At least how many consecutive numbers should be taken?
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
Problem Statement :
At least how many consecutive positive integers should be multiplied so that the product is divisible by $5040$ ?
This problem should be solved using algebra.
My attempt :
Let the multiplication of n consecutive numbers is divisible by 5040 . So according to question :
x*(x+1)(x+2)$ldots$(x+n-1)=5040*m
How can I find the value of n ? Please help me .
abstract-algebra recreational-mathematics puzzle
asked Dec 31 '17 at 19:09
Standard Equation
235110
 |Â
show 11 more comments
up vote
0
down vote
favorite
Problem Statement :
At least how many consecutive positive integers should be multiplied so that the product is divisible by $5040$ ?
This problem should be solved using algebra.
My attempt :
Let the multiplication of n consecutive numbers is divisible by 5040 . So according to question :
x*(x+1)(x+2)$ldots$(x+n-1)=5040*m
How can I find the value of n ? Please help me .
abstract-algebra recreational-mathematics puzzle
asked Dec 31 '17 at 19:09
Standard Equation
235110
1
In the problem statement, perhaps you should specify that the consecutive numbers start at 1. Otherwise any sequence of length 1 whose single term is divisible by 5040 would work.
– Adam Lowrance
Dec 31 '17 at 19:27
2
This is not clear. Are you asking for a number $n$ such that the product of any $n$ consecutive numbers is divisible by $7!$ ?
– lulu
Dec 31 '17 at 19:31
1
Anyway, assuming I have the question right, guessing $n=7$ seems natural. Can you prove that this works? Certainly the answer can not be smaller than $7$.
– lulu
Dec 31 '17 at 19:33
1
I don't understand why you decline to clarify the question. You can see from the comments and from the posted solution that nobody is sure what you mean. I think my interpretation is correct, but I note that it differs from the interpretation used in the posted solution.
– lulu
Dec 31 '17 at 19:36
1
Perhaps language is a barrier here. Could you review the two posted solutions and tell us if either of us are answering the question you had in mind?
– lulu
Dec 31 '17 at 19:46
 |Â
show 11 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Problem Statement :
At least how many consecutive positive integers should be multiplied so that the product is divisible by $5040$ ?
This problem should be solved using algebra.
My attempt :
Let the multiplication of n consecutive numbers is divisible by 5040 . So according to question :
x*(x+1)(x+2)$ldots$(x+n-1)=5040*m
How can I find the value of n ? Please help me .
abstract-algebra recreational-mathematics puzzle
asked Dec 31 '17 at 19:09
Standard Equation
235110
Problem Statement :
At least how many consecutive positive integers should be multiplied so that the product is divisible by $5040$ ?
This problem should be solved using algebra.
My attempt :
Let the multiplication of n consecutive numbers is divisible by 5040 . So according to question :
x*(x+1)(x+2)$ldots$(x+n-1)=5040*m
How can I find the value of n ? Please help me .
abstract-algebra recreational-mathematics puzzle
asked Dec 31 '17 at 19:09
Standard Equation
235110
edited Dec 31 '17 at 19:41
asked Dec 31 '17 at 19:09
Standard Equation
235110
asked Dec 31 '17 at 19:09
Standard Equation
235110
asked Dec 31 '17 at 19:09
Standard Equation
235110
235110
1
In the problem statement, perhaps you should specify that the consecutive numbers start at 1. Otherwise any sequence of length 1 whose single term is divisible by 5040 would work.
– Adam Lowrance
Dec 31 '17 at 19:27
2
This is not clear. Are you asking for a number $n$ such that the product of any $n$ consecutive numbers is divisible by $7!$ ?
– lulu
Dec 31 '17 at 19:31
1
Anyway, assuming I have the question right, guessing $n=7$ seems natural. Can you prove that this works? Certainly the answer can not be smaller than $7$.
– lulu
Dec 31 '17 at 19:33
1
I don't understand why you decline to clarify the question. You can see from the comments and from the posted solution that nobody is sure what you mean. I think my interpretation is correct, but I note that it differs from the interpretation used in the posted solution.
