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Senior Manager  P
Joined: 22 Feb 2018
Posts: 349
What is the sum of all possible 7-digit numbers that can be  [#permalink]

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4 00:00

Difficulty:   65% (hard)

Question Stats: 39% (01:43) correct 61% (02:13) wrong based on 36 sessions

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What is the sum of all possible 7-digit numbers that can be constructed using the digits 1, 2, 3, 4, 5, 6 and 7, if each digit can be used only once in each number?

A) 7! * 8888888
B) 3.5! * 8888888
C) 3.5!/2 * 88888888
D) 7!/2 * 8888888
E) 7 * 8888888

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Intern  B
Joined: 24 Jun 2019
Posts: 2
Re: What is the sum of all possible 7-digit numbers that can be  [#permalink]

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is this question feasible in the GMAT exam?
Senior Manager  P
Joined: 22 Feb 2018
Posts: 349
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1
Yes, rocksailor9,

The question is likely feasible in GMAT exam, if we correlate the sequence property here.

This type of question does not follow directly the Arithmetic properties but indirectly we can apply, i.e. The difference of each side of the AP sequence have the same difference from the mean. e.g. 123, 132, 213, 231, 312, 321 has the mean as 222. Now if you find the difference between each term with respect to mean (123, 132, 213, 222, 231, 312, 321), the first and last must have same difference. Second last of each side must have same difference. The same pattern follows here also. WE CAN APPLY THE SAME PROPERTY TO N NUMBER OF DIGITS. So, we can use the formula to find sum of A.P. sequence.

Sum = n/2 (a+l)
Now, n = Total possible no. of numbers = 7p7 = n! = 7!
a= first number, i.e. smallest one = 1234567
l= last number, i.e. largest one = 7654321,

= (7!/2) (1234567+7654321)
= 7!/2 * 88888888
Ans. D.

Hope this helps.

Bunuel, kindly let me know any other better and quick approach.
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Senior Manager  P
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What is the sum of all possible 7-digit numbers that can be  [#permalink]

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rajatchopra1994 wrote:
Raxit85 wrote:
What is the sum of all possible 7-digit numbers that can be constructed using the digits 1, 2, 3, 4, 5, 6 and 7, if each digit can be used only once in each number?

A) 7! * 8888888
B) 3.5! * 8888888
C) 3.5!/2 * 88888888
D) 7!/2 * 8888888
E) 7 * 8888888

Generally the sum of all the numbers which can be formed by using the n distinct digits, is given by the formula:

(n-1)!*(sum of the digits)*(111…..n times)

Dear Rajatchopra1994,

Kindly do not copy paste exactly, at least, from other members post.

Bunuel, kindly look into matter.

Sum = n/2 (a+l)
n = Total possible no. of numbers = 7p7 = n! = 7!
a= first number, i.e. smallest one = 1234567
l= last number, i.e. largest one = 7654321,

= (7!/2) (1234567+7654321)
= 7!/2 * 88888888 Please Provide Kudos, If you find my explanation good enough _________________
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Intern  Joined: 10 Sep 2018
Posts: 3
Re: What is the sum of all possible 7-digit numbers that can be  [#permalink]

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1
Another approach

in a sequence the sum of last number and first number divided by 2 gives the average of the sequence.
consider all the numbers in a sequence.
so the first number would be 1234567 and the largest 7654321
average=(1234567+7654321)/2=4444444

now,
let sum of the 7 digit numbers formed be n

n/7!=4444444
n=4444444*7! or 8888888/2*7! (D)
Manager  S
Joined: 16 Feb 2015
Posts: 176
Location: United States
Concentration: Finance, Operations
What is the sum of all possible 7-digit numbers that can be  [#permalink]

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one with formula directly,

Hope you understand, as I think you will not  .
Intern  B
Joined: 13 Aug 2019
Posts: 25
Location: India
Concentration: Strategy, Marketing
WE: Analyst (Retail)
Re: What is the sum of all possible 7-digit numbers that can be  [#permalink]

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Raxit85 wrote:
Yes, rocksailor9,

The question is likely feasible in GMAT exam if we correlate the sequence property here.

This type of question does not follow directly the Arithmetic properties but indirectly we can apply, i.e. The difference of each side of the AP sequence have the same difference from the mean. e.g. 123, 132, 213, 231, 312, 321 has the mean as 222. Now if you find the difference between each term with respect to mean (123, 132, 213, 222, 231, 312, 321), the first and last must have the same difference. Second last of each side must have the same difference. The same pattern follows here also. WE CAN APPLY THE SAME PROPERTY TO N NUMBER OF DIGITS. So, we can use the formula to find the sum of A.P. sequence.

