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What is the sum of all possible 3digit numbers that can be constructe
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Updated on: 28 Aug 2019, 22:54
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What is the sum of all possible 3digit numbers that can be constructed using the digits 3, 4 and 5 if each digit can be used only once in each number ? A. 2660 B. 2661 C. 2662 D. 2663 E. 2664
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Originally posted by sugu86 on 26 Apr 2012, 00:59.
Last edited by Bunuel on 28 Aug 2019, 22:54, edited 2 times in total.
Edited the question




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Re: What is the sum of all possible 3digit numbers that can be constructe
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26 Apr 2012, 01:20
sugu86 wrote: What is the sum of all possible 3digit numbers that can be constructed using the digits 3, 4 and 5 if each digit can be used only once in each number ?
A) 2660 B) 2661 C) 2662 D) 2663 E) 2664
Thanks,
Suganth What is the sum of all possible 3digit numbers that can be constructed using the digits 3, 4 and 5 if each digit can be used only once in each number ?A. 2660 B. 2661 C. 2662 D. 2663 E. 2664 Any 3digit number can be written as: 100a+10b+c. # of three digit numbers with digits {3, 4, 5} is 3!=6. These 6 numbers will have 6/3=2 times 3 as hundreds digit (a), 2 times 4 as as hundreds digit, 2 times 5 as hundreds digit. The same with tens and units digits. 100*(2*3+2*4+2*5)+10*(2*3+2*4+2*5)+(2*3+2*4+2*5)=100*24+10*24+24=24*111=2664. Answer: E. Generally the sum of all the numbers which can be formed by using the n distinct digits, is given by the formula:
(n1)!*(sum of the digits)*(111…..n times)In our original question: n=3. sum of digits=3+4+5=12. > (31)!*(12)*(111)=24*111=2664. Hope it's clear.
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Re: What is the sum of all possible 3digit numbers that can be constructe
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20 Jan 2013, 17:33
Another way to approach this problem is to recognize that the way the sequence increases from the min (345) is symmetrical to the way it decreases from the max (543). Therefore if you find the average of the min and max and multiply it by the number of possibilities (3! or 6) then you'll have your answer.
\(\frac{345+543}{2} = 444\)
\(444*3! = 444*6 = 2664\)




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Re: What is the sum of all possible 3digit numbers that can be constructe
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02 Oct 2012, 08:59
sugu86 wrote: What is the sum of all possible 3digit numbers that can be constructed using the digits 3, 4 and 5 if each digit can be used only once in each number ?
A. 2660 B. 2661 C. 2662 D. 2663 E. 2664 the unit digits of all possible 3digit numbers are supposed to have a sum of 3 +3 +4+4+5+5=24, so the sum of numbers should have 4 as a unit digit  2664 is the only possible option.



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Re: What is the sum of all possible 3digit numbers that can be constructe
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02 Oct 2012, 09:36
I think Bunuel got the basic way to solve this kind of question. if the number is 4 digit or the answer has 4 number with last number is 4 then U should follow Bunuel



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Re: What is the sum of all possible 3digit numbers that can be constructe
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06 Oct 2012, 19:53
thaihoang305 wrote: I think Bunuel got the basic way to solve this kind of question. if the number is 4 digit or the answer has 4 number with last number is 4 then U should follow Bunuel Many thanks to Bunuel for his very clear explanations, I am going through all problems with his explanations in forum's PS part. For this very problem I just wanted to find out the fastest way to solve as far as you need to take time into account as well.



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Re: What is the sum of all possible 3digit numbers that can be constructe
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21 Jan 2013, 03:08
Bunuel wrote: sugu86 wrote: What is the sum of all possible 3digit numbers that can be constructed using the digits 3, 4 and 5 if each digit can be used only once in each number ?
A) 2660 B) 2661 C) 2662 D) 2663 E) 2664
Thanks,
Suganth What is the sum of all possible 3digit numbers that can be constructed using the digits 3, 4 and 5 if each digit can be used only once in each number ?A. 2660 B. 2661 C. 2662 D. 2663 E. 2664 Any 3digit number can be written as: 100a+10b+c. # of three digit numbers with digits {3, 4, 5} is 3!=6. These 6 numbers will have 6/3=2 times 3 as hundreds digit (a), 2 times 4 as as hundreds digit, 2 times 5 as hundreds digit. The same with tens and units digits. 100*(2*3+2*4+2*5)+10*(2*3+2*4+2*5)+(2*3+2*4+2*5)=100*24+10*24+24=24*111=2664. Answer: E. Generally the sum of all the numbers which can be formed by using the n distinct digits, is given by the formula:
(n1)!*(sum of the digits)*(111…..n times)In our original question: n=3. sum of digits=3+4+5=12. > (31)!*(12)*(111)=24*111=2664. Hope it's clear. wow on this type of a question i was only able to come up with the computation of possible number of ways of arranging 3 digits,but the rest part gave me problems truly speaking @bunuel i am complete lost on this part( These 6 numbers will have 6/3=2 times 3 as hundreds digit (a), 2 times 4 as as hundreds digit, 2 times 5 as hundreds digit. The same with tens and units digits. 100*(2*3+2*4+2*5)+10*(2*3+2*4+2*5)+(2*3+2*4+2*5)=100*24+10*24+24=24*111=2664.)...but i guess the formula would make it easier..you should add it in the topic of number theory in the gmat math book.. Rgrds Posted from my mobile device



