- #1

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b)Find the sum of all he numbers formed in (a).

I know how to do the part a), but I'm totally stump when I see part b).

Thanks.

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- Start date

- #1

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b)Find the sum of all he numbers formed in (a).

I know how to do the part a), but I'm totally stump when I see part b).

Thanks.

- #2

VietDao29

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ie, something like: ***0, where * denotes a random number.

There will be 6 * 7 * 7 = 294 numbers.

Do the same, and you will come up with 294 numbers ends in 1.

294 numbers ends in 2.

294 numbers ends in 5.

294 numbers ends in 6.

294 numbers ends in 7.

294 numbers ends in 8.

So the sum of all numbers found in a will end in:

294 * (0 + 1 + 2 + 5 + 6 + 7 + 8) = 852

So it will end in 6 (you still carry 852).

There will be again 294 numbers. So again, do the same as #1, and remember that you carried 852.

So the second digit from the right of the sum is:

294 * (0 + 1 + 2 + 5 + 6 + 7 + 8) + 852 = 937

So the second digit from the right of the sum is: 8.

Now do the same for '*0**', '*1**', '*2**', '*5**', '*6**', '*7**', '*8**'

And '1***', '2***', '5***', '6***', '7***', '8***'

Note that '0***'

Viet Dao,

- #3

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So it will end in 6 (you still carry 852).

294 * (0 + 1 + 2 + 5 + 6 + 7 + 8) +852= 9378.

Actually, I don't understand the above steps, why we have to add 852? I am sorry that I am too stupid.

There will be 6 * 7 * 7 = 294 numbers.

a)How many different numbers of 4 digits may be formed with the dgits 0,1,2,5,6,7,8 ifno digit is used more than once in any number?

Also, should it be 6*5*4?

Thanks for your answering. ^_^

- #4

VietDao29

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So the sum will end in 0: 180 * (0 + 1 + 2 + 5 + 6 + 7 + 8) = 522

This is just the way you add the number. For example, when you want to add 123 + 234 + 345.

12

23

34

---

You should first add 3 + 4 + 5 = 12 (you write 2, carry 1).

1

2

3

---

...2

2 + 3 + 4 = 9, Because you carry 1, 9 + 1 = 10, write 0, carry 1.

---

..02

1 + 2 + 3 = 6, 6 + 1 = 7. So 123 + 234 + 345 = 702.

Now you are arrange the number like:

***0

***0

***0

...

***1

***1

...

***2

***2

...

...

***8

***8

You will try to add their first digit from the right first, then add the second digit from the right, then the third and the forth...

-----------

Do the same, and you will have: the second digit of the sum from the right is 2:

180 * (0 + 1 + 2 + 5 + 6 + 7 + 8) + 522 = 574

Viet Dao,

- #5

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I think the solution in this way :

part a)

(7)(6)(5)(4) = 840 ways

part b)

The number of times each digit appear = 4*( 6*5*4) = 480 times

The sum of all numbers that appear in part (a) =

480 *( 0+1+2+5+6+7+8) = 13920

part a)

(7)(6)(5)(4) = 840 ways

part b)

The number of times each digit appear = 4*( 6*5*4) = 480 times

The sum of all numbers that appear in part (a) =

480 *( 0+1+2+5+6+7+8) = 13920

Last edited:

- #6

VietDao29

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??? Why 7 * 6 * 5 * 4? ???Mithal said:I think the solution in this way :

part a)

(7)(6)(5)(4) = 840 ways

Shouldn't it be 6 * 6 * 5 * 4 = 720?

Because you can choose 1 from the 6 digits 1, 2, 5, 6, 7, 8 to be the first digit (

I don't understand...Mithal said:part b)

The number of times each digit appear = 4*( 6*5*4) = 480 times

The sum of all numbers that appear in part (a) =

480 *( 0+1+2+5+6+7+8) = 13920

So assume you are correct on part a (which is

Viet Dao,

- #7

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The number of time each digit appear in the thousands excluding zero =

(6)* (5) * (4) =120 times

The number of times each digit appear in the three remaings ( i.e the hundreds , tens and units)=(5) *(5) *(4)=100 times

1+2+5+6+7+8=29

The sum of the numbers = (1000) *(120) *(29) +(100) *(100) *(29) +(10)*(100) *(29) +(100) *(29) = 3801900

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