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

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31 Jul 2017, 14:05

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54% (09:25) wrong based on 108 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.

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
_________________

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

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31 Jul 2017, 20:55

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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.

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

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

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31 Jul 2017, 22:38

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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

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

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!.

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

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

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07 Aug 2017, 19:26

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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

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

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