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.
We make 4 digit codes and each digit of the code form from 1
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- Max@Math Revolution
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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! = 24Max@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.
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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Number of options for the thousands place = 4. (Any of the 4 digits.)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 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.
Last edited by GMATGuruNY on Tue Jun 11, 2019 10:59 am, edited 1 time in total.
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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
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
- Max@Math Revolution
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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.
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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- talaangoshtari
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Hi sitar,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
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...
talaangoshtari wrote:Hi sitar,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
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.
- talaangoshtari
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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?
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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Alternate Method: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.
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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The set is composed of integers in the following ranges:talaangoshtari wrote:Hi sitar,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
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...
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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