There are 24 different four-digit integers than can be

Question

2345
2354
2435
...
+5432

There are 24 different four-digit integers than can be formed using the digits 2, 3, 4 and 5 exactly once in each integer. The addition problem above shows 4 such integers. What is the sum of all 24 such integers?

A. 24,444
B. 28,000
C. 84,844
D. 90,024
E. 93,324

Bunuel · Accepted Answer

Generally: 1. Sum of all the numbers which can be formed by using the n digits without repetition is: (n-1)!*(sum \ of \ the \ digits)*(111... \ n \ times) . 2. Sum of all the numbers which can be formed by using the n digits ( repetition being allowed ) is: n^{n-1}*(sum \ of \ the \ digits)*(111... \ n \ times) . Similar questions to prctice: the-addition-problem-above-shows-four-of-the-24-different-in-104166.html what-is-the-sum-of-all-3-digit-positive-integers-that-can-be-78143.html find-the-sum-of-all-the-four-digit-numbers-formed-using-the-103523.html what-is-the-sum-of-all-4-digit-numbers-that-can-be-formed-94836.html what-is-the-sum-of-all-3-digit-positive-integers-that-can-be-78143.html find-the-sum-of-all-the-four-digit-numbers-which-are-formed-88357.html find-the-sum-of-all-3-digit-nos-that-can-be-formed-by-88864.html Hope it helps.

gckasi · Answer

Here's how I did it -
If you consider the unit digit, you can have 6 of the numbers ending in 5, 6 ending in 4, 6 ending in 3 and 6 ending in 2. The sum of these would be 84. So we know that the answer will have 4 as the last digit. Eliminate B.

If you consider the tens digit, you can have 6 of the numbers ending in 5, 6 ending in 4, 6 ending in 3 and 6 ending in 2. The sum of these would be 84 again and we have a carry over of 8 from the sum of unit digits . So the sum of tens digit is 84 + 8 = 92. This means that the answer will have 2 in the tens digit. Eliminate A and C.

If you consider the hundreds digit, you can have 6 of the numbers ending in 5, 6 ending in 4, 6 ending in 3 and 6 ending in 2. The sum of these would be 84 again and we have a carry over of 9 from the sum of tens digits . So the sum of hundreds digit is 84 + 9 = 93. This means that the answer will have 3 in the hundreds digit. Eliminate D. Answer is E.

Vips0000 · Answer

There are 4 integers and 4 places. Thus, any integer can be in one place for 6 (3*2*1) combinations. That means, there would be exactly 6 numbers in which 2 is at first place, 6 in which 2 is at second place , 6 in which 2 is at third place and 6 in which 2 is at 4th place.
Same holds good for 3,4,5 also.

Now sum of all numbers, is basically sum of their positional values. eg. 2435 = 2000 +400+30+5

Therefore sum of all such 24 integers = 6* (2000 +3000+4000+5000) + 6* (200 +300+400+500) + 6* (20 +30+40+50)+ 6* (2 +3+4+5)
to avoid too much calculation - 
=1000X +100X+10X+X (where X =6*14 =84)

now we know 1000X is 84000 , eliminate A,B
100 X is 8400, so total 92000 sth, eliminate C,D.

Ans E it is.

bhavinshah5685 · Answer

My approach was different..

I tried to find if there is any pattern in the all 24 Nos.
For that I tried different numbers 1,2 and 3.
with this,you can create 3!=6 Nos without repitation.

if you see the numbers are 
12 3 
13 2 
21 3 
23 1 
31 2 
32 1

here if you see, at Units place, each number comes twice, for tenth place also each numbe comes twice and same for 100th place.
which is 3 diff Nos x 2 times each= total 6 diff numbers

So,I compare it with our number 2,3,4,5.

4 diff Nos x 6 timers each = 24 total diff numbers

So, For unit place 
2*6=12
3*6=18
4*6=24
5*6=30
-----------
84

adding this for all the places, we get last three digit of sum as 324
 which is in our answer choice E

Maxswe · Answer

Nice!

