Finding sum of digits of $m$ [closed]
$begingroup$
If the sequence of 5 positive integers (a,b,c,d,e) satisfy:
$$abcdeleq {a+b+c+d+e} leq 10m$$ then find the sum of digits of m.
I don't know how to approach this question. I know it's not a good way to ask here, but, if you may give any hint regarding this for the approach only, I will try my best to use it and apply.
Thank you.
This is the question. And its answer is given as 9. I don't know how?
elementary-number-theory permutations combinations self-learning decimal-expansion
asked Dec 11 '18 at 16:47
jayant98jayant98
554216
$endgroup$
closed as off-topic by Jam, Surb, Siong Thye Goh, Saad, Alexander Gruber♦ Dec 12 '18 at 5:03
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Siong Thye Goh, Saad, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
|
show 7 more comments
$begingroup$
If the sequence of 5 positive integers (a,b,c,d,e) satisfy:
$$abcdeleq {a+b+c+d+e} leq 10m$$ then find the sum of digits of m.
I don't know how to approach this question. I know it's not a good way to ask here, but, if you may give any hint regarding this for the approach only, I will try my best to use it and apply.
Thank you.
This is the question. And its answer is given as 9. I don't know how?
elementary-number-theory permutations combinations self-learning decimal-expansion
asked Dec 11 '18 at 16:47
jayant98jayant98
554216
$endgroup$
closed as off-topic by Jam, Surb, Siong Thye Goh, Saad, Alexander Gruber♦ Dec 12 '18 at 5:03
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Siong Thye Goh, Saad, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
What is $m$? If it's just some number, the equation will be also valid for $m+1$.
$endgroup$
– Andrei
Dec 11 '18 at 16:53
1
$begingroup$
There is no answer because there is no limit on $m$. If $a=b=c=d=e=1$ we could have $m=1$, or $m=2$ or anything higher. Please correct the question.
$endgroup$
– Ross Millikan
Dec 11 '18 at 16:53
$begingroup$
what is $abcde$? Is it five-digit number or a product?
$endgroup$
– Vasya
Dec 11 '18 at 16:57
$begingroup$
@zenith: No, $abcde$ has to be the product. In your case the left side is at least $10000$ and will be larger than the middle.
$endgroup$
– Ross Millikan
Dec 11 '18 at 16:57
1
$begingroup$
I claim the problem is incorrect, and my answer is the best I can do to rescue it. The point is that products tend to be greater than sums.
$endgroup$
– Ross Millikan
Dec 11 '18 at 17:42
|
show 7 more comments
$begingroup$
If the sequence of 5 positive integers (a,b,c,d,e) satisfy:
$$abcdeleq {a+b+c+d+e} leq 10m$$ then find the sum of digits of m.
I don't know how to approach this question. I know it's not a good way to ask here, but, if you may give any hint regarding this for the approach only, I will try my best to use it and apply.
Thank you.
This is the question. And its answer is given as 9. I don't know how?
elementary-number-theory permutations combinations self-learning decimal-expansion
asked Dec 11 '18 at 16:47
jayant98jayant98
554216
$endgroup$
If the sequence of 5 positive integers (a,b,c,d,e) satisfy:
$$abcdeleq {a+b+c+d+e} leq 10m$$ then find the sum of digits of m.
I don't know how to approach this question. I know it's not a good way to ask here, but, if you may give any hint regarding this for the approach only, I will try my best to use it and apply.
Thank you.
This is the question. And its answer is given as 9. I don't know how?
elementary-number-theory permutations combinations self-learning decimal-expansion
elementary-number-theory permutations combinations self-learning decimal-expansion
asked Dec 11 '18 at 16:47
jayant98jayant98
554216
asked Dec 11 '18 at 16:47
jayant98jayant98
554216
edited Dec 11 '18 at 17:20
jayant98
asked Dec 11 '18 at 16:47
jayant98jayant98
554216
asked Dec 11 '18 at 16:47
jayant98jayant98
554216
asked Dec 11 '18 at 16:47
jayant98jayant98
554216
554216
closed as off-topic by Jam, Surb, Siong Thye Goh, Saad, Alexander Gruber♦ Dec 12 '18 at 5:03
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Siong Thye Goh, Saad, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Jam, Surb, Siong Thye Goh, Saad, Alexander Gruber♦ Dec 12 '18 at 5:03
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Siong Thye Goh, Saad, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
What is $m$? If it's just some number, the equation will be also valid for $m+1$.
$endgroup$
– Andrei
Dec 11 '18 at 16:53
1
$begingroup$
There is no answer because there is no limit on $m$. If $a=b=c=d=e=1$ we could have $m=1$, or $m=2$ or anything higher. Please correct the question.
