Digit sum with given constraints
Number Theory
Level
4
Find the sum of all 3-digit natural numbers that contains at least one odd digit and one even digit.
The answer is 370775.
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1 solution
We have to find sum of all the 3 digit numbers with all even and all odd digits and subtract it from the sum of all 3 digit numbers
_ _ _ let this be a 3 digit number
So with 1 in hundred place 1,3,5,7,9 will repeat 25 times
because of similarity sum of all odd digit 3 digit numbers is 2500(1+3+5+7+9)+250(1+3+5+7+9)+25(1+3+5+7+9)=69375
similarly for even digits it is 2500(0+2+4+6+8)+250(0+2+4+6+8)+25(0+2+4+6+8)=55500
But we have to subtract the numbers with 0 in hundreds place, as they are 2 digit numbers 50(2+4+6+8)+5(2+4+6+8)=1100
now sum of all the 3 digit numbers is n/2*[2a+(n-1)d]=494550
so final sum is = 494550-(69375+55500-1100)=370775