– lulu
Dec 31 '17 at 19:36
1
Perhaps language is a barrier here. Could you review the two posted solutions and tell us if either of us are answering the question you had in mind?
– lulu
Dec 31 '17 at 19:46
 |Â
show 11 more comments
1
In the problem statement, perhaps you should specify that the consecutive numbers start at 1. Otherwise any sequence of length 1 whose single term is divisible by 5040 would work.
– Adam Lowrance
Dec 31 '17 at 19:27
2
This is not clear. Are you asking for a number $n$ such that the product of any $n$ consecutive numbers is divisible by $7!$ ?
– lulu
Dec 31 '17 at 19:31
1
Anyway, assuming I have the question right, guessing $n=7$ seems natural. Can you prove that this works? Certainly the answer can not be smaller than $7$.
– lulu
Dec 31 '17 at 19:33
1
I don't understand why you decline to clarify the question. You can see from the comments and from the posted solution that nobody is sure what you mean. I think my interpretation is correct, but I note that it differs from the interpretation used in the posted solution.
– lulu
Dec 31 '17 at 19:36
1
Perhaps language is a barrier here. Could you review the two posted solutions and tell us if either of us are answering the question you had in mind?
– lulu
Dec 31 '17 at 19:46
1
1
In the problem statement, perhaps you should specify that the consecutive numbers start at 1. Otherwise any sequence of length 1 whose single term is divisible by 5040 would work.
– Adam Lowrance
Dec 31 '17 at 19:27
In the problem statement, perhaps you should specify that the consecutive numbers start at 1. Otherwise any sequence of length 1 whose single term is divisible by 5040 would work.
– Adam Lowrance
Dec 31 '17 at 19:27
2
2
This is not clear. Are you asking for a number $n$ such that the product of any $n$ consecutive numbers is divisible by $7!$ ?
– lulu
Dec 31 '17 at 19:31
This is not clear. Are you asking for a number $n$ such that the product of any $n$ consecutive numbers is divisible by $7!$ ?
– lulu
Dec 31 '17 at 19:31
1
1
Anyway, assuming I have the question right, guessing $n=7$ seems natural. Can you prove that this works? Certainly the answer can not be smaller than $7$.
– lulu
Dec 31 '17 at 19:33
Anyway, assuming I have the question right, guessing $n=7$ seems natural. Can you prove that this works? Certainly the answer can not be smaller than $7$.
– lulu
Dec 31 '17 at 19:33
1
1
I don't understand why you decline to clarify the question. You can see from the comments and from the posted solution that nobody is sure what you mean. I think my interpretation is correct, but I note that it differs from the interpretation used in the posted solution.
– lulu
Dec 31 '17 at 19:36
I don't understand why you decline to clarify the question. You can see from the comments and from the posted solution that nobody is sure what you mean. I think my interpretation is correct, but I note that it differs from the interpretation used in the posted solution.
– lulu
Dec 31 '17 at 19:36
1
1
Perhaps language is a barrier here. Could you review the two posted solutions and tell us if either of us are answering the question you had in mind?
– lulu
Dec 31 '17 at 19:46
Perhaps language is a barrier here. Could you review the two posted solutions and tell us if either of us are answering the question you had in mind?
– lulu
Dec 31 '17 at 19:46
 |Â
show 11 more comments
3 Answers
3
active
oldest
votes
up vote
6
down vote
accepted
The question is phrased poorly and it is not clear what is intended. I believe the following is relevant:
Claim: the product of any $7$ consecutive natural numbers is divisible by $7!$ and no number smaller than $7$ has this property.
Proof: It is clear that no number smaller than $7$ works as $6!$ is not divisible by $7$, let alone $7!$, and it is (visibly) the product of $6$ natural numbers.