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Re: What is the sum of all possible 3digit numbers that can be constructe
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27 Aug 2013, 23:42
I like that approach because it has precise formula, but could you please clarify this part: Bunuel wrote: (n1)!*(sum of the digits)*(111…..n times)[/b]
So when to use 111 and when to use different number? You stated 111.....n times, isn't it should be 3 times in our case?
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Re: What is the sum of all possible 3digit numbers that can be constructe
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28 Aug 2013, 09:26



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Re: What is the sum of all possible 3digit numbers that can be constructe
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05 Apr 2018, 16:58
sugu86 wrote: What is the sum of all possible 3digit numbers that can be constructed using the digits 3, 4 and 5 if each digit can be used only once in each number ?
A. 2660 B. 2661 C. 2662 D. 2663 E. 2664 The possible 3digit numbers are: 345, 354, 435, 453, 543, 534 Summing just the units digits of these 6 numbers, we get: 5 + 5 + 4 + 4 + 3 + 3 = 10 + 8 + 6 = 24 The only answer with a units digit of 4 is 2,664. Answer: E
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Re: What is the sum of all possible 3digit numbers that can be constructe
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22 Aug 2019, 11:25
when we need to find the sum of digits such as in this case we can either form cases because the number of digits are less.
else the concept to find the sum is
ABC. when C is 3,4,5 so AB can be arranged in 2! ways.
the sum of digits is 3+4+5 and since it is in the units digits it takes a place value of 1. so it becomes (3+4+5)*1*2!
similarly in 10s digit, we fix 3,4,5 and the rest of the digits can be arranged in 2! ways.
so the place value is 10 and sum becomes (3+4+5)*10*2!
similarly for hundreds place it becomes (3+4+5)*100*2!
to find the sum of all digits we add these three, it becomes 100(3+4+5)2! + 10(3+4+5)2! + 1(3+4+5)2! taking 2! and sum of digits as common.
2!(3+4+5)*111
we can also generalise a formula from this, 3,4,5 > no of digits(n) = 3
the formula becomes (n1)! (111....n times) ( sum of digits)
so here n = 3. putting the values in the formula, (31)! (111) (3+4+5).
This gives us the answer as 2664.
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What is the sum of all possible 3digit numbers that can be constructe
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22 Aug 2019, 13:40
HullDown wrote: What is the sum of all the possible three digit numbers that can be constructed using the digits 3,4 and 5 if each digit is used only once in each number?
A) 2664 B) 2550 C) 2320 D) 3567 E) 1456 Each no. 3, 4, 5 will occur twice at any place. So in all the places it will be 2*(3+4+5) = 24. In the units place 4 will come and carry is 2. In the tens place it will be 24+2 = 26 so 6 and carry 2. then at hundredth place it will be again 24+2 = 26 so 6 and carry 2. So the sum will be 2664



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What is the sum of all possible 3digit numbers that can be constructe
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22 Aug 2019, 13:57
HullDown wrote: What is the sum of all the possible three digit numbers that can be constructed using the digits 3,4 and 5 if each digit is used only once in each number?
A) 2664 B) 2550 C) 2320 D) 3567 E) 1456 add least and greatest: 345+543=888 888/2=444 average 444*3!=2664



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Re: What is the sum of all possible 3digit numbers that can be constructe
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22 Aug 2019, 18:02
HullDown wrote: What is the sum of all the possible three digit numbers that can be constructed using the digits 3,4 and 5 if each digit is used only once in each number?
A) 2664 B) 2550 C) 2320 D) 3567 E) 1456 Number of 3 digit numbers formed using 3,4,5 = 3! with each digit appearing twice at each place. Sum of all 6 such numbers = 2*111*(3+4+5) = 222*12= 2664 IMO E Posted from my mobile device
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Re: What is the sum of all possible 3digit numbers that can be constructe
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