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We make 4 digit codes and each digit of the code form from 1

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We make 4 digit codes and each digit of the code form from 1

by Max@Math Revolution » Tue Jan 26, 2016 9:10 pm
We make 4 digit codes and each digit of the code form from 1, 2, 3, and 4. If none of each digit use more than once, eg, 1234 can be a code but 1124 cannot be a code, what is the sum of all the possible codes?

A. 56,660
B. 58,660
C. 60,660
D. 66,660
E. 68,660


* A solution will be posted in two days.
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by GMATinsight » Wed Jan 27, 2016 1:15 am
Max@Math Revolution wrote:We make 4 digit codes and each digit of the code form from 1, 2, 3, and 4. If none of each digit use more than once, eg, 1234 can be a code but 1124 cannot be a code, what is the sum of all the possible codes?

A. 56,660
B. 58,660
C. 60,660
D. 66,660
E. 68,660


* A solution will be posted in two days.
Total Numbers that can be made by using digits 1,2,3,4 without repetition of any digit in a number = 4*3*2*1 = 4! = 24

Now Since every digit has equal chances to appear at any place out of 4 places so each digit will be used at each of the four places an equal number of times

i.e. each digit will be used at Unit digit = 24/4 = 6 times

i.e. 6 numbers will have 1 as unit digit, 6 numbers will have 2 as unit digit, 6 numbers will have 3 as unit digit and 6 numbers will have 4 as unit digit

i.e. Sum of Unit digits of the numbers = 6*(1+2+3+4) = 60

i.e. Number will be _ _ _ 0 with 6 as carry over on ten's place

Now sum of the tens digit will again be 60 and adding carry over will make it 66

i.e. Number will be _ _ 6 0 with 6 as carry over on Hundred's place

Now sum of the Hundreds digit will again be 60 and adding carry over will make it 66

i.e. Number will be _ 6 6 0 with 6 as carry over on Thousand's place

Now sum of the Thousand's digits will again be 60 and adding carry over will make it 66

i.e. Number will be 66 6 6 0
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by GMATGuruNY » Wed Jan 27, 2016 3:30 am
Max@Math Revolution wrote:We make 4 digit codes and each digit of the code form from 1, 2, 3, and 4. If none of each digit use more than once, eg, 1234 can be a code but 1124 cannot be a code, what is the sum of all the possible codes?

A. 56,660
B. 58,660
C. 60,660
D. 66,660
E. 68,660
Number of options for the thousands place = 4. (Any of the 4 digits.)
Number of options for the hundreds place = 3. (Any of the 3 remaining digits.)
Number of options for the tens place = 2. (Any of the 2 remaining digits.)
Number of options for the units place = 1. (Only 1 digit left.)
To combine these options, we multiply:
Total number of codes = 4*3*2*1 = 24.

Each digit will appear in each position 24/4 = 6 times.
Thus, in each position, there will be six 1's, six 2's, six 3's, and six 4's.
Sum of the digits in each position = 6(1+2+3+4) = 60.

Sum of the thousands place = 60*1000 = 60,000.
Sum of the hundreds place =60*100 =6,000.
Sum of the tens place = 60*10 = 600.
Sum of the units place = 60*1 = 60.

Sum of all the codes = 60000 + 6000 + 600 + 60 = 66,660.

The correct answer is D.
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by sitar » Wed Jan 27, 2016 4:17 am
One more way -
Average of set = median of set = (1234+4321)/2 = 5555/2
Possible number of ways in which these 4 numbers can be arranged = 4*3*2*1 = 24
So sum of all of these numbers = average of the set multiplied by number of way of arrangement = 5555/2 * 24
= 66,660
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by Max@Math Revolution » Thu Jan 28, 2016 5:50 pm
We make 4 digit codes and each digit of the code form from 1, 2, 3, and 4. If none of each digit use more than once, eg, 1234 can be a code but 1124 cannot be a code, what is the sum of all the possible codes?

