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There are 24 different fourdigit integers than can be [#permalink]
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04 Nov 2012, 10:39
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2345 2354 2435 +5432 There are 24 different fourdigit 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
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Re: There are 24 different fourdigit integers than can be [#permalink]
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04 Nov 2012, 20:12
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Maxswe wrote: 2345 2354 2435
+5432
There are 24 different fourdigit 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. Ans E it is.
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Re: There are 24 different fourdigit integers than can be [#permalink]
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04 Nov 2012, 21:05
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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.



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Re: There are 24 different fourdigit integers than can be [#permalink]
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05 Nov 2012, 03:42
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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



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Re: There are 24 different fourdigit integers than can be [#permalink]
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05 Nov 2012, 09:18
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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 ndigit numbers then the sum of all such numbers is = (n1)! *(sum of n digits)*(11....n times)
In our example: n= 4 (n1) ! = ( 41 ) ! = 3 ! = 6 sum of digits = 14 (5+4+3+2) 111..n times = here n = 4 >>1111
So 6*14* 1111 = 93,324



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Re: There are 24 different fourdigit integers than can be [#permalink]
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06 Nov 2012, 03:25
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Maxswe wrote: 2345 2354 2435
+5432
There are 24 different fourdigit 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: \((n1)!*(sum \ of \ the \ digits)*(111... \ n \ times)\). 2. Sum of all the numbers which can be formed by using the \(n\) digits ( repetition being allowed) is: \(n^{n1}*(sum \ of \ the \ digits)*(111... \ n \ times)\). Similar questions to prctice: theadditionproblemaboveshowsfourofthe24differentin104166.htmlwhatisthesumofall3digitpositiveintegersthatcanbe78143.htmlfindthesumofallthefourdigitnumbersformedusingthe103523.htmlwhatisthesumofall4digitnumbersthatcanbeformed94836.htmlwhatisthesumofall3digitpositiveintegersthatcanbe78143.htmlfindthesumofallthefourdigitnumberswhichareformed88357.htmlfindthesumofall3digitnosthatcanbeformedby88864.htmlHope it helps.
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Re: There are 24 different fourdigit integers than can be [#permalink]
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06 Nov 2012, 08:44
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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.



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Re: There are 24 different fourdigit integers than can be [#permalink]
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06 Jan 2013, 00:52
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Maxswe wrote: 2345 2354 2435
+5432
There are 24 different fourdigit 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



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Re: There are 24 different fourdigit integers than can be [#permalink]
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29 Dec 2013, 18:28
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Maxswe wrote: 2345 2354 2435
+5432
There are 24 different fourdigit 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 * (n1)! * 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



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Re: There are 24 different fourdigit integers than can be [#permalink]
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04 Apr 2015, 10:41
Wasn't aware of this Formula .. Thanks bunuel
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There are 24 different fourdigit integers than can be formed using... [#permalink]
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16 Oct 2015, 10:46
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 (notso) Straightforward Math. Consider the column on the righthand 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.



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Re: There are 24 different fourdigit integers than can be formed using... [#permalink]
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16 Oct 2015, 10:59
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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
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Re: There are 24 different fourdigit integers than can be formed using... [#permalink]
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16 Oct 2015, 11:54
Digit at each position will be (2+3+4+5)=14 each will be appeared 3! =6 times so 14*6=84 will be values at each digit position.
840 840 84 =Last three digits will be 324 so optionE



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There are 24 different fourdigit integers than can be formed using... [#permalink]
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16 Oct 2015, 16:38
For a lack of better approach I just calculated the average and multiplied it by 24.
Largest number is 5432 Smallest number is 2345 Sum is 7777.
\(\frac{7777}{2}*24=7777*12=93,324\)
Calculation is actually easier than the numbers suggest.
I'm not sure why this works. Obviously, the numbers are not an evenly spaced set. But I could see that all the differences from the mean cancel out since the same digest are used in alternating order and the digest are consecutive integers.
I'd appreciate some clarification on this one. Thanks!



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Re: There are 24 different fourdigit integers than can be [#permalink]
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09 Nov 2015, 10:52
since there are 24 numbers total, that means that there are 6 numbers when 2 is the digit for thousands, 6 numbers when 2 is the digit for hundreds, 6 when 2 is the digit for tens, and 2 when the digit is for units 6 numbers when 3 is the digit for thousands, 6 numbers when 3 is the digit for hundreds, 6 when 3 is the digit for tens, and 3 when the digit is for units 6 numbers when 4 is the digit for thousands, 6 numbers when 4 is the digit for hundreds, 6 when 4 is the digit for tens, and 4 when the digit is for units 6 numbers when 5 is the digit for thousands,6 numbers when 5 is the digit for hundreds, 6 when 5 is the digit for tens, and 5 when the digit is for units
thus, we have 2k*6 + 3k*6+4k*6+5k*6 + 200*6 + 300*6+400*6 +500*6 + 20*6 + 30*6 + 40*6 + 50*6 + 2*6 + 3*6 + 4*6+5*6 or 6(2k+3k+4k+5k) + 6(200+300+400+500) + 6(20+30+40+50) +6(2+3+4+5) 6*14k + 6*1400 + 6*140 +6*14 = we see that 14k*6 = 84k. We can eliminate A, B, and C right away. 84k +8400 = 92++k we can eliminate D, and thus E must be the answer, without calculating the rest.



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There are 24 different fourdigit integers than can be [#permalink]
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19 Dec 2015, 11:41
I got very easy to the answer choice by applying this technique:
so we can have 2 in the thousands digit, thus, the number of ways to arrange the digits in such ways that 2 is in thousands is 6. now, we have 2000 * 6 = 12,000. then we have 6 ways in which 3 is in place of thousands, so 3000*6 = 18,000. we have 6 ways in which 4 is in place of thousands, so 4000*6=24,000 we have 6 ways in which 5 is in place of thousands, so 5000*6=30,000.
now, add everything: 12k+18k+24k+30k=84k. this is not the end, but if we know how to get at least here, we can eliminate A, B, and C. thus, we increase our chance of getting the right answer to 50%.
now, the same technique apply with the hundreds digit. the sum is 8400. same thing with the tenths digit = 840 same thing with the units digit = 84.
so, the total sum is 84,000+8,400+840+84 = 93,324. E.
oh wow, solving the second time this question..using the same technique.. :D



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Re: There are 24 different fourdigit integers than can be [#permalink]
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30 Dec 2017, 11:31
Another Approach
((Lower Limit + Upper Limit) /2 ) *24 ((2345+5432)/2)*24=93,334 Cheers



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Re: There are 24 different fourdigit integers than can be [#permalink]
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23 Apr 2018, 07:53
as 24 combination is possible with four digit, each digit gets six numbers. if we add 6 two's, 6 three's, 6 four's, and 6 five's. Which adds up to 84. So the last digit will be 4. Eliminate B. no the tenth digit will be 84 + 8 = (9) 2. Eliminate C the hundredth digit will be 84+ 9 = (9)3. Eliminate D. So E is the answer
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Re: There are 24 different fourdigit integers than can be [#permalink]
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23 Apr 2018, 13:28
Quote: There are 24 different fourdigit 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 2,345+5,432=7,777 7,777/2=3,888.5 24*3,888.5=93,324 E




Re: There are 24 different fourdigit integers than can be
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