To see that $7$ has the desired property, let $m$ denote the start of the consecutive block and we remark that $$binom m+67=frac (m+6)(m+5)cdots (m)7!in mathbb N$$ is an integer, hence the desired result.
answered Dec 31 '17 at 19:41
lulu
35.7k14274
I have edited the question . Please see the update and let me know
– Standard Equation
Dec 31 '17 at 19:42
I think you missed the word consecutive.
– Mohammad Zuhair Khan
Dec 31 '17 at 19:42
@MohammadZuhairKhan What do you mean? $(m+6)cdots (m)$ is the product of $7$ consecutive integers starting with $m$.
– lulu
Dec 31 '17 at 19:43
No I meant that in the Claim section you missed the word consecutive. Your proof is perfect.
– Mohammad Zuhair Khan
Dec 31 '17 at 19:44
@MohammadZuhairKhan Oh! Of course you are right. Thanks. Corrected.
– lulu
Dec 31 '17 at 19:47
 |Â
show 3 more comments
up vote
1
down vote
I personally believe that you need only one number: $5040$, but I assume that is not the kind of answer you want.
Algebraically:
$n = 5040$
The next best is four consecutive integers:
Let $x$ be the smallest integer:
$x(x+1)(x+2)(x+3)= 5040$
I would prefer graphing it but here goes the algebraic way:
$x^4 + 6 x^3 + 11 x^2 + 6 x = 5040$
Solving this gives:
$x = 7, x= -10, x= frac 32 i (sqrt(31) +i), x = -frac 32 i (sqrt(31) - i)$
$x = 7$ is a valid answer as $7*8*9*10 = 5040$
$therefore$ Four is the (second) least number of consecutive positive integers whose product is $5040$.
$n = 7$ is the least value iff (if and only if) $x = 1$.
answered Dec 31 '17 at 19:34
Mohammad Zuhair Khan
782422
The answer is n = 7 .
– Standard Equation
Dec 31 '17 at 19:46
$n = 4$ is also valid, as is $n = 1$.
– Mohammad Zuhair Khan
Dec 31 '17 at 19:48
add a comment |Â
up vote
0
down vote
The least product which is divisible by $5040$ is $5040$ itself. So, the problem is saying to determine a set of consecutive integer numbers whose product (product of all the numbers in that set) is $5040$.and we have to determine the number of the elements in that set.
Let, the starting number be $x$.
and there are $n$ consecutive numbers we need to multiply to make $5040$.
So, $$x(x+1)(x+2)(x+3).....(x+n-2)(x+n-1)=5040...................(1)$$
Again we know, the product of n consecutive numbers is must divisible by $n!$ Which is the least number.
So,$$ n! = 5040…..........(2)$$
$$implies n=7$$
If we put $n=7$ in equation (1) then we get, $x=1$.
So,the answer is we need to multiply at least $7$ consecutive numbers starting from $1$ to get $5040$
answered Aug 13 at 6:24
Rakibul Islam Prince
945
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
The question is phrased poorly and it is not clear what is intended. I believe the following is relevant:
Claim: the product of any $7$ consecutive natural numbers is divisible by $7!$ and no number smaller than $7$ has this property.
Proof: It is clear that no number smaller than $7$ works as $6!$ is not divisible by $7$, let alone $7!$, and it is (visibly) the product of $6$ natural numbers.
To see that $7$ has the desired property, let $m$ denote the start of the consecutive block and we remark that $$binom m+67=frac (m+6)(m+5)cdots (m)7!in mathbb N$$ is an integer, hence the desired result.
answered Dec 31 '17 at 19:41
lulu
35.7k14274
I have edited the question . Please see the update and let me know
– Standard Equation
Dec 31 '17 at 19:42
I think you missed the word consecutive.
– Mohammad Zuhair Khan
Dec 31 '17 at 19:42
@MohammadZuhairKhan What do you mean? $(m+6)cdots (m)$ is the product of $7$ consecutive integers starting with $m$.
– lulu
Dec 31 '17 at 19:43
No I meant that in the Claim section you missed the word consecutive. Your proof is perfect.
