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What is the sum of all integers greater than 10000 formed by using the

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Bunuel
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What is the sum of all integers greater than 10000 formed by using the [#permalink] New post   04 Dec 2020, 06:45
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What is the sum of all integers greater than 10000 formed by using the digits 0, 2, 4, 6, 8 no digit being repeated in any number?

A. 133,020
B. 133,320
C. 5,199,960
D. 5,333,280
E. 52,000,060

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Find the sum of all numbers greater than 10000 formed by using the digits 0,2,4,6,8 no digit being repeated in any number.


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chetan2u
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What is the sum of all integers greater than 10000 formed by using the [#permalink] New post   04 Dec 2020, 08:05
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Bunuel wrote:
What is the sum of all integers greater than 10000 formed by using the digits 0, 2, 4, 6, 8 no digit being repeated in any number?

A. 133,020
B. 133,320
C. 5,199,960
D. 5,333,280
E. 52,000,060




Let the number be in the form ABCDE
Sum of all the numbers that can be formed by using the n digits without any digit repeated is: \((n-1)!*(sum \ of \ the \ digits)*(111... \ n \ times)\).
First let us take all ways without any restriction
\(=> (5-1)!(0+2+4+6+8)*11111=24*20*11111=5333280\)
Now, to get numbers less than 10000, let us subtract ways in which A is 0, so the number becomes BCDE
\(=> (4-1)!(2+4+6+8)*1111=6*20*1111=133320\)

Our answer = \(5333280-133320=5,199,960\)

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Re: What is the sum of all integers greater than 10000 formed by using the [#permalink] New post   05 Dec 2020, 12:10
[quote="chetan2u"][quote="Bunuel"] why did you remove 0 ?? of course if we were to remove another number result would change but I did not get the logic behind it.
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Re: What is the sum of all integers greater than 10000 formed by using the [#permalink] New post   05 Dec 2020, 22:06
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korelgundem wrote:
chetan2u wrote:
Bunuel wrote:
why did you remove 0 ?? of course if we were to remove another number result would change but I did not get the logic behind it.


We removed the cases where the first digit is 0, because all those numbers will become 4-digit and we are looking for 5-digit numbers.
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Re: What is the sum of all integers greater than 10000 formed by using the [#permalink] New post   05 Dec 2020, 22:22
chetan2u wrote:
Bunuel wrote:
What is the sum of all integers greater than 10000 formed by using the digits 0, 2, 4, 6, 8 no digit being repeated in any number?

A. 133,020
B. 133,320
C. 5,199,960
D. 5,333,280
E. 52,000,060




Let the number be in the form ABCDE
Sum of all the numbers that can be formed by using the n digits without any digit repeated is: \((n-1)!*(sum \ of \ the \ digits)*(111... \ n \ times)\).
First let us take all ways without any restriction
\(=> (5-1)!(0+2+4+6+8)*11111=24*20*11111=5333280\)
Now, to get numbers less than 10000, let us subtract ways in which A is 0, so the number becomes BCDE
\(=> (4-1)!(2+4+6+8)*1111=6*20*1111=133320\)

Our answer = \(5333280-133320=5,199,960\)

C
..Hi Chetan....can you please explain how did you derive the sum as (n-1)!*(sum of the digits)*(111... n times)?

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Re: What is the sum of all integers greater than 10000 formed by using the [#permalink] New post   05 Dec 2020, 22:56
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AntrikshR wrote:
chetan2u wrote:
Bunuel wrote:
What is the sum of all integers greater than 10000 formed by using the digits 0, 2, 4, 6, 8 no digit being repeated in any number?

A. 133,020
B. 133,320
C. 5,199,960
D. 5,333,280
E. 52,000,060




Let the number be in the form ABCDE
Sum of all the numbers that can be formed by using the n digits without any digit repeated is: \((n-1)!*(sum \ of \ the \ digits)*(111... \ n \ times)\).
First let us take all ways without any restriction
\(=> (5-1)!(0+2+4+6+8)*11111=24*20*11111=5333280\)
Now, to get numbers less than 10000, let us subtract ways in which A is 0, so the number becomes BCDE
\(=> (4-1)!(2+4+6+8)*1111=6*20*1111=133320\)

Our answer = \(5333280-133320=5,199,960\)

C
..Hi Chetan....can you please explain how did you derive the sum as (n-1)!*(sum of the digits)*(111... n times)?

Posted from my mobile device


Hi

Let us take a smaller number, say 3-digit numbers, ABC using digits 1,2 and 3.
Now for A, the other 2 places can be filled in 2! Ways, which is (3-1)! or (n-1)!
Next A will take all the values 1, 2 and 3, so when we sum such numbers we are basically getting each of 1, 2 and 3, how many times - (n-1)! times
So (1+2+3)*(n-1)!
But the same will happen at tens and ones place also, that is at B and C also.
Now A at 100s place means 100*A=100(sum of digits)(n-1)!
B at 10s place means 10B=10(sum of digits)(n-1)!
And for C, it remains (sum of digits)(n-1)!
You have to add all of them =>
(100+10+1)(sum of digits)(n-1)!=(1..n times)(sum of digits)(n-1)!
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Re: What is the sum of all integers greater than 10000 formed by using the [#permalink] New post   06 Dec 2020, 01:02
Thanks for this Chetan. This was a tough one. I was hoping to get a ball park figure when I started the question (I did not know your approach) and got about $5 million. Option C and D got me.
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Re: What is the sum of all integers greater than 10000 formed by using the [#permalink]
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