Help with this math problem

Question

For any positive integer n, let O(n) be the sum of the odd digits of n and E(n) be the sum of the even digits of n. Let

X = O(1)+O(2)+…+O(100)

and

Y = E(1)+E(2)+…+E(100)

Find X and Y.

Answer 1

know 1+2+...+n=[n(n+1)]/2

so,,, 1+2+...+100=50*101

E(100)=2+4+...+100=2(1+2+...+50)=50*51

O(100)=1+3+...+99=[1+2+...+100]-[2+4+...+100]=50*101-50*51=50*50

E(99)=49*50

O(99)=50*50

E(98)=49*50

O(98)=49*49

E(97)=48*49

O(97)=49*49

E(96)=48*49

O(96)=48*48

for O(1)+O(3)+...+O(100)

have X(100)=2*(1²+2²+...+50²)=2*[50*51*101}/6,

,,,,,formula [n*(n+1)*(2n+1)]/6=1²+2²+...+n²

for Y(100)

Y(100)=2*(1*2+2*3+4*5+5*6+...+49*50)+50*51

=2*([1²+1]+[2²+2]+[3²+3}+...[49²+49])+50*51

Y(100)=2*[(49*50*99)/6]+2*[(49*50)/2]+50*51