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# A four digit number using the digits 3,4,6 and 7 without repetition

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Manager
Joined: 15 Dec 2015
Posts: 120
GMAT 1: 660 Q46 V35
GPA: 4
WE: Information Technology (Computer Software)
A four digit number using the digits 3,4,6 and 7 without repetition [#permalink]

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31 Jul 2017, 14:05
11
00:00

Difficulty:

95% (hard)

Question Stats:

47% (01:38) correct 53% (07:57) wrong based on 135 sessions

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A four digit number using the digits 3,4,6 and 7 without repetition is formed.All the possible numbers so formed are added and the sum is equal to P, which of the following is the value of P?

A.133200
B.123320
C.133220
D.133320
E.None of these.
Math Expert
Joined: 02 Aug 2009
Posts: 5890
Re: A four digit number using the digits 3,4,6 and 7 without repetition [#permalink]

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31 Jul 2017, 18:58
5
3
DH99 wrote:
A four digit number using the digits 3,4,6 and 7 without repetition is formed.All the possible numbers so formed are added and the sum is equal to P, which of the following is the value of P?

A.133200
B.123320
C.133220
D.133320
E.None of these.

hi..
what will be the SUM of digits in each..
in each position, each of 3,4,6,7 will stay for 3! ways...WHY?
say 3 is in units digit... how many such number are possible..
thousands digit.. any of 3
humdreds digit.. any of remaining 2
tens digit remaining one
units digit.. digit 3

so $$3*2*1=3!=6$$..

so SUM is $$6*(3+4+6+7)=120..$$

so number = 120*thousand+120*hundred+120*ten+120*one=120*1000+120*100+120*10+120*1=133320
D
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VP
Joined: 07 Dec 2014
Posts: 1018
A four digit number using the digits 3,4,6 and 7 without repetition [#permalink]

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31 Jul 2017, 20:55
1
1
DH99 wrote:
A four digit number using the digits 3,4,6 and 7 without repetition is formed.All the possible numbers so formed are added and the sum is equal to P, which of the following is the value of P?

A.133200
B.123320
C.133220
D.133320
E.None of these.

least value=3,467
greatest value=7,643
3,467+7,643=11,110
11,110/2=5,555
4*6*5,555=133,320
D
Manager
Joined: 15 Dec 2015
Posts: 120
GMAT 1: 660 Q46 V35
GPA: 4
WE: Information Technology (Computer Software)
A four digit number using the digits 3,4,6 and 7 without repetition [#permalink]

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31 Jul 2017, 22:38
1
chetan2u wrote:
DH99 wrote:
A four digit number using the digits 3,4,6 and 7 without repetition is formed.All the possible numbers so formed are added and the sum is equal to P, which of the following is the value of P?

A.133200
B.123320
C.133220
D.133320
E.None of these.

hi..
what will be the SUM of digits in each..
in each position, each of 3,4,6,7 will stay for 3! ways...WHY?
[highlight]say 3 is in units digit... how many such number are possible..

thousands digit.. any of 3
humdreds digit.. any of remaining 2
tens digit remaining one
units digit.. digit 3
[/highlight]
so $$3*2*1=3!=6$$..

so SUM is $$6*(3+4+6+7)=120..$$

so number = 120*thousand+120*hundred+120*ten+120*one=120*1000+120*100+120*10+120*1=133320
D

Hi chetan2u: Can you please elaborate the highlighted part.Thank you
Manager
Joined: 18 Jun 2017
Posts: 60
Re: A four digit number using the digits 3,4,6 and 7 without repetition [#permalink]

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02 Aug 2017, 05:05
gracie wrote:
DH99 wrote:
A four digit number using the digits 3,4,6 and 7 without repetition is formed.All the possible numbers so formed are added and the sum is equal to P, which of the following is the value of P?

