What is the sum of all the numbers formed by taking the digits 3, 5, 6

Question

What is the sum of all positive numbers formed by taking the digits 3, 5, 6, 7 and 8, exactly once?

(A) 5241698
(B) 5894162
(C) 6581426
(D) 7733256
(E) 7799158

stne · Accepted Answer

Total number of positive numbers taking the above digits only once  = 5! = 120

Each of the digits will appear in  10,000^{th}  Place,  1000^{th}  Place ,  100^{th}  Place,  10^{th}  Place and unit place

# of times the each digit will appear  = \frac{120}{5} =24

Hence  (240000 +24000+ 2400+ 240+24 ) ( 3 +5 +6 +7 +8 )

= 666664 (29) = 7733256

Ans D

Ritaj1 · Answer

The solution posted by stne is very elegant. However, when i was solving it, I thought it implies sum of all numbers not just 5 digits.

addy10024 · Answer

I am not sure if my weird method is the way to do it.

I saw 5 factorial is 120

So from answer choices, I saw which are multiples of 4 and 3,

Option D was the only one

(Most probably I arrived at the answer by fluke only)

DylanD · Answer

Interesting. I used max/min and number properties to get an answer... Only took me 10 minutes

Permutation I realized each digit would occur 24 times at each slot Min Sum < Answer < Max Sum Noticing how spread the answers are, I thought to round each 3xxxx, 5xxxx,... down to 30,000, 50,000... and then up to 40,000, 60,000... The answer will fall somewhere in the middle. (3+5+6+7+8) * (24 #s / slot) * (10,000) < Sum < (4+6+7+8+9) * (24 #s / slot) * (10,000) (29)*(24)*(10,000) < Sum < (34)*(24)*(10,000) 6,760,000 < Sum < 8,160,000 Number Property: Unit Analysis Noting that the unit-digit of D and E differ, I then went back to summing the unit-digit based on the above logic (2 9 )*(2 4 )*(1) = ...3 6 .0 ANSWER: D

GMATCoachBen · Answer

Agreed,  stne  has a great solution above.

Ritaj1  The reason we know it has to be 5-digit numbers is this wording in the question:

"by taking the digits 3, 5, 6, 7 and 8,  exactly once "

Kinshook · Answer

What is the sum of all positive numbers formed by taking the digits 3, 5, 6, 7 and 8, exactly once?

Total such numbers = 5! = 120

Each digits appears at one place value 120/5 = 24 times

Sum of all such numbers = (3+5+6+7+8)*24*11111 = 7733256

IMO D

suman94 · Answer

The question needs to be "sum of all 5 digit numbers". By "all the numbers", it makes it a little ambiguous.

Coolguy22 · Answer

This is exactly what i thought first. But then when i read the question stem carefully, it read "exactly once" which implies all digits should be used

PriChats · Answer

Most elegant answer is what's posted above but I found it easier to visualize like this when I was solving the question above:

We know that it's a 5 digit non repeating number so for any place value we have 5 choices - 8,7,6,5,3. 
Now imagine a number which has it's one's digit fixed as digit 8 and has 4 other choices in hand:
xxxx8, for this number we will have 4! possibilities, which means 24 cases. What happens for 8 will happen for all the other numbers as well.

Therefore, 24 cases of when 8 is at unit's place, 24 cases for when 6 is at unit's place and so on.
this results in => 24*(8+7+6+5+3) = 24 * 29

Now if we fix ten's place, the same thing will happen: xxx8x : again we will have 4! cases and sum of all cases on ten's place would be 29, therefore 24*29 will again appear.

Now, consider a three digit number 123 which is nothing but (1*100 + 2*10 + 1*3)

Similarly the above cases would become : 
(29*24)   * (1+10+100+1000+10000)

Therefore, the total sum would be 29*24*11111 = 696 * 11111, 
here, you can easily calculate the product by shifting and adding digits of 696 - 5 times like this :
 696
 6960
 69600
 696000
6960000
+---------
7733256

Adit_ · Answer

Keep first digit fixed and each time you will have 4! numbers = 24.
80*24 = 1920.

Similarly for all digits once added up you get:
6960k
Options A,B,C ELIMINATED.

Now, Take the maximum possible combo for 2nd to last digit by keeping first digit as 3 then:
5,6,7,8 and arranging them keeping one of them fixed is 6 ways.
5k*6=30
Similarly adding them up you get 156k
Multiply by 5 to get absolte max possible even : 780k

Total sum = 780k+6960k = 7740k. But this value is assuming the max values from 2nd to last when thats not the case when u take other digits. We took the peak values in this case. So answer is definitely smaller than 7740k.

Answer:  Option D

What is the sum of all the numbers formed by taking the digits 3, 5, 6

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What is the sum of all the numbers formed by taking the digits 3, 5, 6

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