A 100 Streak Goal

The sum of digits at the unit place from 1 to 100 will be $10 \times \frac{9 \times (9+1)}{2} = 450$ as the digits in the unit place (i.e., 0 to 9) repeat 10 times.

The sum of digits at tens place will be $(10 \times 0) + (10 \times 1) +\ldots + (10 \times9)$

= $10 \times \frac{9 \times (9+1)}{2} = 450$ .

The only number in the hundreds place is 1.So the total no. of problems solved = 450 + 450 + 1 = $\boxed{901}$

We borrow the old Gauss trick: if you sum the number of problems solved in day $x$ and in day $99-x$ , when $x$ ranges from $1$ to $49$ , you get $18$ . So the total number of solved problems is just $50\cdot 18 + 1 = 901$ .

Hello all,for total number of problems from days [1,100],

as for

[1-10] = 1+2+3+4+5+6+7+8+9+(1+0) = 46

[11-20] = 2+3+4+5+6+7+8+9+10+(2+0) = 56

[21-30] = 3+4+5+6+7+8+9+10+11+(3+0) = 66

[31-40] = 4+5+6+7+8+9+10+11+12+(4+0) = 76

[41-50] = 5+6+7+8+9+10+11+12+13+(5+0) = 86

[51-60] = 6+7+8+9+10+11+12+13+14+(6+0) = 96

[61-70] = 7+8+9+10+11+12+13+14+15+(7+0) = 106

[71-80] = 8+9+10+11+12+13+14+15+16+(8+0) = 116

[81-90] = 9+10+11+12+13+14+15+16+17+(9+0)= 126

[91-100] = 10+11+12+13+14+15+16+17+18+(1+0+0)=127

Since already calculated to 100 days, therefore total problems = 46+56+66+76+86+96+106+116+126+127 = 901

thanks....

WE first take the sum of the digits in the unit's place of all the numbers,the result will be $(1+2+......+9)*10 = 450$ .

Now we take the sum of the digits along the tens place and the sum will be = $10*0+ 10*2+10*3+......10*9 = 450$ .

Clearly the only number at the hundredth place is $1$ .

So the sum of all the digits is $450 + 450+1 = \boxed{901}$ .

very easy, the problem can be made simpler by adding all digits from 1to 99 i.e all 1's + 2's +...+9's so all nos comes 20 times hence sum of digit of 1-99 is=20(1+2+3+...+9)=20(45)=900 and then add fo 100 so answer is 900+1=901

I pretty much just bashed it out. The first one through nine produces forty-five. From there, the eleven through twenty produces fifty-five. Each subsequent group produces ten more than the last. Thus, their total is $45+55+65...+135 \Longrightarrow 900$ . But then you have to add one to account for the one-hundred. Thus, the desired answer is $\boxed{901}$ .

sum=0

for i in range(101):

while i>0:

    temp=int(i)%10

    print(i)

    sum+=temp

    print(i,temp,sum)

    i/=10

print(sum)

46 (sum of digits in 1 to 10). This will keep increasing by 10. 46+56+66+.... + 126 + 126 + 1 = 901

I solved it using C: int main( ) { int j,i=0,sum=0,k,bSum=0;

for(j=0;j<=9;j++)
{
    for(k=0;k<=9;k++)
    {
        sum=k+i;
        bSum+=sum;
    }
    i++;
}
printf("%d",bSum);

} This provides the sum from 1 to 99, then you add 1 to that to obtain the final result.