A.133200
B.123320
C.133220
D.133320
E.None of these.

least value=3,467
greatest value=7,643
3,467+7,643=11,110
11,110/2=5,555
4*6*5,555=133,320
D

4*6*5,555=133,320
How do you come to the conclusion on the multiplication factors 4*6
Intern
Joined: 15 Jun 2013
Posts: 47
Schools: Ivey '19 (I)
GMAT 1: 690 Q49 V35
GPA: 3.82
Re: A four digit number using the digits 3,4,6 and 7 without repetition [#permalink]

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02 Aug 2017, 11:50
DH99 wrote:
chetan2u wrote:
DH99 wrote:
A four digit number using the digits 3,4,6 and 7 without repetition is formed.All the possible numbers so formed are added and the sum is equal to P, which of the following is the value of P?

A.133200
B.123320
C.133220
D.133320
E.None of these.

hi..
what will be the SUM of digits in each..
in each position, each of 3,4,6,7 will stay for 3! ways...WHY?
[highlight]say 3 is in units digit... how many such number are possible..

thousands digit.. any of 3
humdreds digit.. any of remaining 2
tens digit remaining one
units digit.. digit 3
[/highlight]
so $$3*2*1=3!=6$$..

so SUM is $$6*(3+4+6+7)=120..$$

so number = 120*thousand+120*hundred+120*ten+120*one=120*1000+120*100+120*10+120*1=133320
D

Hi chetan2u: Can you please elaborate the highlighted part.Thank you

Why? it is part of combinatorics. You have 3 digits (after one of 3,4, 6,7 is chosen) and all possible combinations for these digits in one number is 3!.
VP
Joined: 07 Dec 2014
Posts: 1018
A four digit number using the digits 3,4,6 and 7 without repetition [#permalink]

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02 Aug 2017, 13:05
FB2017 wrote:
gracie wrote:
DH99 wrote:
A four digit number using the digits 3,4,6 and 7 without repetition is formed.All the possible numbers so formed are added and the sum is equal to P, which of the following is the value of P?

A.133200
B.123320
C.133220
D.133320
E.None of these.

least value=3,467
greatest value=7,643
3,467+7,643=11,110
11,110/2=5,555
4*6*5,555=133,320
D

4*6*5,555=133,320
How do you come to the conclusion on the multiplication factors 4*6

hi FB2017,
I put 3 in the first position and counted 6 possible arrangements for 4,6,7.
hence, 4*6
I hope this helps
gracie
Manager
Joined: 07 Jun 2017
Posts: 103
Re: A four digit number using the digits 3,4,6 and 7 without repetition [#permalink]

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04 Aug 2017, 03:22
gracie wrote:

hi FB2017,
I put 3 in the first position and counted 6 possible arrangements for 4,6,7.
hence, 4*6
I hope this helps
gracie

Dear Gracie,
I understand why you time 6
but I don't understand why you time 4?
Thank you very much
Intern
Joined: 17 Mar 2017
Posts: 9
GMAT 1: 510 Q47 V15
GMAT 2: 600 Q49 V22
A four digit number using the digits 3,4,6 and 7 without repetition [#permalink]

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07 Aug 2017, 03:13
each number will be represented 6 times (for the unit, tenth , hundred etc..)

Sum of unit Digit: 6 (7+6+4+3)=120
Sum of tenth Digit: 6 (7+6+4+3)=120
Sum of hundred Digit: 6 (7+6+4+3)=120
...

..120000
+..12000
+....1200
+......120
---------
...133320
VP
Joined: 07 Dec 2014
Posts: 1018
A four digit number using the digits 3,4,6 and 7 without repetition [#permalink]

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07 Aug 2017, 10:23
pclawong wrote:
gracie wrote:

hi FB2017,
I put 3 in the first position and counted 6 possible arrangements for 4,6,7.
hence, 4*6
I hope this helps
gracie

Dear Gracie,
I understand why you time 6
but I don't understand why you time 4?
Thank you very much

hi pclawong,
I multiplied by 4 because there are 4 digits,
each giving 6 possible arrangements of the other 3
I hope this helps
gracie
Intern
Joined: 11 Oct 2016
Posts: 2
GMAT 1: 720 Q49 V38
GPA: 3.5
Re: A four digit number using the digits 3,4,6 and 7 without repetition [#permalink]

