Bunuel wrote:
sugu86 wrote:
What is the sum of all possible 3-digit 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 3-digit 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 3-digit 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:
(n-1)!*(sum of the digits)*(111…..n times)In our original question: n=3. sum of digits=3+4+5=12. --> (3-1)!*(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