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Algebra

New kid on the Block

 Joined: 11 May 2009 Post: 4
3 Jun 2009 11:57:22 IST
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find the sum of all four digit numbers that can be formed with the digits o,1,2,3...... answer wit
Engineering Entrance , JEE Main , JEE Advanced , Mathematics , Algebra

find the sum of all four digit numbers that can be formed with the digits o,1,2,3......answer with steps please

Cool goIITian

Joined: 29 May 2009
Posts: 84
3 Jun 2009 14:48:29 IST
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(1+2+3+4+5+6+7+8+9)*10*10*9  =  Addition of all units places

(1+2+3+4+5+6+7+8+9)*10*10*9  =  Addition of all tens places

(1+2+3+4+5+6+7+8+9)*10*10*9  =  Addition of all hundreds places

(1+2+3+4+5+6+7+8+9)*10*10*9  =  Addition of all thousands places

Now,

Addition of all units places = 40500

Addition of all tens places = 40500

Addition of all hundreds places = 40500

Addition of all thousands places = 45000

So the ans wer is

494955

If everything is correct than I am right but check it because I solved it quite quickly with less precautions.

(Sorry if it wrong and please inform me, I willcorrect it.)

Cool goIITian

Joined: 29 May 2009
Posts: 84
6 Jun 2009 16:09:06 IST
0 people liked this

Sorr there was a minute mistake in the above post.

(1+2+3+4+5+6+7+8+9)*10*10*9  =  Addition of all units places

(1+2+3+4+5+6+7+8+9)*10*10*9  =  Addition of all tens places

(1+2+3+4+5+6+7+8+9)*10*10*9  =  Addition of all hundreds places

(1+2+3+4+5+6+7+8+9)*10*10*9  =  Addition of all thousands places

Now,

Addition of all units places = 40500

Addition of all tens places = 40500

Addition of all hundreds places = 40500

Addition of all thousands places = 45000

So the ans wer is

49495500

If everything is correct than I am right but check it because I solved it quite quickly with less precautions.

(Sorry if it wrong and please inform me, I willcorrect it.)

New kid on the Block

Joined: 17 Mar 2009
Posts: 9
6 Jun 2009 18:11:23 IST
0 people liked this

the total no. formed with 0,1,2,3 will be

4! =24 ,to find the sum of these 24 no., we will find the sum of digits at unit's,ten's,hundred's and thousand's places in all these no.

consider the digits in the unit's places in all these no. Each of the digit 0,1,2,3 occurs in 3! (=6) times in the unit's place.

So,total for the digits in the unit's place in all the no.

=(0+1+2+3) X 3! =36

Since each of the digit 0,1,2,3 occurs in 3! (=6) times in any one of the remaining places.

So the sum of digits in the digits in the ten's,hundred's and thousand's places in all the no.

=(0+1+2+3) X 3! =36

Hence,the sum of all the no.=36(100+101+102+103)=39996

[I wish my ans. will be liked by you]

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