– Mohammad Zuhair Khan
Dec 31 '17 at 19:44
@MohammadZuhairKhan Oh! Of course you are right. Thanks. Corrected.
– lulu
Dec 31 '17 at 19:47
 |Â
show 3 more comments
up vote
6
down vote
accepted
The question is phrased poorly and it is not clear what is intended. I believe the following is relevant:
Claim: the product of any $7$ consecutive natural numbers is divisible by $7!$ and no number smaller than $7$ has this property.
Proof: It is clear that no number smaller than $7$ works as $6!$ is not divisible by $7$, let alone $7!$, and it is (visibly) the product of $6$ natural numbers.
To see that $7$ has the desired property, let $m$ denote the start of the consecutive block and we remark that $$binom m+67=frac (m+6)(m+5)cdots (m)7!in mathbb N$$ is an integer, hence the desired result.
answered Dec 31 '17 at 19:41
lulu
35.7k14274
I have edited the question . Please see the update and let me know
– Standard Equation
Dec 31 '17 at 19:42
I think you missed the word consecutive.
– Mohammad Zuhair Khan
Dec 31 '17 at 19:42
@MohammadZuhairKhan What do you mean? $(m+6)cdots (m)$ is the product of $7$ consecutive integers starting with $m$.
– lulu
Dec 31 '17 at 19:43
No I meant that in the Claim section you missed the word consecutive. Your proof is perfect.
– Mohammad Zuhair Khan
Dec 31 '17 at 19:44
@MohammadZuhairKhan Oh! Of course you are right. Thanks. Corrected.
– lulu
Dec 31 '17 at 19:47
 |Â
show 3 more comments
up vote
6
down vote
accepted
up vote
6
down vote
accepted
The question is phrased poorly and it is not clear what is intended. I believe the following is relevant:
Claim: the product of any $7$ consecutive natural numbers is divisible by $7!$ and no number smaller than $7$ has this property.
Proof: It is clear that no number smaller than $7$ works as $6!$ is not divisible by $7$, let alone $7!$, and it is (visibly) the product of $6$ natural numbers.
To see that $7$ has the desired property, let $m$ denote the start of the consecutive block and we remark that $$binom m+67=frac (m+6)(m+5)cdots (m)7!in mathbb N$$ is an integer, hence the desired result.
answered Dec 31 '17 at 19:41
lulu
35.7k14274
The question is phrased poorly and it is not clear what is intended. I believe the following is relevant:
Claim: the product of any $7$ consecutive natural numbers is divisible by $7!$ and no number smaller than $7$ has this property.
Proof: It is clear that no number smaller than $7$ works as $6!$ is not divisible by $7$, let alone $7!$, and it is (visibly) the product of $6$ natural numbers.
To see that $7$ has the desired property, let $m$ denote the start of the consecutive block and we remark that $$binom m+67=frac (m+6)(m+5)cdots (m)7!in mathbb N$$ is an integer, hence the desired result.
answered Dec 31 '17 at 19:41
lulu
35.7k14274
edited Dec 31 '17 at 19:47
answered Dec 31 '17 at 19:41
lulu
35.7k14274
answered Dec 31 '17 at 19:41
lulu
35.7k14274
answered Dec 31 '17 at 19:41
lulu
35.7k14274
35.7k14274
I have edited the question . Please see the update and let me know
– Standard Equation
Dec 31 '17 at 19:42
I think you missed the word consecutive.
– Mohammad Zuhair Khan
Dec 31 '17 at 19:42
@MohammadZuhairKhan What do you mean? $(m+6)cdots (m)$ is the product of $7$ consecutive integers starting with $m$.
– lulu
Dec 31 '17 at 19:43
No I meant that in the Claim section you missed the word consecutive. Your proof is perfect.
– Mohammad Zuhair Khan
Dec 31 '17 at 19:44
@MohammadZuhairKhan Oh! Of course you are right. Thanks. Corrected.
