A difficult math problem?

Question

Consider all five-digit numbers formed by using the digits 1, 2, 3, 4, and 5 once and only once in each number formed. What is the sum of all of these five-digit numbers?

Answer 1

There are 5! = 120 numbers. Each digit will be in each position the identical number of times (120/5 = 24).

So the answer is

24*(1+2+3+4+5)*(10,000 + 1,000 + 100 + 10 + 1)

24*15*11,111

3,999,960

Answer 2

We have 5! (5 * 4 * 3 * 2 * 1) numbers.
We have five places: A B C D E.
Of the 5! numbers, each place has each number (1,2,3,4,5) one fifth of the time.
Consider the first digit ("A"):
It is 1 1/5 of the time.
It is 2 1/5 of the time.
It is 3 1/5 of the time.
It is 4 1/5 of the time.
It is 5 1/5 of the time.
So, the sum of the First digits of the 5! numbers is:

5! [ 1* (1/5) + 2 * (1/5) + 3 * (1/5) + 4 * (1/5) + 5 * (1/5)]
= 5! [ (1 + 2 + 3 + 4 + 5) * (1/5)]

The value of the sum of the first column, the 10,000's column, is 5! [ (1 + 2 + 3 + 4 + 5) * (1/5)] * 10^4

Likewise, the other four columns will have the same sum
{ 5! [ 1* (1/5) + 2 * (1/5) + 3 * (1/5) + 4 * (1/5) + 5 * (1/5)] },
so the value for the sum of the numbers is:

5! [ (1 + 2 + 3 + 4 + 5) * (1/5)] * 10^4 {Ten Thousands Column}
+5! [ (1 + 2 + 3 + 4 + 5) * (1/5)] * 10^3 {Thousands Column}
+5! [ (1 + 2 + 3 + 4 + 5) * (1/5)] * 10^2 {Hundreds Column}
+5! [ (1 + 2 + 3 + 4 + 5) * (1/5)] * 10^1 {Tens Column}
+5! [ (1 + 2 + 3 + 4 + 5) * (1/5)] * 10^0 {Units Column}
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Answer 3

Each digit will appear in one of the position among 4*3*2*1 = 24 times. So, there will be 24 numbers with 1 as the last digit. This will be the case for each digit and for each position.
Thus, the sum of these numbers is
24 * 1 + 24 * 2 + 24 * 3 + 24 * 4 + 24 * 5 +
24 * (10 + 20 + 30 + 40 + 50) +
24 * 1500 +
24 * 15000 +
24 * 150000
= 24 * 15 * (1 + 10 + 100 + 1000 + 10000)
= 24 * 15 * 11111
= 360 * 11111
= 3999960

Thus, the general case for the case of n, instead of 5 is
(n-1)! * [n + (n-1) + ... + 2 + 1] * [10^(n-1) + 10^(n-2) + ... + 10^1 + 10^0]
or
(n-1)! * n(n+1)/2 * sum_k=1->n (10^(n-1))

Answer 4

You should first find how many five digit numbers which can be produced. 5*4*3*2*1=120.

For all numbers the sum must be 5+4+3+2+1=15

sum of all these five digit numbers is:
15*120=1800.

Answer 5

All these 5-digit numbers are 5!=120
(because in the first place you can put 5 numbers, in th 2nd 4, which is 5-1, the previous, etc.)

So each of those digits is in each place (ten thousands, thousands, hundreds, tens etc.) 120/5=24 times.
So their sum is:
24*1 + 24*2 + ... + 24*5 +
+10*24*1 + 10*24*2 +...+10*24*5 (tens)
+100*24*1 + 100*24*2 +...+100*24*5 (hundreds)
+1000*24*1 + 1000*24*2 +...+1000*24*5 (thousands)
+10000*24*1 + 10000*24*2 +...+10000*24*5
=24*(1+2+3+4+5)+ 10*24*(1+2+3+4+5)
+ ... + 10000*24*(1+2+3+4+5)=
=24*15 + 10*24*15 +100*24*15 + 1000*24*15
+10000*24*15=
24*15*(1+10+100+1000+10000)=3999960
So their sum is 3,999,960

Answer 6

Well, there are 5*4*3*2*1 = 120 of these numbers
and 1+2+3+4+5 = 15 is the sum of each one,
so the sum of all of them is 120*15 = 1800.

Answer 7

Break it up: There are going to be 5! such numbers, and each digit will be used equally in the ones, tens, hundreds, thousands, and ten thousands place.

How about 4!(55555+44444+33333+22222+11111)? Think about it this way: The 5 will be in the ten thousands place 4! times (in 1/5 of the total numbers). Etc...

Answer 8

actually, it's not that hard. you just ned to make a chart to figure out the numbers. there are a lot of numbers though. i just worked it out and tried to calculate it on my calculator but the number was way to big. so the only problem would be adding it. it takes a while to get all the #s

Answer 9

undertaking one million ? i^2 = (one million/3) n^3 + (one million/2) n^2 + (one million/6) n subsequently a hundred ? i^2 = a hundred [ (one million/3) (fifty 9)^3 + (one million/2) (fifty 9)^2 + (one million/6) (fifty 9) ] undertaking 2 ? i^3 = (one million/4) n^4 + (one million/2) n^3 + (one million/4) n^2 subsequently = 10 ? i^2 (i-40 9) = 10 ? i^3 - 490 ? i^2 = 10 [ (one million/4) (a hundred)^4 + (one million/2) (a hundred)^3 + (one million/4) (a hundred)^2 ] - 10 [ (one million/4) (fifty 9)^4 + (one million/2) (fifty 9)^3 + (one million/4) (fifty 9)^2 ] - 490 [ (one million/3) (a hundred)^3 + (one million/2) (a hundred)^2 + (one million/6) (a hundred) ] + 490 [ (one million/3) (fifty 9)^3 + (one million/2) (fifty 9)^2 + (one million/6) (fifty 9) ]

Answer 10

i got 166,665 when i did the problem