Latest M4maths Aptitude Question SOLUTION: find the sum of all number greater than 10000 that can be formed using digits 0,2,3,4,5 if repetition is not allowed.

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(#M40038023) M4MATHS Numerical Ability question sum Keep an EYE

find the sum of all number greater than 10000 that can be formed using digits 0,2,3,4,5 if repetition is not allowed.

Asked In M4maths RAKESH (9 years ago)

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sarvan  (9 years ago)

toal no. formed which is greater than 100000 =4*4*3*2*1
=96;
no. started with only 4 no. i.e 2,3,4,5
sum of no=6(2+3+4+5)(4*10^4+3*10^3+3*10^2+3*10+3);
=6*14*43333
=3639972

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Divyasri  (9 years ago)

0,2,3,4,5
numbers must be greater than 10,000 and repitition not allowed
so, we consider only 5 digit numbers
_ _ _ _ _
first digit do not start with 0
so, for first digit we have 4! ways
2nd digit also have 4! ways
(2 numbers are over)
3rd digit-3! ways
4th digit-2! ways
5th-1!
finally 4!*4!*3!*2!*1! ways
=24*24*6*2*1
=6912

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manoj kumar  (9 years ago)

first num does not starts with 0 so
4*5*4*3*2*1=480(since repetation is not allowed)

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RITESH KUMAR  (9 years ago)

to find out sum of n-digit numbers formed by digit using x,y,z...upto n digits
=(n-1)!*(x+y+z)*(111...n times )
so let fix 2 at ten thousand place then we have 0,3,4,5 to make 4 digit numbers
so total sum of numbers starting with 2 will be
2*24*10000+3!(0+3+4+5)*1111=48000+6*12*1111=127992
similary for 5
5*24*10000+3!(0+3+4+2)*1111=120000+6*9*1111=179994
similarly for 3
3*24*10000+3!(0+2+4+5)*1111=72000+6*11*1111=145321
similarly for 4
4*24*10000+3!(0+3+2+5)*1111=96000+6*10*1111=162660
so total sum=127992+179994+145321+162660=615697

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