The addition problem above shows four of the 24 different integers tha

1,2,3,4 can be arranged in 4! = 24 ways

The units place of all the integers will have six 1's, six 2's, six 3's and six 4's
Likewise,
The tens place of all the integers will have six 1's, six 2's, six 3's and six 4's
The hundreds place of all the integers will have six 1's, six 2's, six 3's and six 4's
The thousands place of all the integers will have six 1's, six 2's, six 3's and six 4's

Addition always start from right(UNITS) to left(THOUSANDS);

Units place addition; 6(1+2+3+4) = 60.
Unit place of the result: 0
carried over to tens place: 6

Tens place addition; 6(1+2+3+4) = 60 + 6(Carried over from Units place) = 66
Tens place of the result: 6
carried over to hunderes place: 6

Hundreds place addition; 6(1+2+3+4) = 60 + 6(Carried over from tens place) = 66
Hundreds place of the result: 6
carried over to thousands place: 6

Thousands place addition; 6(1+2+3+4) = 60 + 6(Carried over from hundreds place) = 66
Thousands place of the result: 6
carried over to ten thousands place: 6

Ten thousands place of the result: 0+6(Carried over from thousands place) = 6

Result: 66660

Ans: "E"

This can be solved much easier by realizing that, since the number of four term permutations is 4!, and that summing the a sequence to its reverse gives

1234 +4321 = 5555

1243 +3421 = 5555

we may see that there are 4!/2 pairings we can make, giving us

5555(12) = 66660

The logic is no different from 'no repetition allowed' question. The only thing different is the number of numbers you can make.

How many numbers can you make using the four digits 1, 2, 3 and 4 if repetition is allowed?
You can make 4*4*4*4 = 256 numbers (there are 4 options for each digit)

1111
1112
1121
... and so on till 4444

By symmetry, each digit will appear equally in each place i.e. in unit's place, of the 256 numbers, 64 will have 1, 64 will have 2, 64 will have 3 and 64 will have 4.
Same for 10s, 100s and 1000s place.

Sum = 1000*(64*1 + 64*2 + 64*3 + 64*4) + 100*(64*1 + 64*2 + 64*3 + 64*4) + 10*(64*1 + 64*2 + 64*3 + 64*4) + 1*(64*1 + 64*2 + 64*3 + 64*4)
= (1000 + 100 + 10 + 1)(64*1 + 64*2 + 64*3 + 64*4)
= 1111*64*10 = 711040

or use the formula given by Bunuel above:
Sum of all the numbers which can be formed by using the digits (repetition being allowed) is:\(n^{n-1}\)*Sum of digits*(111...n times)
=\(4^3*(1+2+3+4)*(1111) = 711040\) (Same calculation as above)

For those who could not memorize the formular, you can guess the answer in 30 secs:
Since we have 24 numbers, we will have 6 of 1 thousand something, 6 of 2 thousand something, 6 of 3 thousand something, and 6 of 4 thousand something
So,
6x1(thousand something) = 6 (thousand something)
6x2(thousand something) = 12 (thousand something)
6x3(thousand something) = 18 (thousand something)
6x4(thousand something) = 24 (thousand something)
Add them all 6+12 +18 + 24 = 60 (thousand something)
----> E

Could you tell me the way to calculate the sum which the repetition is allowed? For example: from 1,2,3,4. how can we calculate the sum of four digit number that formed from 1,2,3,4 and repetition is allowed?

Each digit will come at the respective place i.e units,tens, hundreds , thousands
So calculate sum of each digit for the all the places
for 4=4000+400+40+4
3=3000+300+30+3
2=2000+200+20+2
1=1000+100+10+1
Now calculate the sum of the sums of these digits =11110
Now we know that each digit is used 6 times therefore we have to multiply with 6
6*11110=66660
Hence E is our answer .

How about using the Average*no. of terms formula where the average is found by adding the smallest number and the largest number divided by 2. It worked for this question. My question is will it work for all such types of questions?