– lulu
Dec 31 '17 at 19:47
 |Â
show 3 more comments
I have edited the question . Please see the update and let me know
– Standard Equation
Dec 31 '17 at 19:42
I think you missed the word consecutive.
– Mohammad Zuhair Khan
Dec 31 '17 at 19:42
@MohammadZuhairKhan What do you mean? $(m+6)cdots (m)$ is the product of $7$ consecutive integers starting with $m$.
– lulu
Dec 31 '17 at 19:43
No I meant that in the Claim section you missed the word consecutive. Your proof is perfect.
– Mohammad Zuhair Khan
Dec 31 '17 at 19:44
@MohammadZuhairKhan Oh! Of course you are right. Thanks. Corrected.
– lulu
Dec 31 '17 at 19:47
I have edited the question . Please see the update and let me know
– Standard Equation
Dec 31 '17 at 19:42
I have edited the question . Please see the update and let me know
– Standard Equation
Dec 31 '17 at 19:42
I think you missed the word consecutive.
– Mohammad Zuhair Khan
Dec 31 '17 at 19:42
I think you missed the word consecutive.
– Mohammad Zuhair Khan
Dec 31 '17 at 19:42
@MohammadZuhairKhan What do you mean? $(m+6)cdots (m)$ is the product of $7$ consecutive integers starting with $m$.
– lulu
Dec 31 '17 at 19:43
@MohammadZuhairKhan What do you mean? $(m+6)cdots (m)$ is the product of $7$ consecutive integers starting with $m$.
– lulu
Dec 31 '17 at 19:43
No I meant that in the Claim section you missed the word consecutive. Your proof is perfect.
– Mohammad Zuhair Khan
Dec 31 '17 at 19:44
No I meant that in the Claim section you missed the word consecutive. Your proof is perfect.
– Mohammad Zuhair Khan
Dec 31 '17 at 19:44
@MohammadZuhairKhan Oh! Of course you are right. Thanks. Corrected.
– lulu
Dec 31 '17 at 19:47
@MohammadZuhairKhan Oh! Of course you are right. Thanks. Corrected.
– lulu
Dec 31 '17 at 19:47
 |Â
show 3 more comments
up vote
1
down vote
I personally believe that you need only one number: $5040$, but I assume that is not the kind of answer you want.
Algebraically:
$n = 5040$
The next best is four consecutive integers:
Let $x$ be the smallest integer:
$x(x+1)(x+2)(x+3)= 5040$
I would prefer graphing it but here goes the algebraic way:
$x^4 + 6 x^3 + 11 x^2 + 6 x = 5040$
Solving this gives:
$x = 7, x= -10, x= frac 32 i (sqrt(31) +i), x = -frac 32 i (sqrt(31) - i)$
$x = 7$ is a valid answer as $7*8*9*10 = 5040$
$therefore$ Four is the (second) least number of consecutive positive integers whose product is $5040$.
$n = 7$ is the least value iff (if and only if) $x = 1$.
answered Dec 31 '17 at 19:34
Mohammad Zuhair Khan
782422
The answer is n = 7 .
– Standard Equation
Dec 31 '17 at 19:46
$n = 4$ is also valid, as is $n = 1$.
– Mohammad Zuhair Khan
Dec 31 '17 at 19:48
add a comment |Â
up vote
1
down vote
I personally believe that you need only one number: $5040$, but I assume that is not the kind of answer you want.
Algebraically:
$n = 5040$
The next best is four consecutive integers:
Let $x$ be the smallest integer:
$x(x+1)(x+2)(x+3)= 5040$
I would prefer graphing it but here goes the algebraic way:
$x^4 + 6 x^3 + 11 x^2 + 6 x = 5040$
Solving this gives:
$x = 7, x= -10, x= frac 32 i (sqrt(31) +i), x = -frac 32 i (sqrt(31) - i)$
$x = 7$ is a valid answer as $7*8*9*10 = 5040$
$therefore$ Four is the (second) least number of consecutive positive integers whose product is $5040$.
