What is the sum of all possible 3-digit numbers that can be constructe

the unit digits of all possible 3-digit 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.

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.

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

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I like that approach because it has precise formula, but could you please clarify this part:



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?

111... n times mean that if we have 2 digits it should be 11, if 3 digits 111, if 4 digits it should be 1,111.

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Hope it helps.

The possible 3-digit 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

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
(n-1)! (111....n times) ( sum of digits)

so here n = 3.
putting the values in the formula,
(3-1)! (111) (3+4+5).

This gives us the answer as 2664.

If you like my solution, do give me kudos!

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

add least and greatest:
345+543=888
888/2=444 average
444*3!=2664

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

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Here is a different approach,

Identify the numbers as a series

a1=345 (smallest possible number with {3,4,5})
an=543 (largest possible number with the {3,4,5})

How many numbers can be formed using the 3,4,5 without repetition?
3*2*1 ways = 6 numbers

Therefore 'n'= 6

a1=345
a6=543

Sum of numbers= n/2(first number+last number)
S= n/2(a+l)
= 6/2(345+543)
= 3*888
= 2664

Hope this helps, this should be easy if you're thorough with APs

Possible 3-digit numbers containing 3, 4 and 5 such that each digit appears only once are:

345 , 354, 435, 453, 534, 543

Sum: 2,664

Answer: E

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