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05 Jul 2011, 10:45
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
(N/A)
Question Stats:
88% (01:03) correct
12%
(01:53)
wrong
based on 34
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What is the sum of all the possible 3 digit numbers that can be constructed using the digit 3, 4 and 5, if each digit can be used only once in each number?
Ans:
OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/what-is-the- ... 31362.html
Ans:
2664
OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/what-is-the- ... 31362.html
05 Jul 2011, 21:16
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@enigma123 - Here is the solution to your question:
Question1: What is the sum of all the possible 3 digit numbers that can be constructed using the digit 3,4 and 5, if each digit can be used only once in each number?
(no options given)
Solution1: There is a logical method of doing such questions and I would encourage you to understand that but for this one, one could use brute force since its very straight forward. Let's look at the brute force method first and then we will go on to the logical method.
In how many ways can you make 3 digit numbers using 3,4 and 5 if you cannot repeat the numbers?
In 3! ways (arranging 3 digits in 3 places)
These numbers are
345
354
435
453
534
543
Sum: 2664
This will help us understand the logical method too. Let's look at that now:
Look at the 6 numbers above. In the unit's digits, we have two 3s, two 4s and two 5s. Makes sense, right? We have a total of 6 numbers and 3 digits. So each digit will be in the unit's place in 2 numbers (since no digit is special!). Similarly, each digit will be in the ten's place in 2 numbers and in hundred's place in 2 numbers.
The addition will be:
Unit's digits: \(3*2 + 4*2 + 5*2\)
Ten's digits: \(3*2 + 4*2 + 5*2\)
Hundred's digits: \(3*2 + 4*2 + 5*2\)
Total Sum of the numbers: \(3*2 + 4*2 + 5*2 + 10(3*2 + 4*2 + 5*2) + 100(3*2 + 4*2 + 5*2) = 2(3+4+5)(1+10+100) = 2*12*111 = 2664\)
Let's see how to use this in another question.
Question2: What is the sum of all the possible 6 digit numbers that can be constructed using the digit 1 to 6 if each digit can be used only once in each number?
How many numbers will be there? 6! because 6 digits can be arranged in 6 places in 6! ways i.e. 720 ways
So you will have 720 numbers to add.
In 720/6 = 120 numbers, unit's digit will be 1, in another 120 numbers it will be 2 and so on...
Total Sum = 120*(1+2+3+4+5+6)*111111
Question1: What is the sum of all the possible 3 digit numbers that can be constructed using the digit 3,4 and 5, if each digit can be used only once in each number?
(no options given)
Solution1: There is a logical method of doing such questions and I would encourage you to understand that but for this one, one could use brute force since its very straight forward. Let's look at the brute force method first and then we will go on to the logical method.
In how many ways can you make 3 digit numbers using 3,4 and 5 if you cannot repeat the numbers?
In 3! ways (arranging 3 digits in 3 places)
These numbers are
345
354
435
453
534
543
Sum: 2664
This will help us understand the logical method too. Let's look at that now:
Look at the 6 numbers above. In the unit's digits, we have two 3s, two 4s and two 5s. Makes sense, right? We have a total of 6 numbers and 3 digits. So each digit will be in the unit's place in 2 numbers (since no digit is special!). Similarly, each digit will be in the ten's place in 2 numbers and in hundred's place in 2 numbers.
The addition will be:
Unit's digits: \(3*2 + 4*2 + 5*2\)
Ten's digits: \(3*2 + 4*2 + 5*2\)
Hundred's digits: \(3*2 + 4*2 + 5*2\)
Total Sum of the numbers: \(3*2 + 4*2 + 5*2 + 10(3*2 + 4*2 + 5*2) + 100(3*2 + 4*2 + 5*2) = 2(3+4+5)(1+10+100) = 2*12*111 = 2664\)
Let's see how to use this in another question.
Question2: What is the sum of all the possible 6 digit numbers that can be constructed using the digit 1 to 6 if each digit can be used only once in each number?
How many numbers will be there? 6! because 6 digits can be arranged in 6 places in 6! ways i.e. 720 ways
So you will have 720 numbers to add.
In 720/6 = 120 numbers, unit's digit will be 1, in another 120 numbers it will be 2 and so on...
Total Sum = 120*(1+2+3+4+5+6)*111111
General Discussion
05 Jul 2011, 11:07
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The total such numbers possible = 6. Each of 3,4 and 5 will appear twice in the hundredth, tenth and unit place.
Sum = 2( ( 300 + 400+500) + (30+40+50) +(3+4+5)) = 2664
Sum = 2( ( 300 + 400+500) + (30+40+50) +(3+4+5)) = 2664
