What is sum of all rearrangements of the 4-digit number 3214?

Question

Competition Mode Question

A. 66,660
B. 65,024
C. 60,048
D. 55,554
E. 55,552

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freedom128 · Accepted Answer

3214 has 4!=24 arrangements.

In all arrangements, the thousands digit is occupied by each number (1,2,3,4)  six times . 
So are the the hundreds digit, the tens digit and the unit digit.

Let's sum all arrangements 
* Sum of 1000s digit only = 6*(1+2+3+4)*1000
* Sum of 100s digit only = 6*(1+2+3+4)*100
* Sum of 10s digit only = 6*(1+2+3+4)*10
* Sum of unit digit only = 6*(1+2+3+4)*1

Sum of all arrangements  
= 6*(1+2+3+4)*(1000+100+10+1)
= 6*10*1111
= 66660

FINAL ANSWER IS (A)

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nick1816 · Answer

Solution 1.

The unit digit of sum of all rearrangements of the 4-digit number 3214 is same as that of 6*(3+2+1+4), that is 0, as the probability of each digit appera at unit place is same.

A

Solution 2.

Total possible rearrangements = 4! = 24

Number of times 3,2,1 or 4 appear ar each position = 24/4 = 6

Sum of all rearrangements = 1111*  = 66660

Archit3110 · Answer

total ways to arrange 3214 ; 4! ; 24
and units digit sum ; 3+2+1+4 ; 0 ; which times 6 as the units digit would be placed 6 times ; would give as 0 in unit place
so option A is the one with 0 at unit place
 66,660

What is sum of all rearrangements of the 4-digit number 3214?

A. 66,660
B. 65,024
C. 60,048
D. 55,554
E. 55,552

unraveled · Answer

What is sum of all rearrangements of the 4-digit number 3214?

A. 66,660
B. 65,024
C. 60,048
D. 55,554
E. 55,552

Total arrangements would be 4! since there are 4 digits. But it would be highly cumbersome. Let's try checking units digit.
All the digits occur 6 times at units place hence sum of all the units digits 6*4 + 6*3 + 6*2 + 6*1 = 4 + 8 + 2 + 6 = 20 i.e. final units digits is '0'.

Detailed Solution:
In the rearrangements every digit would take place thousand's place. So there are four ways(4 digits) that thousand's place can be taken:
1. 1xxx - rest of the digits(2,3,4) can combine in 3! ways and each of them takes place the unit digit twice. As 2,3,4 sum to 9, unit digit of all the arrangements with 1 at thousand's place is the unit digit of 9*2 = 18 i.e. 8

2. 2xxx - rest of the digits(1,3,4) can combine in 3! ways and each of them takes place the unit digit twice. As 1,3,4 sum to 8, unit digit of all the arrangements with 2 at thousand's place is the unit digit of 8*2 = 16 i.e. 6

3. 3xxx - rest of the digits(1,2,4) can combine in 3! ways and each of them takes place the unit digit twice. As 1,2,4 sum to 7, unit digit of all the arrangements with 1 at thousand's place is the unit digit of 7*2 = 14 i.e. 4

4. 4xxx - rest of the digits(1,2,3) can combine in 3! ways and each of them takes place the unit digit twice. As 1,2,3 sum to 6, unit digit of all the arrangements with 1 at thousand's place is the unit digit of 6*2 = 12 i.e. 2
 
Adding all the resultant units digits 8+6+4+2 = 20, we have '0' as final unit's digits
Only option A has '0' as unit digit.

Answer A.

Note: A 10 second answer would be that digits in 3214 repeat uniformly(all occur at each place equally) and sum to 10 the unit digit of all the arrangements will always be '0'.

MayankSingh · Answer

What is sum of all rearrangements of the 4-digit number 3214?

Digits= { 1,2,3,4}
Now, Four digit number be XYZW

Let X=1, then Number with X=1 as thousand digit= 3x2x1 = 6
Similarly for X=2, 3, 4 total Numbers = 6
Sum of thousand digits of all possible numbers= 6 X (1+2+3+4) x 10^3 = 60 X 10^3

Similarly for hundred digit = 60 X 10^2
Tens digit = 60 x 10
Unit digit = 60

So total Sum = 60 X (10^3 +10^2+10+1)= 60 X 1111= 66660

Ans A. 66,660

exc4libur · Answer

1234, 1243, 1324, 1343...4123, 4132..
thousands: 6*(1000+2000..4000)=60000
60000<ans<=66660
each unit will repeat 6x
6(1+2+3+4)=6(10)=units=0

ans (A)

RaghavKhanna · Answer

When the digit 1 is in thousand's place,
1 _ _ _ 
No, of possible permutations = 6

When 1 is in hundred's place,
No. Of possible permutations = 6

When 1 is in ten's place,
No.of possible permutations = 6

When 1 is in one's place,
No. Of possible permutations = 6

Similarly, we will have the same no. Of permutations for each digit.