Sum of terms = Average * Number of terms
holds for all sets of numbers

Average = (First term + Last term)/2
holds for an Arithmetic Progression only

Yes, you are right. But what was the arithmetic progression in this question?
And do such types of questions always have arithmetic progression so that I can use the above-mentioned formula for such questions?

There isn't and hence, I haven't used this formula. Note that Avg concept will not work when the digits are say 1, 2, 4, 6.
It works in this case because of the symmetry of the digits 1, 2, 3 and 4.

We have 6 numbers with 1XXX, 6 with 2XXX 6 with 3XXX and 6 with 4XXX.

6*1000+6*2000+6*3000+6*4000= 60000. Hence, the solution must be higher than 60000. Only E fits.

A slightly different approach:

We are looking to sum a four digit integer where each digit is not equal to zero. Therefore, taking each digit's place holder, we have a template of (1000 + 100 + 10 + 1) or (1,111).

We know that this 1,111 is repeated 24 times; therefore (1111 * 24) = 26,664.

We are unsure which digit of (1,2,3,4) goes where for any specific number within the set of 24 possibilities, but we know that the average number of (1,2,3,4) is 2.5

26,664 * 2.5 = 66,660

The solutions above are great. However, I believe this can be answered using simple logic and leveraging your answer choices.

The question tells us that there are 24 integers. Given there are 4 unique integers, we know there will be 6 of each (given the symmetry).

Here's where common sense comes handy: we know that 6 unique numbers will begin with the digit 4; therefore, they will have a minimum sum of 4*6000, or 24,000. Similarly, 6 unique numbers will begin with 3 and will have a minimum sum of 3000*6, or 18,000. Therefore, we have a minimum total sum of:

6*4,000 + 6*3,000 + 6*2,000 + 6*1,000 = 24,000 + 18,000 + 12,000 + 6,000 = 60,000. Logically, our answer will be greater than 60,000. Only one answer choice remains --> E.

Method 1: Sum Each Column of Digits

We need positive integers having 4 digits.

S = __ __ __ __

Since no repetition allowed, we can make 4*3*2*1 = 24 such positive integers.


Now imagine writing these 24 numbers one below the other to add.

1234
1243
...
...
x 24 combinations

When we add them, noticing the symmetry we know that there will be 6 1's in units digits, 6 2's, 6 3's and 6 4's. So units digits will add up to (1+2+3+4)*6.

Similarly, tens digits will add up (1+2+3+4)*6*10
Similarly, hundreds digits will add up (1+2+3+4)*6*100
Similarly, thousands digits will add up (1+2+3+4)*6*100)

Adding all of them:
(1+2+3+4)*6 + (1+2+3+4)*6*10 + (1+2+3+4)*6*100 + (1+2+3+4)*6*1000 = (1+2+3+4)*6 * (1 + 10 + 100 + 1000) = 10 * 6 * 1111 = 66,660

ANSWER: E

Method 2: Direct Formula

Sum of all n digit numbers formed by n non-zero digits without repetition is:

(n−1)!∗(sum of the digits)∗(111... n times)

= (4-3)! * (1 + 2 + 3 + 4) * (1111)
= 6 * 10 * 1111
= 66,660

Answer: E

Since we're adding 24 numbers, we know that:
Six numbers will be in the form 1 _ _ _
Six numbers will be in the form 2 _ _ _
Six numbers will be in the form 3 _ _ _
Six numbers will be in the form 4 _ _ _

Let's first see what the sum is when we say all 24 numbers are 1000, 2000, 3000 or 4000
The sum = (6)(1000) + (6)(2000) + (6)(3000) + (6)(4000)
= 6(1000 + 2000 + 3000 + 4000)
= 6(10,000)
= 60,000

Since the 24 numbers are actually greater than 1000, 2000, etc, we know that the actual sum must be greater than 60,000

Answer: E

Cheers,
Brent

We can just take the thousands digit.

We know that 1 will appear 6 times in thousands place, and so will 2,3 and 4
So just adding thousands place gives us

6000+12,000+18,000+24,000 = 60,000

We havnt even considered hundreds and tens and units places and already sum is 60,000

So sum > 60,000
only E > 60,000

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