A. 56,660 B. 58,660 C. 60,660 D. 66,660 E. 68,660

--> (1,000*6+2,000*6+3,000*6+4,000*6)+(100*6+200*6+300*6+400*6)
+(10*6+20*6+30*6+40*6)+(1*6+2*6+3*6+4*6)
=10,000*6+1,000*6+100*6+10*6
=66,660

Therefore, the answer is D.
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by talaangoshtari » Thu Jan 28, 2016 9:15 pm
sitar wrote:One more way -
Average of set = median of set = (1234+4321)/2 = 5555/2
Possible number of ways in which these 4 numbers can be arranged = 4*3*2*1 = 24
So sum of all of these numbers = average of the set multiplied by number of way of arrangement = 5555/2 * 24
= 66,660
Hi sitar,

How did you find the median of the set? For example, I consider this set:

1234
1243
1324
1342
1423
1432

median of set = (1324+1342)/2 = 2666/2 = 1333
1333*24 = 31992

I do not think that this set is evenly spaced actually...
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by sitar » Fri Jan 29, 2016 3:00 am
talaangoshtari wrote:
sitar wrote:One more way -
Average of set = median of set = (1234+4321)/2 = 5555/2
Possible number of ways in which these 4 numbers can be arranged = 4*3*2*1 = 24
So sum of all of these numbers = average of the set multiplied by number of way of arrangement = 5555/2 * 24
= 66,660
Hi sitar,

How did you find the median of the set? For example, I consider this set:

1234
1243
1324
1342
1423
1432

median of set = (1324+1342)/2 = 2666/2 = 1333
1333*24 = 31992

I do not think that this set is evenly spaced actually...

Hi Talaangoshtari,

Median of set is middle number(if odd) and average of middle 2 number(if even number in set)
Median of a set of consecutive numbers is average of smallest and biggest number in the set. Smallest possible number with 4 digit code is 1234. Biggest possible number with 4 digit is 4321. Thus median = average of 1234 and 4321.
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by talaangoshtari » Fri Jan 29, 2016 3:14 am
Yes, I know... But the problem is that this set is not evenly spaced. Therefore we cannot say that the mean is equal to the median.

1243 - 1234 = 9
1324 - 1243 = 81
1342 - 1324 = 18
1423 - 1324 = 99
1432 - 1423 = 9

Why do you consider this set the set of consecutive numbers?
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by GMATinsight » Fri Jan 29, 2016 4:14 am
Max@Math Revolution wrote:We make 4 digit codes and each digit of the code form from 1, 2, 3, and 4. If none of each digit use more than once, eg, 1234 can be a code but 1124 cannot be a code, what is the sum of all the possible codes?

A. 56,660
B. 58,660
C. 60,660
D. 66,660
E. 68,660


* A solution will be posted in two days.
Alternate Method:

The biggest Number using these digits = 4321
The Second Biggest Number Using these digits = 4312
The Third Biggest Number Using these digits = 4231
and so on...


The smallest Number using these digits = 1234
The Second smallest Number Using these digits = 1243
The Third smallest Number Using these digits = 1324
and so on...

Sum of Biggest and Smallest Numbers = 4321+1234 = 5555
Sum of Second Biggest and Second Smallest Numbers = 4312+1243 = 5555
Sum of Third Biggest and Third Smallest Numbers = 4231+1324 = 5555
and so on...

I.e. each pair of numbers from extreme ends is ending with a sum = 5555
Total Such numbers = 4*3*2*1 = 4! = 24
i.e. Total Pairs = 24/2 = 12

i.e. Sum of all pairs = 12*5555 = 66660

Answer: Option D
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by GMATGuruNY » Fri Jan 29, 2016 5:03 am
talaangoshtari wrote:
sitar wrote:One more way -
Average of set = median of set = (1234+4321)/2 = 5555/2
Possible number of ways in which these 4 numbers can be arranged = 4*3*2*1 = 24
So sum of all of these numbers = average of the set multiplied by number of way of arrangement = 5555/2 * 24
= 66,660
Hi sitar,

How did you find the median of the set? For example, I consider this set:

1234
1243
1324
1342
1423
1432

median of set = (1324+1342)/2 = 2666/2 = 1333
1333*24 = 31992

I do not think that this set is evenly spaced actually...
The set is composed of integers in the following ranges:
1234...1432
2134...2431
3124...3421
4123...4321.

Each range contains the same number of integers.
Thus, the median of the set is equal to the average of the two integers in red:
(2431 + 3124)/2 = 5555/2.

The set is SYMMETRICAL ABOUT THE MEDIAN:
...2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241...

For any set symmetrical about the median:
average = median
sum = (number of integers)(average).

As shown in my post above, the set is composed of 24 integers.
Thus:
sum = (24)(5555/2) = (12)(5555) = (12)(5000 + 500 + 50 + 5) = 60000 + 6000 + 600 + 60 = 66660.

The correct answer is D.

For a similar problem that can be solved with the same line of reasoning, check my post here:
https://www.beatthegmat.com/on-consecuti ... 85395.html
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