$endgroup$
– Ross Millikan
Dec 11 '18 at 16:53
$begingroup$
what is $abcde$? Is it five-digit number or a product?
$endgroup$
– Vasya
Dec 11 '18 at 16:57
$begingroup$
@zenith: No, $abcde$ has to be the product. In your case the left side is at least $10000$ and will be larger than the middle.
$endgroup$
– Ross Millikan
Dec 11 '18 at 16:57
1
$begingroup$
I claim the problem is incorrect, and my answer is the best I can do to rescue it. The point is that products tend to be greater than sums.
$endgroup$
– Ross Millikan
Dec 11 '18 at 17:42
|
show 7 more comments
1
$begingroup$
What is $m$? If it's just some number, the equation will be also valid for $m+1$.
$endgroup$
– Andrei
Dec 11 '18 at 16:53
1
$begingroup$
There is no answer because there is no limit on $m$. If $a=b=c=d=e=1$ we could have $m=1$, or $m=2$ or anything higher. Please correct the question.
$endgroup$
– Ross Millikan
Dec 11 '18 at 16:53
$begingroup$
what is $abcde$? Is it five-digit number or a product?
$endgroup$
– Vasya
Dec 11 '18 at 16:57
$begingroup$
@zenith: No, $abcde$ has to be the product. In your case the left side is at least $10000$ and will be larger than the middle.
$endgroup$
– Ross Millikan
Dec 11 '18 at 16:57
1
$begingroup$
I claim the problem is incorrect, and my answer is the best I can do to rescue it. The point is that products tend to be greater than sums.
$endgroup$
– Ross Millikan
Dec 11 '18 at 17:42
1
1
$begingroup$
What is $m$? If it's just some number, the equation will be also valid for $m+1$.
$endgroup$
– Andrei
Dec 11 '18 at 16:53
$begingroup$
What is $m$? If it's just some number, the equation will be also valid for $m+1$.
$endgroup$
– Andrei
Dec 11 '18 at 16:53
1
1
$begingroup$
There is no answer because there is no limit on $m$. If $a=b=c=d=e=1$ we could have $m=1$, or $m=2$ or anything higher. Please correct the question.
$endgroup$
– Ross Millikan
Dec 11 '18 at 16:53
$begingroup$
There is no answer because there is no limit on $m$. If $a=b=c=d=e=1$ we could have $m=1$, or $m=2$ or anything higher. Please correct the question.
$endgroup$
– Ross Millikan
Dec 11 '18 at 16:53
$begingroup$
what is $abcde$? Is it five-digit number or a product?
$endgroup$
– Vasya
Dec 11 '18 at 16:57
$begingroup$
what is $abcde$? Is it five-digit number or a product?
$endgroup$
– Vasya
Dec 11 '18 at 16:57
$begingroup$
@zenith: No, $abcde$ has to be the product. In your case the left side is at least $10000$ and will be larger than the middle.
$endgroup$
– Ross Millikan
Dec 11 '18 at 16:57
$begingroup$
@zenith: No, $abcde$ has to be the product. In your case the left side is at least $10000$ and will be larger than the middle.
$endgroup$
– Ross Millikan
Dec 11 '18 at 16:57
1
1
$begingroup$
I claim the problem is incorrect, and my answer is the best I can do to rescue it. The point is that products tend to be greater than sums.
$endgroup$
– Ross Millikan
Dec 11 '18 at 17:42
$begingroup$
I claim the problem is incorrect, and my answer is the best I can do to rescue it. The point is that products tend to be greater than sums.
$endgroup$
– Ross Millikan
Dec 11 '18 at 17:42
|
show 7 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Hint: Presumably $m$ must be the minimum value that makes the right inequality true. Unless some of $a,b,c,d,e$ are $1$, the product will be too large for the left inequality to hold. Even if one of them is $1$, the product will be at least $16$ and that will be too large. Work on justifying how small the sum must be.
If you are willing to count on the problem setter to have made sure there is a unique solution, you can just observe that $a=b=c=d=e=1$ allows $m=1$, assert that the sum of digits of $m$ must be $1$, and declare victory. Really you should prove that any $a,b,c,d,e$ that satisfies the left must allow $m=1$
Note that $a=b=c=d=1$ allows $e$ to be anything, which means we can force the minimum $m$ to be as large as we want. The problem is badly flawed.
answered Dec 11 '18 at 16:59
Ross MillikanRoss Millikan
295k23198371
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Presumably $m$ must be the minimum value that makes the right inequality true. Unless some of $a,b,c,d,e$ are $1$, the product will be too large for the left inequality to hold. Even if one of them is $1$, the product will be at least $16$ and that will be too large. Work on justifying how small the sum must be.
