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

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

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New post 31 Jul 2017, 14:05
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Question Stats:

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

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New post 31 Jul 2017, 18:58
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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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A four digit number using the digits 3,4,6 and 7 without repetition [#permalink]

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New post 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
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A four digit number using the digits 3,4,6 and 7 without repetition [#permalink]

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New post 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
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Re: A four digit number using the digits 3,4,6 and 7 without repetition [#permalink]

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New post 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
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Re: A four digit number using the digits 3,4,6 and 7 without repetition [#permalink]

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New post 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!.
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A four digit number using the digits 3,4,6 and 7 without repetition [#permalink]

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New post 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
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Re: A four digit number using the digits 3,4,6 and 7 without repetition [#permalink]

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New post 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
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A four digit number using the digits 3,4,6 and 7 without repetition [#permalink]

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New post 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
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A four digit number using the digits 3,4,6 and 7 without repetition [#permalink]

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New post 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
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Re: A four digit number using the digits 3,4,6 and 7 without repetition [#permalink]

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

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New post 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
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Re: A four digit number using the digits 3,4,6 and 7 without repetition [#permalink]

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New post 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
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A four digit number using the digits 3,4,6 and 7 without repetition [#permalink]

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New post 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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Re: A four digit number using the digits 3,4,6 and 7 without repetition [#permalink]

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New post 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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Re: A four digit number using the digits 3,4,6 and 7 without repetition [#permalink]

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New post 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.

Answer D
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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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