$n = 7$ is the least value iff (if and only if) $x = 1$.
answered Dec 31 '17 at 19:34
Mohammad Zuhair Khan
782422
The answer is n = 7 .
– Standard Equation
Dec 31 '17 at 19:46
$n = 4$ is also valid, as is $n = 1$.
– Mohammad Zuhair Khan
Dec 31 '17 at 19:48
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I personally believe that you need only one number: $5040$, but I assume that is not the kind of answer you want.
Algebraically:
$n = 5040$
The next best is four consecutive integers:
Let $x$ be the smallest integer:
$x(x+1)(x+2)(x+3)= 5040$
I would prefer graphing it but here goes the algebraic way:
$x^4 + 6 x^3 + 11 x^2 + 6 x = 5040$
Solving this gives:
$x = 7, x= -10, x= frac 32 i (sqrt(31) +i), x = -frac 32 i (sqrt(31) - i)$
$x = 7$ is a valid answer as $7*8*9*10 = 5040$
$therefore$ Four is the (second) least number of consecutive positive integers whose product is $5040$.
$n = 7$ is the least value iff (if and only if) $x = 1$.
answered Dec 31 '17 at 19:34
Mohammad Zuhair Khan
782422
I personally believe that you need only one number: $5040$, but I assume that is not the kind of answer you want.
Algebraically:
$n = 5040$
The next best is four consecutive integers:
Let $x$ be the smallest integer:
$x(x+1)(x+2)(x+3)= 5040$
I would prefer graphing it but here goes the algebraic way:
$x^4 + 6 x^3 + 11 x^2 + 6 x = 5040$
Solving this gives:
$x = 7, x= -10, x= frac 32 i (sqrt(31) +i), x = -frac 32 i (sqrt(31) - i)$
$x = 7$ is a valid answer as $7*8*9*10 = 5040$
$therefore$ Four is the (second) least number of consecutive positive integers whose product is $5040$.
$n = 7$ is the least value iff (if and only if) $x = 1$.
answered Dec 31 '17 at 19:34
Mohammad Zuhair Khan
782422
edited Dec 31 '17 at 19:53
answered Dec 31 '17 at 19:34
Mohammad Zuhair Khan
782422
answered Dec 31 '17 at 19:34
Mohammad Zuhair Khan
782422
answered Dec 31 '17 at 19:34
Mohammad Zuhair Khan
782422
782422
The answer is n = 7 .
– Standard Equation
Dec 31 '17 at 19:46
$n = 4$ is also valid, as is $n = 1$.
– Mohammad Zuhair Khan
Dec 31 '17 at 19:48
add a comment |Â
The answer is n = 7 .
– Standard Equation
Dec 31 '17 at 19:46
$n = 4$ is also valid, as is $n = 1$.
– Mohammad Zuhair Khan
Dec 31 '17 at 19:48
The answer is n = 7 .
– Standard Equation
Dec 31 '17 at 19:46
The answer is n = 7 .
– Standard Equation
Dec 31 '17 at 19:46
$n = 4$ is also valid, as is $n = 1$.
– Mohammad Zuhair Khan
Dec 31 '17 at 19:48
$n = 4$ is also valid, as is $n = 1$.
– Mohammad Zuhair Khan
Dec 31 '17 at 19:48
add a comment |Â
up vote
0
down vote
The least product which is divisible by $5040$ is $5040$ itself. So, the problem is saying to determine a set of consecutive integer numbers whose product (product of all the numbers in that set) is $5040$.and we have to determine the number of the elements in that set.
Let, the starting number be $x$.
and there are $n$ consecutive numbers we need to multiply to make $5040$.
So, $$x(x+1)(x+2)(x+3).....(x+n-2)(x+n-1)=5040...................(1)$$
Again we know, the product of n consecutive numbers is must divisible by $n!$ Which is the least number.
So,$$ n! = 5040…..........(2)$$
$$implies n=7$$
If we put $n=7$ in equation (1) then we get, $x=1$.
