Latest Elitmus Aptitude Question SOLUTION: A five digit number is formed by using digits 1,3,5,7 and 9. without repeating any of them. what is the sum all such possible numbers? a) 6666600 b) 6666660 c) 6666666 d) none of t

5*4*3*2*1=120/5=24
so sum of 5 digit number
24*10000(1+3+5+7+9)+24*1000(1+3+5+7+9)+24*100*(1+3+5+7+9)+24*10*(1+3+5+7+9)+24*1*(1+3+5+7+9)
after solving we get
6000000+600000+60000+6000+600
=6666600
so ans a is correct

For the 1st digit we can choose any of the 5 numbers, for the 2nd digit we then have 4 numbers to choose from (because we can't choose the one we've already chosen), for the 3rd digit we have 3 choices, then 2 for the 4th digit, and 1 for the 5th digit. The total unique arrangements is then: 5*4*3*2*1 (commonly written 5!) = 120 different 5-digit numbers.

Out of these 120, our 5 possible values will each appear an equal number of times in the units place, the tens place, the hundreds place, etc. In other words, we will have the same amount of numbers starting with 5 as we will numbers starting with 1, 3, 7 and 9, the same amount of numbers ending in 1 as numbers ending in 3, 5, 7 and 9, etc. This applies to every place value in our 5-digit numbers.

120/5 = 24, so there will be 24 numbers starting with 1, and 24 starting with 3, and 24 starting with 5, etc...

This helps us because clearly we don't want to actually add up all of these 120 numbers. We need a shortcut if we're going to find the sum.

We can make use of the fact that in any 5-digit number we have one digit representing 10,000, another representing 1000, another for 100, and 10 and 1. For example, in 53971, the 5 represents 5*10,000, the 3 is 3*1000, etc...

So we know each of our 5 values will appear in each position 24 times. This means that we can sum them up like this:

Sum of ten thousands digits: 24*10,000*(1 + 3 + 5 + 7 + 9)
+
Sum of thousands digits: 24*1,000*(1 + 3 + 5 + 7 + 9)
+
Sum of hundreds digits: 24*100*(1 + 3 + 5 + 7 + 9)
+
Sum of tens digits: 24*10*(1 + 3 + 5 + 7 + 9)
+
Sum of units digits: 24*1*(1 + 3 + 5 + 7 + 9)

= 6,666,600

formula :- sum of all the no which can be formed by the given no.in these case 1,3,5,7 and 9
how many possible no. formed from these no.=5!=120
now, the sum =120/2(13579+97531)
=6666600

formula is

Sum of all such possible numbers=(Sum of all digits)*(n-1)!*(1111....n terms)
=>(1+3+5+7+9)*(5-1)!*(11111)
=>25*4!*11111
=>25*24*11111
=>600*11111
=6666600

Option is a)

Ans should be ...(a
600*1+600*10+600*100+600*1000+600*10000=6666600

6666600 is correct ans

let us take an example of 3 numbers that is 1,3,5

number of different no from above is 3!=6

135
315
531
351
513
153

we can see that from above that every no is repeating 2 times in vertical line of example that is

no of numbers made by arranging/no used in arranging = 6/3 = 2

we can see in above every vertical line has a sum of 6*(1+3+5)

hence no of diff no formed by 1,3,5,9,7 = 5! = 120

no of repeatations of 1,3,57,9 in diff numbers = 120/5 =24

hence sum of every vertical line = 24*(1+3+5+7+9) = 600

our unit digit is 0 taking carryover of 60 and adding it to 600 that is 660 ,
hence our tens digit comes out to be 0

hence a is correct

Hey NITHA VARGHESE,

Your solution and way of explanation is damn good. Thank you. Appreciate it.