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# We make 4 digit codes and each digit of the code form from 1, 2, 3, an

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Math Revolution GMAT Instructor
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We make 4 digit codes and each digit of the code form from 1, 2, 3, an  [#permalink]

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26 Jan 2016, 21:07
2
15
00:00

Difficulty:

55% (hard)

Question Stats:

63% (02:23) correct 37% (02:23) wrong based on 117 sessions

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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

* A solution will be posted in two days.

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Re: We make 4 digit codes and each digit of the code form from 1, 2, 3, an  [#permalink]

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27 Jan 2016, 01:14
3
5
MathRevolution 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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##### General Discussion
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Joined: 02 Aug 2009
Posts: 7108
Re: We make 4 digit codes and each digit of the code form from 1, 2, 3, an  [#permalink]

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26 Jan 2016, 22:00
1
1
MathRevolution 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.

Hi,

its a 4 digit code from four different digits..
so each digit will come equal number of times in each digits place when all codes are attached.

so in thousnds place, hundreds place, tens and units place, the sum would be = (1+2+3+4)*x..
x is what we have to find ...
lets see how many times 1 will be in thousands place..
when thousands is taken by 1, the remaining three places can be occupied in 3! ways.. 6 ways..

so the number becomes=..
th....h....tens....units
60..60...60.......60....
so when we carry over the numbers..
units digit=0
tens=60+6, so tens digit=6
hundreds=60+6, so hundreds digit=6
thousands=60+6=66...
our sum=66660...
D

2) the sum has to be div by 11and 6,
as each number 1,2,3,4 is used equal number of times at four places..
only 66660 fits in
D
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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Re: We make 4 digit codes and each digit of the code form from 1, 2, 3, an  [#permalink]

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28 Jan 2016, 17:38
1
1
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

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Re: We make 4 digit codes and each digit of the code form from 1, 2, 3, an  [#permalink]

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21 Jun 2017, 00:46
Numbers to be used: 1 , 2 , 3 , 4
total number of possibilites= 4 . 3 . 2 . 1 = 24 (no repetition of any number)
We need to calculate sum of these 24 combinations of numbers.

Since there are 4 numbers (1,2,3,4) and 24 combinations of numbers, then each number will be used 24/4 =6 times in the total sum.
xxxx
yyyy (x+y+z) = 6 x 4 + 6 x 3 + 6 x 2 + 6 x 1 = 60
+zzzz at units place = 0 (6 will be carried)
at tens place = 0+6 = 6 ( 6 is from previous carry, and 6 again will be carried)
at 100s place = 6 with 6 carried forward
at 1000s place =6 with 6 carried forward
at 10,000s place = 6 from previous carry

ANS = 66660
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Re: We make 4 digit codes and each digit of the code form from 1, 2, 3, an  [#permalink]

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05 Sep 2017, 08:20
I feel it is a very tough question that cannot be solved within 2 min.
Has it appeared on the GMAT or is it that only I am finding it difficult.
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Re: We make 4 digit codes and each digit of the code form from 1, 2, 3, an  [#permalink]

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08 Sep 2017, 04:20
souvonik2k wrote:
I feel it is a very tough question that cannot be solved within 2 min.
Has it appeared on the GMAT or is it that only I am finding it difficult.

formula =1111*sum(a+b+c+d)*n-1! (in this case n-3)
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Re: We make 4 digit codes and each digit of the code form from 1, 2, 3, an  [#permalink]

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09 Jul 2018, 19:03
1
MathRevolution 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

Since we are forming a code using 4 distinct digits, the number of codes that can be formed is 4! = 24. However, for each digit at a certain place value, it can be at that place value 6 times. For example, at the thousands place, 1 can only appear 6 times. That is because 2, 3 and 4 will also appear 6 times each at the the thousands place so that there are a total of 24 numbers. Therefore, the following must be true:

1, 2, 3 and 4 each appears:

i) at the thousands place 6 times with a total value of 6 x (1+2+3+4) x 1000 = 60 x 1000 = 60,000.

ii) at the hundreds place 6 times with a total value of 6 x (1+2+3+4) x 100 = 60 x 100 = 6,000.

iii) at the tens place 6 times with a total value of 6 x (1+2+3+4) x 10 = 60 x 10 = 600.

iv) at the ones place 6 times with a total value of 6 x (1+2+3+4) x 1 = 60 x 1 = 60.

So the sum of these 24 codes (or numbers) must be 60,000 + 6,000 + 600 + 60 = 66,660.

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Re: We make 4 digit codes and each digit of the code form from 1, 2, 3, an  [#permalink]

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09 Jul 2018, 22:29
Can somebody please tell me if my approach is correct?

total combinations possible - 4! = 24; with one number appearing only once, I can say that we will get 12 complementing pairs (eg. 1234 and 4321)

we add 1234 and 4321 to get 5555 and multiply it by 12, we get 66660.

Re: We make 4 digit codes and each digit of the code form from 1, 2, 3, an &nbs [#permalink] 09 Jul 2018, 22:29
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