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07 Aug 2017, 17:35
hey gracie, why did you divide 11110/2?
VP
Joined: 07 Dec 2014
Posts: 1018
A four digit number using the digits 3,4,6 and 7 without repetition [#permalink]

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07 Aug 2017, 19:26
1
aspreneur wrote:
hey gracie, why did you divide 11110/2?

hi aspreneur,

to find the arithmetic mean of the the lowest and highest value combinations,
to multiply it by the number of possible combinations
I hope this helps
gracie
Intern
Joined: 09 Apr 2017
Posts: 2
Location: India
WE: Consulting (Real Estate)
Re: A four digit number using the digits 3,4,6 and 7 without repetition [#permalink]

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26 Aug 2017, 10:20
DH99 wrote:
A four digit number using the digits 3,4,6 and 7 without repetition is formed.All the possible numbers so formed are added and the sum is equal to P, which of the following is the value of P?

A.133200
B.123320
C.133220
D.133320
E.None of these.

Here is a formula:
(n-1)!*(Sum of n digits)*(1111...n times)
Thus, in this case: n = 4, Sum of 4 digits = 20; therefore Total Sum = 3!*20*1111 which is equal to 133320
Senior Manager
Joined: 12 Feb 2015
Posts: 259
A four digit number using the digits 3,4,6 and 7 without repetition [#permalink]

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22 May 2018, 19:47
Total possible numbers are 4!=24
Each of the 4 digits repeat 6 times in units, tens, hundreds and thousands place:-
(3333 + 4444 + 6666 + 7777) * 6

= 1111 (3 + 4 + 6 + 7) * 6

= 1111 (20) * 6

= 22220 * 6

= 133320 (Ans D)
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Manish

"Only I can change my life. No one can do it for me"

Senior Manager
Joined: 12 Feb 2015
Posts: 259
Re: A four digit number using the digits 3,4,6 and 7 without repetition [#permalink]

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22 May 2018, 19:58
Yogesh_24 wrote:
DH99 wrote:
A four digit number using the digits 3,4,6 and 7 without repetition is formed.All the possible numbers so formed are added and the sum is equal to P, which of the following is the value of P?

A.133200
B.123320
C.133220
D.133320
E.None of these.

Here is a formula:
(n-1)!*(Sum of n digits)*(1111...n times)
Thus, in this case: n = 4, Sum of 4 digits = 20; therefore Total Sum = 3!*20*1111 which is equal to 133320

Here is the derivation of the formula:-
Total possible numbers are 4!=24
Each of the 4 digits repeat 6 times in units, tens, hundreds and thousands place:-
(3333 + 4444 + 6666 + 7777) * 6

= 1111 * (3 + 4 + 6 + 7) * 6

= 1111 * (20) * 6 [your formula: (n-1)!*(Sum of n digits)*(1111...n times) where n = 4, Sum of 4 digits = 20]

= 22220 * 6

= 133320 (Ans D)
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Manish

"Only I can change my life. No one can do it for me"

Manager
Joined: 01 Aug 2017
Posts: 135
Location: India
Schools: ISB '20 (S), IIMA (S)
GMAT 1: 500 Q47 V15
GPA: 3.4
WE: Information Technology (Computer Software)
Re: A four digit number using the digits 3,4,6 and 7 without repetition [#permalink]

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22 May 2018, 22:03
DHAR wrote:
A four digit number using the digits 3,4,6 and 7 without repetition is formed.All the possible numbers so formed are added and the sum is equal to P, which of the following is the value of P?

A.133200
B.123320
C.133220
D.133320
E.None of these.

If all possible n digit numbers using n distinct non-zero digits are formed, sum of all the numbers so formed
=(n−1)! × (sum of the all the digits in numbers ) × (111 ... n times).

It is a 4 digit number . n =4

Just substitute the value.
= (4-1)! * (3+4+6+7) * (1111)
= 3! * 20 * 1111
= 6*20*1111
= 133320.

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Re: A four digit number using the digits 3,4,6 and 7 without repetition   [#permalink] 22 May 2018, 22:03
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