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There are 24 different four-digit integers than can be [#permalink]
04 Nov 2012, 09:39

3

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

35% (medium)

Question Stats:

66% (02:20) correct
34% (01:39) wrong based on 74 sessions

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?

Re: There are 24 different four-digit integers than can be [#permalink]
04 Nov 2012, 19:12

1

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Maxswe wrote:

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 I'll give the explanation later

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.

Re: There are 24 different four-digit integers than can be [#permalink]
04 Nov 2012, 20:05

1

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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.

Re: There are 24 different four-digit integers than can be [#permalink]
05 Nov 2012, 02:42

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 123 132 213 231 312 321

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

Re: There are 24 different four-digit integers than can be [#permalink]
05 Nov 2012, 08:18

2

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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

Re: There are 24 different four-digit integers than can be [#permalink]
06 Nov 2012, 02:25

3

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Expert's post

2

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Maxswe wrote:

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

I'll give the explanation later

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).

Re: There are 24 different four-digit integers than can be [#permalink]
06 Nov 2012, 07:44

2

This post received KUDOS

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.

Re: There are 24 different four-digit integers than can be [#permalink]
05 Jan 2013, 23:52

1

This post received KUDOS

Maxswe wrote:

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

I'll give the explanation later

We have 4 digits and 4 places without repetition. Each digit will occupy 1000, 100, 10 and units place 6 times. Sum = 6[2222+3333+4444+5555] = 93,324 Answer: E

Re: There are 24 different four-digit integers than can be [#permalink]
29 Dec 2013, 17:28

1

This post received KUDOS

Maxswe wrote:

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

I'll give the explanation later

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

gmatclubot

Re: There are 24 different four-digit integers than can be
[#permalink]
29 Dec 2013, 17:28