If you are willing to count on the problem setter to have made sure there is a unique solution, you can just observe that $a=b=c=d=e=1$ allows $m=1$, assert that the sum of digits of $m$ must be $1$, and declare victory. Really you should prove that any $a,b,c,d,e$ that satisfies the left must allow $m=1$
Note that $a=b=c=d=1$ allows $e$ to be anything, which means we can force the minimum $m$ to be as large as we want. The problem is badly flawed.
answered Dec 11 '18 at 16:59
Ross MillikanRoss Millikan
295k23198371
$endgroup$
add a comment |
$begingroup$
Hint: Presumably $m$ must be the minimum value that makes the right inequality true. Unless some of $a,b,c,d,e$ are $1$, the product will be too large for the left inequality to hold. Even if one of them is $1$, the product will be at least $16$ and that will be too large. Work on justifying how small the sum must be.
If you are willing to count on the problem setter to have made sure there is a unique solution, you can just observe that $a=b=c=d=e=1$ allows $m=1$, assert that the sum of digits of $m$ must be $1$, and declare victory. Really you should prove that any $a,b,c,d,e$ that satisfies the left must allow $m=1$
Note that $a=b=c=d=1$ allows $e$ to be anything, which means we can force the minimum $m$ to be as large as we want. The problem is badly flawed.
answered Dec 11 '18 at 16:59
Ross MillikanRoss Millikan
295k23198371
$endgroup$
add a comment |
$begingroup$
Hint: Presumably $m$ must be the minimum value that makes the right inequality true. Unless some of $a,b,c,d,e$ are $1$, the product will be too large for the left inequality to hold. Even if one of them is $1$, the product will be at least $16$ and that will be too large. Work on justifying how small the sum must be.
If you are willing to count on the problem setter to have made sure there is a unique solution, you can just observe that $a=b=c=d=e=1$ allows $m=1$, assert that the sum of digits of $m$ must be $1$, and declare victory. Really you should prove that any $a,b,c,d,e$ that satisfies the left must allow $m=1$
Note that $a=b=c=d=1$ allows $e$ to be anything, which means we can force the minimum $m$ to be as large as we want. The problem is badly flawed.
answered Dec 11 '18 at 16:59
Ross MillikanRoss Millikan
295k23198371
$endgroup$
Hint: Presumably $m$ must be the minimum value that makes the right inequality true. Unless some of $a,b,c,d,e$ are $1$, the product will be too large for the left inequality to hold. Even if one of them is $1$, the product will be at least $16$ and that will be too large. Work on justifying how small the sum must be.
If you are willing to count on the problem setter to have made sure there is a unique solution, you can just observe that $a=b=c=d=e=1$ allows $m=1$, assert that the sum of digits of $m$ must be $1$, and declare victory. Really you should prove that any $a,b,c,d,e$ that satisfies the left must allow $m=1$
Note that $a=b=c=d=1$ allows $e$ to be anything, which means we can force the minimum $m$ to be as large as we want. The problem is badly flawed.
answered Dec 11 '18 at 16:59
Ross MillikanRoss Millikan
295k23198371
edited Dec 12 '18 at 17:25
answered Dec 11 '18 at 16:59
Ross MillikanRoss Millikan
295k23198371
answered Dec 11 '18 at 16:59
Ross MillikanRoss Millikan
295k23198371
answered Dec 11 '18 at 16:59
Ross MillikanRoss Millikan
295k23198371
295k23198371
add a comment |
add a comment |
1
$begingroup$
What is $m$? If it's just some number, the equation will be also valid for $m+1$.
$endgroup$
– Andrei
Dec 11 '18 at 16:53
1
$begingroup$
There is no answer because there is no limit on $m$. If $a=b=c=d=e=1$ we could have $m=1$, or $m=2$ or anything higher. Please correct the question.
$endgroup$
– Ross Millikan
Dec 11 '18 at 16:53
$begingroup$
what is $abcde$? Is it five-digit number or a product?
$endgroup$
– Vasya
Dec 11 '18 at 16:57
$begingroup$
@zenith: No, $abcde$ has to be the product. In your case the left side is at least $10000$ and will be larger than the middle.
$endgroup$
– Ross Millikan
Dec 11 '18 at 16:57
1
$begingroup$
I claim the problem is incorrect, and my answer is the best I can do to rescue it. The point is that products tend to be greater than sums.
$endgroup$
– Ross Millikan
Dec 11 '18 at 17:42