So,the answer is we need to multiply at least $7$ consecutive numbers starting from $1$ to get $5040$
answered Aug 13 at 6:24
Rakibul Islam Prince
945
add a comment |Â
up vote
0
down vote
The least product which is divisible by $5040$ is $5040$ itself. So, the problem is saying to determine a set of consecutive integer numbers whose product (product of all the numbers in that set) is $5040$.and we have to determine the number of the elements in that set.
Let, the starting number be $x$.
and there are $n$ consecutive numbers we need to multiply to make $5040$.
So, $$x(x+1)(x+2)(x+3).....(x+n-2)(x+n-1)=5040...................(1)$$
Again we know, the product of n consecutive numbers is must divisible by $n!$ Which is the least number.
So,$$ n! = 5040…..........(2)$$
$$implies n=7$$
If we put $n=7$ in equation (1) then we get, $x=1$.
So,the answer is we need to multiply at least $7$ consecutive numbers starting from $1$ to get $5040$
answered Aug 13 at 6:24
Rakibul Islam Prince
945
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The least product which is divisible by $5040$ is $5040$ itself. So, the problem is saying to determine a set of consecutive integer numbers whose product (product of all the numbers in that set) is $5040$.and we have to determine the number of the elements in that set.
Let, the starting number be $x$.
and there are $n$ consecutive numbers we need to multiply to make $5040$.
So, $$x(x+1)(x+2)(x+3).....(x+n-2)(x+n-1)=5040...................(1)$$
Again we know, the product of n consecutive numbers is must divisible by $n!$ Which is the least number.
So,$$ n! = 5040…..........(2)$$
$$implies n=7$$
If we put $n=7$ in equation (1) then we get, $x=1$.
So,the answer is we need to multiply at least $7$ consecutive numbers starting from $1$ to get $5040$
answered Aug 13 at 6:24
Rakibul Islam Prince
945
The least product which is divisible by $5040$ is $5040$ itself. So, the problem is saying to determine a set of consecutive integer numbers whose product (product of all the numbers in that set) is $5040$.and we have to determine the number of the elements in that set.
Let, the starting number be $x$.
and there are $n$ consecutive numbers we need to multiply to make $5040$.
So, $$x(x+1)(x+2)(x+3).....(x+n-2)(x+n-1)=5040...................(1)$$
Again we know, the product of n consecutive numbers is must divisible by $n!$ Which is the least number.
So,$$ n! = 5040…..........(2)$$
$$implies n=7$$
If we put $n=7$ in equation (1) then we get, $x=1$.
So,the answer is we need to multiply at least $7$ consecutive numbers starting from $1$ to get $5040$
answered Aug 13 at 6:24
Rakibul Islam Prince
945
answered Aug 13 at 6:24
Rakibul Islam Prince
945
answered Aug 13 at 6:24
Rakibul Islam Prince
945
answered Aug 13 at 6:24
Rakibul Islam Prince
945
945
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2586933%2fat-least-how-many-consecutive-numbers-should-be-taken%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
In the problem statement, perhaps you should specify that the consecutive numbers start at 1. Otherwise any sequence of length 1 whose single term is divisible by 5040 would work.
– Adam Lowrance
Dec 31 '17 at 19:27
2
This is not clear. Are you asking for a number $n$ such that the product of any $n$ consecutive numbers is divisible by $7!$ ?
– lulu
Dec 31 '17 at 19:31
1
Anyway, assuming I have the question right, guessing $n=7$ seems natural. Can you prove that this works? Certainly the answer can not be smaller than $7$.
– lulu
Dec 31 '17 at 19:33
1
I don't understand why you decline to clarify the question. You can see from the comments and from the posted solution that nobody is sure what you mean. I think my interpretation is correct, but I note that it differs from the interpretation used in the posted solution.
– lulu
Dec 31 '17 at 19:36
1
Perhaps language is a barrier here. Could you review the two posted solutions and tell us if either of us are answering the question you had in mind?
– lulu
Dec 31 '17 at 19:46