Elitmus
Exam
Numerical Ability
Number System
What is the sum of all 3 digit nos formed by {1,3,5,7,9} with nos not repeated
Read Solution (Total 7)
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- Total nos are 5*4*3=60
Now smallest no. formed using (1,3,5,7,9) is 135 & largest no is 975.
Now sum of series when first and last no.is known = n/2(a+l)
n=no.of terms
a=first term
l=last term
So here we have sum = 60/2(135+975)
=33300 - 8 years agoHelpfull: Yes(15) No(0)
- sum of all digit is= 111(three digit) * (ways) * (sum of all digit)
=111 * 12 * (1+3+5+7+9)
=111 * 12 * 25
=33300 - 9 years agoHelpfull: Yes(13) No(0)
- 3 digits so if last digit is kept constant then remaining two number can be arranged in 4p2 ways , so 4p2 number of ways for one constant number , total numbers is 5 {1,3,5,7,9} so total arrangements will be 5*4p2 ,since one number will be repeated 4p2 times in each row addition of each row will be 9* 12+7*12+5*12+3*12+1*12=300, so each row will be having 300 sum ,total 3 rows because 3 digit number , so total sum is 33300
- 9 years agoHelpfull: Yes(3) No(1)
- Ans will be : 111 ( 3 digits ) * (1+3+5+7+9) * 12 (no of time digit repeats )
= 111 * 25 * 12
=33300 - 9 years agoHelpfull: Yes(2) No(2)
- total number of 3 digits from the given numbers 1,3,5,7,9=5*4*3=60 numbers
if first digit is 1 then three digit number is like{135,137,139,153,157,159,173,175,179,193,195,197}these are 12 numbers whose first digit 1 as like, numbers start with 3,5,7,9 are 12
sum=1+3+5+7+9=25
12*100*25+12*10*25+1*12*25
=30000+3000+300
=33300 - 9 years agoHelpfull: Yes(2) No(0)
- [(1+3+5)+(3+5+7)+(5+7+9)+(7+9+1)+(9+1+3)]*111*2!
=75*111*2
=16650 - 9 years agoHelpfull: Yes(1) No(3)
- 33300
You can do it by using permutations - 9 years agoHelpfull: Yes(1) No(2)
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