Sum = n/2 (a+l)
Now, n = Total possible no. of numbers = 7p7 = n! = 7!
a= first number, i.e. smallest one = 1234567
l= last number, i.e. largest one = 7654321,

= (7!/2) (1234567+7654321)
= 7!/2 * 88888888
Ans. D.

Thanks for the solution but I have one doubt,

as you mentioned below sequence is in AP {123, 132, 213, 231, 312, 321} but here the difference between two terms (as d in AP) is not common.
As in, $$132-123 = 9$$, while $$213-132 = 81$$ and $$231-213= 18$$,
then how can we say that the given sequence is in AP and apply the formula for the sum of the terms in AP?

Please correct if I am missing something here.

Regards,
Senior Manager  P
Joined: 22 Feb 2018
Posts: 349
What is the sum of all possible 7-digit numbers that can be  [#permalink]

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rajatchopra1994 wrote:

one with formula directly,

Hope you understand, as I think you will not  .

Copied or not can be very much seen from the timing of your post editing. I am sure that you understood my message. Leave it.

Best of Luck for your GMAT preparation!!!
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Originally posted by Raxit85 on 20 Nov 2019, 22:48.
Last edited by Raxit85 on 20 Nov 2019, 22:55, edited 1 time in total.
Senior Manager  P
Joined: 22 Feb 2018
Posts: 349
Re: What is the sum of all possible 7-digit numbers that can be  [#permalink]

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Harsht7 wrote:
Raxit85 wrote:
Yes, rocksailor9,

The question is likely feasible in GMAT exam if we correlate the sequence property here.

This type of question does not follow directly the Arithmetic properties but indirectly we can apply, i.e. The difference of each side of the AP sequence have the same difference from the mean. e.g. 123, 132, 213, 231, 312, 321 has the mean as 222. Now if you find the difference between each term with respect to mean (123, 132, 213, 222, 231, 312, 321), the first and last must have the same difference. Second last of each side must have the same difference. The same pattern follows here also. WE CAN APPLY THE SAME PROPERTY TO N NUMBER OF DIGITS. So, we can use the formula to find the sum of A.P. sequence.

Sum = n/2 (a+l)
Now, n = Total possible no. of numbers = 7p7 = n! = 7!
a= first number, i.e. smallest one = 1234567
l= last number, i.e. largest one = 7654321,

= (7!/2) (1234567+7654321)
= 7!/2 * 88888888
Ans. D.

Thanks for the solution but I have one doubt,

as you mentioned below sequence is in AP {123, 132, 213, 231, 312, 321} but here the difference between two terms (as d in AP) is not common.
As in, $$132-123 = 9$$, while $$213-132 = 81$$ and $$231-213= 18$$,
then how can we say that the given sequence is in AP and apply the formula for the sum of the terms in AP?

Please correct if I am missing something here.

Regards,

Hi Harsht7,
We need to take difference with respect to mean, i.e. 222-123 = 99, 321-222 = 99, 222-132 = 90, 312-222 = 90, So, difference between first and last will be same, difference between second and second last will be same and so on.

Hope this helps.

Regards
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Manager  S
Joined: 16 Feb 2015
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Re: What is the sum of all possible 7-digit numbers that can be  [#permalink]

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Raxit85 wrote:
rajatchopra1994 wrote:

one with formula directly,

Hope you understand, as I think you will not  .

Copied or not can be very much seen from the timing of your post editing. Leave it, i am sure that you understood my message. Best of Luck for your GMAT preparation!!!

I have given the formula, not You... Editing was done for other users to explain my approach. Not for you.

Hope you are happy enough after removing my approaches
Best of Luck for your GMAT preparation!!!!!
Director  P
Joined: 16 Jan 2019
Posts: 507
Location: India
Concentration: General Management
WE: Sales (Other)
Re: What is the sum of all possible 7-digit numbers that can be  [#permalink]

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This is a fairly common question type in the GMAT and there is even a formula to solve this

1. Sum of all the numbers which can be formed by using the $$n$$ digits without repetition is: (n-1)!*(Sum of the digits)*(1111....n times)

2. Sum of all the numbers which can be formed by using the n digits with repetition being allowed is: $$n^{n-1}$$*(Sum of the digits)*(1111....n times)

In this question, we are not allowed repetition and so the answer is

$$(7-1)!*(1+2+..+7)*11111111$$

That is, $$6!*28*1111111 = 6!*7*4*1111111 = 7!*4*1111111 = \frac{7!}{2}*2*4*1111111 = \frac{7!}{2}*8888888$$ Re: What is the sum of all possible 7-digit numbers that can be   [#permalink] 20 Nov 2019, 23:22
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