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?
Sum of all 3-digit numbers
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- money9111
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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.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?
You would then do this:
24 *100 + 24* 10 + 24 *1
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- money9111
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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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I will clarify to you the problem approach.I guess you didn't understand it clearly.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.
The no. of possible such 3-digit numbers are 3! i.e. 6 numbers.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?
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 [spoiler]2400+240+24 = 2664.[/spoiler]
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.
Last edited by harsh.champ on Sat Feb 06, 2010 1:59 pm, edited 2 times in total.
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huh?
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Hmm, I begin by writing out: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.
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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- shashank.ism
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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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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?
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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You know that each digit will be in each place twice because of the combinations left for the other numbers.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.
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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The average of all possible numbers will not always be the average of the minimum and maximum number.Thouraya wrote:If I want to follow the strategy I used above, what is it that I am doing wrong?
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|.
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
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