– user10183910
1 hour ago
add a comment |
$begingroup$
If "the number is 169 and 12s and 17s are the numbers used." then you have the Diophantine equation 12x+ 17y= 169. 12 divides into 17 once with remainder 5: 17- 12= 5. 5 divides into 12 twice with remainder 2: 12- 2(5)= 2. Finally, 2 divides into 5 once with remainder 1: 5- 2(2)= 1. Replace that (2) by 12- 2(5): 5- 2(12- 2(5))= 5(5)- 2(12)= 1. Replace that (5) by 17- 12: 5(17- 12)- 2(12)= 5(17)- 7(12)= 1. Multiply by 169: 845(17)- 1183(12)= 169.
So one solution is x= -1183, y= 845. But adding any multiple of 17 to x and subtracting that same multiple of 12 from y gives another solution: 12(-1183+ 17k)+ 17(845- 12k)= -14196+ 12(17)k+ 14365- 17(12)k= 169. In particular, taking k= 70 gives positive values for both x and y: x= -1183+ 17(70)= 7 and y= 845- 12(70)= 5. 169= 7(12)+ 5(17).
answered 2 hours ago
user247327user247327
10.5k1515
$endgroup$
$begingroup$
got confused. i didnt get the replacing part
$endgroup$
– user10183910
2 hours ago
$begingroup$
why have to considered k=70?
$endgroup$
– user10183910
1 hour ago
add a comment |
$begingroup$
If "the number is 169 and 12s and 17s are the numbers used." then you have the Diophantine equation 12x+ 17y= 169. 12 divides into 17 once with remainder 5: 17- 12= 5. 5 divides into 12 twice with remainder 2: 12- 2(5)= 2. Finally, 2 divides into 5 once with remainder 1: 5- 2(2)= 1. Replace that (2) by 12- 2(5): 5- 2(12- 2(5))= 5(5)- 2(12)= 1. Replace that (5) by 17- 12: 5(17- 12)- 2(12)= 5(17)- 7(12)= 1. Multiply by 169: 845(17)- 1183(12)= 169.
So one solution is x= -1183, y= 845. But adding any multiple of 17 to x and subtracting that same multiple of 12 from y gives another solution: 12(-1183+ 17k)+ 17(845- 12k)= -14196+ 12(17)k+ 14365- 17(12)k= 169. In particular, taking k= 70 gives positive values for both x and y: x= -1183+ 17(70)= 7 and y= 845- 12(70)= 5. 169= 7(12)+ 5(17).
answered 2 hours ago
user247327user247327
10.5k1515
$endgroup$
If "the number is 169 and 12s and 17s are the numbers used." then you have the Diophantine equation 12x+ 17y= 169. 12 divides into 17 once with remainder 5: 17- 12= 5. 5 divides into 12 twice with remainder 2: 12- 2(5)= 2. Finally, 2 divides into 5 once with remainder 1: 5- 2(2)= 1. Replace that (2) by 12- 2(5): 5- 2(12- 2(5))= 5(5)- 2(12)= 1. Replace that (5) by 17- 12: 5(17- 12)- 2(12)= 5(17)- 7(12)= 1. Multiply by 169: 845(17)- 1183(12)= 169.
So one solution is x= -1183, y= 845. But adding any multiple of 17 to x and subtracting that same multiple of 12 from y gives another solution: 12(-1183+ 17k)+ 17(845- 12k)= -14196+ 12(17)k+ 14365- 17(12)k= 169. In particular, taking k= 70 gives positive values for both x and y: x= -1183+ 17(70)= 7 and y= 845- 12(70)= 5. 169= 7(12)+ 5(17).
answered 2 hours ago
user247327user247327
10.5k1515
answered 2 hours ago
user247327user247327
10.5k1515
answered 2 hours ago
user247327user247327
10.5k1515
answered 2 hours ago
user247327user247327
10.5k1515
10.5k1515
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got confused. i didnt get the replacing part
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– user10183910
2 hours ago
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why have to considered k=70?
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– user10183910
1 hour ago
add a comment |
$begingroup$
got confused. i didnt get the replacing part
$endgroup$
– user10183910
2 hours ago
$begingroup$
why have to considered k=70?
$endgroup$
– user10183910
1 hour ago
$begingroup$
got confused. i didnt get the replacing part
$endgroup$
– user10183910
2 hours ago
$begingroup$
got confused. i didnt get the replacing part
$endgroup$
– user10183910
2 hours ago
$begingroup$
why have to considered k=70?
$endgroup$
– user10183910
1 hour ago
$begingroup$
why have to considered k=70?
$endgroup$
– user10183910
1 hour ago
add a comment |
user10183910 is a new contributor. Be nice, and check out our Code of Conduct.
user10183910 is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
You can relatively easy determine all solutions over the integers. If there is only one solution over the positive integers, you can recover the representation.
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– Peter
3 hours ago
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Not sure this is clear. For instance, If you are told you get $5$ using $2's$ and $1's$ you could have $5=2+1+1+1$ or $5=2+2+1$ , for example.
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– lulu
3 hours ago
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suppose the number is 169 and 12s and 17s are the numbers used.. The answer I need is its the addition of 5times17 and 7times 12.. Is there a formula to deduce this
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– user10183910
3 hours ago