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Sum of all 3-digit numbers

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money9111 Legendary Member
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Sum of all 3-digit numbers

Post Sat Feb 06, 2010 1:19 pm
Here's another:

What is the sum of all possible 3-digit numbers that can be constructed using the digits 3,4,5, if each digit can be used only once in each number?

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BarryLi Senior | Next Rank: 100 Posts Default Avatar
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Post Wed Mar 02, 2011 1:51 am
money9111 wrote:
how do you know that each can be in each place twice? did you write them out? how do you know to add 3*2, 4*2, and 5*2? it's not so much the math that gets me as much as it's knowing how to set it up.. and go about solving.
You know that each digit will be in each place twice because of the combinations left for the other numbers.

For example, 3 will be in the hundreds place twice because the digits 4 & 5 can only form two different numbers: 45 & 54. Hence the two numbers that have 3 in the hundreds column are 345 & 354.

This example can be extended to the other digits and other place values.

As a result, (3+3+4+4+5+5)*100 + (3+3+4+4+5+5)*10 + (3+3+4+4+5+5)*1 = 24*100 + 24*10 + 24*1, which is what everyone else has typed up.

Knowing how to set up the question is the more difficult part of solving a problem, not the math. This is especially true with the more difficult quantitative questions imo.

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Thouraya Master | Next Rank: 500 Posts Default Avatar
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Post Wed Mar 02, 2011 1:55 am
If I want to follow the strategy I used above, what is it that I am doing wrong?

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Post Sat Feb 06, 2010 1:23 pm
money9111 wrote:
Here's another:

What is the sum of all possible 3-digit numbers that can be constructed using the digits 3,4,5, if each digit can be used only once in each number?
each digit can be in ones place twice, the tens place twice and the hundreds place twice. You would add 3*2 + 4*2 + 5*2 = 24.

You would then do this:

24 *100 + 24* 10 + 24 *1

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Post Sat Feb 06, 2010 1:26 pm
how do you know that each can be in each place twice? did you write them out? how do you know to add 3*2, 4*2, and 5*2? it's not so much the math that gets me as much as it's knowing how to set it up.. and go about solving.

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harsh.champ Legendary Member
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Post Sat Feb 06, 2010 1:32 pm
money9111 wrote:
how do you know that each can be in each place twice? did you write them out? how do you know to add 3*2, 4*2, and 5*2? it's not so much the math that gets me as much as it's knowing how to set it up.. and go about solving.
I will clarify to you the problem approach.I guess you didn't understand it clearly.
Quote:
What is the sum of all possible 3-digit numbers that can be constructed using the digits 3,4,5, if each digit can be used only once in each number?
The no. of possible such 3-digit numbers are 3! i.e. 6 numbers.

Now,each no. can only occur for 2 times in units/tens/hundreds place.
so, for units place = 2x(3+4+5) = 24
for tens place= 2x10(3+4+5)=240
for hundreds place = 2x100(3+4+5)=2400

Hence,the sum of the digits will be 2400+240+24 = 2664.



Alternatively,since only 6 numbers are there you can even write it down and solve the question
345
354
435
453
534
543

Quickly,add the no.s and you will get the result.

But if your problem approach is clear,I found the formal method to quicker in this question.

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Last edited by harsh.champ on Sat Feb 06, 2010 1:59 pm; edited 2 times in total

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money9111 Legendary Member
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Post Sat Feb 06, 2010 1:38 pm
huh?

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Post Sat Feb 06, 2010 1:42 pm
money9111 wrote:
how do you know that each can be in each place twice? did you write them out? how do you know to add 3*2, 4*2, and 5*2? it's not so much the math that gets me as much as it's knowing how to set it up.. and go about solving.
Hmm, I begin by writing out:

345
354
435
453
543
534

This gives you a visual. If you would find it faster to just add that up, then go about it that way. What you see from writing it out like that though is each digit is in each place twice. Knowing this you can add 3 4 and 5 and multiply by 2 since each digit is in each place twice. This gives you 24. You have 24 in the hundreds place the tens place and the ones place. That is why I multiplied 24 by 100, 10 and 1.

I would recommend doing the approach that seems natural. You can add those numbers in less than a minute, so knowing the Manhattan way isn't necessary for this problem.

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Post Sat Feb 06, 2010 1:56 pm
Quote:
money9111 wrote:
how do you know that each can be in each place twice? did you write them out? how do you know to add 3*2, 4*2, and 5*2? it's not so much the math that gets me as much as it's knowing how to set it up.. and go about solving
see here we have three numbers 3,4,5 now if we take 1no. say 3 we have 2 left nos. i.e. 4 & 5 which can be placed in two ways as 45 and 54
similarly taking 4 and 5 at hundreds place , you can easily observe each number is going to repeat 2 times at 100's 10's and unit place
so the sum of all numbers which can be formed= (3+4+5)x2x100+ (3+4+5)x2x10+ (3+4+5)x2x1 =24 x(100+10+1) = 24x111 = 2664.

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Post Wed Mar 02, 2011 1:13 am
What is the sum of all possible 3-digit numbers that can be constructed using the digits 3,4,5, if each digit can be used only once in each number?

Here's how I solved it:

Number of integers: 543-345/1+1=199
Average: 345+543/2=444
Sum= 444*199=88,356

What's the correct answer?

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BarryLi Senior | Next Rank: 100 Posts Default Avatar
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Post Wed Mar 02, 2011 2:02 am
Thouraya wrote:
If I want to follow the strategy I used above, what is it that I am doing wrong?
The average of all possible numbers will not always be the average of the minimum and maximum number.

For example, 1, 2, and 5.

Min: 125
Max: 521
Average of the Min and Max: 323



List of Numbers:
125
152
215
251
512
521

Average of the List of Numbers:296


I believe the method using the min and max only worked in the original poster's case because the absolute difference of the 1st and 3rd digit from the median digit are equal, i.e. |4-3| = |4-5|.

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Post Wed Mar 02, 2011 4:39 am
You're right, the mistake I did here is that I assume that the integers are consecutive, while they are not..and the theory that average of sequence is equal to average of smallest and biggest only holds true in CONSECUTIVE integers..

Do you have an alternative approach to the one being presented above? I usually like to follow systematic approaches, rather than figure the pattern or think of the "logic" behind it. Any other approaches would greatly be appreciated. Thanks

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