Here is another way to solve it using this formula

If there are n distinct numbers that are used to make all possible n-digit numbers then the sum of all such numbers is
= (n-1)! *(sum of n digits)*(11....n times)

In our example:
n= 4
(n-1) ! = ( 4-1 ) ! = 3 ! = 6
sum of digits = 14 (5+4+3+2)
111..n times = here n = 4 >>1111

So 6*14* 1111 = 93,324

KevinBrink · Answer

Here is how I did it in 0:28 seconds. First consider the thousands digit, all the numbers are placed 6 times here.
So the sum of the thousands digits is 6*5 + 6*4 + 6*3 + 6*2 = 84 --> cancel A, B
The sum of the hundreds digits is also 84 --> 84000 + 8400 = 92,400 --> cancel C, D
Hence the answer is E.

geneticsgene · Answer

We have 4 digits and 4 places without repetition. Each digit will occupy 1000, 100, 10 and units place 6 times. Sum = 6  = 93,324
Answer: E

jlgdr · Answer

Is good to remember this formula Sum of terms * (n-1)! * 1111 (1 for each digit) So we'll have (14)(6)(1111) Answer is 90,324 Hence, E Normally, you want to just quickly check the units digit before doing the whole multiplication Problem is that almost all of the answer choices had 4 at the end So had to do it anyways Hope it helps Cheers! J

adityadon · Answer

Wasn't aware of this Formula .. 
Thanks bunuel

AlexGenkins1234 · Answer

Fast Solution:

First stage: we will look at the numbers that contain only 1 digit and the rest is zeros, e.g.: 5000,0200 (=200) ,0040 (=40)

- Let's look at the thousand digit:
 * We have: 4 options 5000,4000,3000,2000 = (5+4+3+2)*1000= 14*1000
 * We will use each option 3! times (because _X_,__,__,__ - you have 3! options to place the rest 3 numbers)
 * that means that the total is :14*1000*3!
- We can do the same for the hundreds digit, tens digits and ones.

Second stage: We will sum every number

14*3!*1000 +
14*3!*100 +
14*3!*10 +
14*3!*1 
-------------------
14*3!*1111 -> 14*6*1111=93,324

Long Solution  (taken from kaplan)

Backsolving and Picking Numbers aren’t feasible here, so we’ll have to use some (not-so) Straightforward Math. Consider the column on the right-hand side, representing the ones digit of each integer. There are 24 integers and 4 different digits, so each of the 4 digits must appear 6 times in the ones column. Thus, the sum of the ones place for the 24 integers is (6 × 2) + (6 × 3) + (6 × 4) + (6 × 5), or 6 × 14 = 84. Now we know that the correct choice must end with a 4, so eliminate choice (B).

The same pattern holds for the tens column EXCEPT that we have to add "8" to represent the amount carried over from the 84 in the ones column. So, the tens column must add up to 84 + 8 = 92. Now we see that we must have a 2 in the tens place, so eliminate choices (A) and (C).

For the hundreds place, the sum will be 84 + the 9 carried over from the 92 in the tens column; 84 + 9 = 93, so we must have a 3 in the hundreds place. Answer Choice (E) is the only possibility.

Alternatively, if the ones column adds up to 84, the tens column will be the same thing multiplied by 10, or 840. Similarly, the hundreds column will be 8,400 and the thousands column will be 84,000. Adding up these four numbers, we get 84 + 840 + 8,400 + 84,000 = 93,324.

Skywalker18 · Answer

All the 4 digits will appear in each of units , tens , hundred and thousands position six times .
 _ _ _ _

Sum per position = 6x2 + 6x3 + 6x4 + 6x5 = 84

Since , to get the sum we multiply the sum of digits by the place value
i.e the units column will be multplied by 1 , tens columns by 10 and so on
= 84 + 840 + 8400 + 84000
=93324
Answer E

Argh123 · Answer

Average ~ 7700 x ( no of int) 12 = 92400 hence answer is more 92400 ( E.)

Jsaint · Answer

Is there a reason not to take the average of (2222+7777)/2=+/- 3888...then multiply by 24?

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