Adding everything, 1000*6*(1+2+3+4) + 100*6*(1+2+3+4) + 10*6*(1+2+3+4) + 1*6*(1+2+3+4)
= 66660

General formula = (111.. n times) * (n-1)! * (Sum of digits)

Where, n = no. Of digits

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gracie · Answer

What is sum of all rearrangements of the 4-digit number 3214?

A. 66,660
B. 65,024
C. 60,048
D. 55,554
E. 55,552

mean of all integers=(1234+4321)/2=2777.5
sum of all integers=2777.5*4!=66,660
A

BenjaminS · Answer

Let explain what I think.

We have 4 digits (3,2,1,4). and then each digit will be the unit of the new arrangement at 6 times.
For example,
The last digit is 4 >> there will be 6 numbers to arrange. This is 3214, 3124, 2134, 2314, 1234, 1324.
etc.

So the last digit of all possible number is 3, 2, 1 and 4
Then the sum of all arrangement is 3*6+2*6+1*6+4*6 ( = 0)

The answer is A. (Hope it correct and help)

monikakumar · Answer

What is sum of all rearrangements of the 4-digit number 3214?

A. 66,660
B. 65,024
C. 60,048
D. 55,554
E. 55,552
sum of the digits =10
so 1000*10=10000
100*10=1000
10*10=100
1*10=10 (units place)
permutation of 3214=24
from this it can be seen that every digit will appear 6 times in each digit place
so 6(11110)=66660
Ans A

ScottTargetTestPrep · Answer

The number of ways to arrange the digits 1, 2, 3, and 4 is 4! = 24. Of these 24 numbers, each of the fours digits 1, 2, 3, and 4 appears as the units digit 6 times; therefore, the sum of all the units digits is (1 + 2 + 3 + 4) x 6 = 60. Using the same analogy, the four digits each appear as the tens digit 6 times (yielding a sum of 600), as the hundreds digit 6 times (yielding a sum of 6000), and as the thousands digits 6 times (yielding a sum of 60,000). Therefore, the sum of all 24 numbers is:

60 + 600 + 6000 + 60,000 = 66,660

Answer: A

rainingrelics · Answer

Similarly, if there were five digits, each place would be occupied by the number 24 times right?
Or if there were 3 digits, each place would be occupied by each number 2 times.
Please let me know if I've understood this properly.

freedom128 · Answer

Yes your understanding is correct provided that the 3 or 5 numbers are unique (ie. 123, 56789). Please also think if you have 1123. What are the sum?

rainingrelics · Answer

Thanks for the question. 
For 1123, considering we start out with 1 in the thousandth place 1_ _ _, hundreds place can be occupied by 3 numbers, tens place by 2 numbers and lastly ones place by 1 number.
But if we start out with 2 or 3 in the thousandth place - 2_ _ _ or 3_ _ _ only one place can have 3 numbers, because after that 1 will be repeated resulting in the same number. 2131 or 3121 or 3112.

Hence total arrangments of 1123 are 12. But I'm a little confused as to go about finding the sum of all the arrangements.

freedom128 · Answer

Question (Modified) :
What is sum of all rearrangements of the 4-digit number 1123 ?

Answer:
1123 has 4!/2! = 12 arrangements.

In all twelve arrangements, the thousands digit is occupied by:
No. 1... 6 times (2 out of 4 nos in 1123) 
No. 2... 3 times (1 out of 4 nos in 1123)
No. 3... 3 times (1 out of 4 nos in 1123)
So are the the hundreds digit, the tens digit and the unit digit.

Let's sum all arrangements
* Sum of 1000s digit only = (6*1 +3*2 +3*3)*1000
* Sum of 100s digit only = (6*1 +3*2 +3*3)*100
* Sum of 10s digit only = (6*1 +3*2 +3*3)*10
* Sum of unit digit only = (6*1 +3*2 +3*3)*1

Sum of all arrangements
= (6*1 +3*2 +3*3)*(1000+100+10+1)
= 21*1111
= 23331

smithca35801 · Answer

Maybe it was a fluke, so can someone let me know if my method is valid? It took 21 secs instead of the avg. 1:43.

Since there are 4 answer choices with distinct singles digits, solve for that first. There are six combinations for each number in the singles digit. So,

(_ _ _ 4)*6 ends in 4
(_ _ _ 3)*6 ends in 8
(_ _ _ 2)*6 ends in 2
(_ _ _ 1)*6 ends in 6

Those sum to 20, so the sum of all combinations must also end in 0.

freedom128 · Answer

It's a yes for given answer choices. But if 2 or more answer choices had unit digit of 0 (which GMAC usually does) , then the method would only be a preliminary (yet useful) step to eliminate wrong choices.

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smithca35801 · Answer

Thanks for the input!

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What is sum of all rearrangements of the 4-digit number 3214?

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What is sum of all rearrangements of the 4-digit number 3214?

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