Daniel wants to get a 100-day streak on Brilliant.org. He plans to do so in the following manner:

On the first day, he does $1$ problem. On the second day, he does $2$ problems. On the third day, he does $3$ problems. This pattern continues on until the $10$ th day. On the $10$ th day, he does $1+0=1$ problem. On the $11$ th day, he does $1+1=2$ problems. In general, on the $\overline{ab}$ th day, he does $a+b$ problems. Finally, on the $100$ th day, he does $1+0+0=1$ problem, completing his streak.

In total, how many problems did he do?

The answer is 901.

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I thought, for example, in the 47th day you do 4+7=12, 1+2=3 problems. Apparently not =\

Klahrinz William Catubig
- 7 years, 2 months ago

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The same at first!

Ramasubramaniyan Gunasridharan
- 7 years, 2 months ago

4+7=11 actually :D I did the same mistake too

Anuran Sonowal
- 7 years, 1 month ago

I think it would have been really funny if this was submitted this after you had solved 901 problems. Cool problem nonetheless!

Trevor B.
- 7 years, 2 months ago

Oh for second step did mistake

U Z
- 6 years, 4 months ago

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jack if u could elaborate on the gauss trick that would be helpful since i don't knw what that is

Vasudev Kp
- 7 years, 2 months ago

Nice trick!

Wow, very clever! I didn't think of that.

Daniel Liu
- 7 years, 2 months ago

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Cool problem. Don't you think that the guy in the cartoon picture at the top of the problem looks just like Daniel Chiu? It's startlingly similar.

Finn Hulse
- 7 years, 2 months ago

very clever!

Eloy Machado
- 7 years, 2 months ago

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....

5 Helpful
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There's an easier way to do this

Joshua Ong
- 7 years, 2 months ago

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yes bro,i'm just go on with long ones,hehe....wating time....

MOHD NAIM MOHD AMIN
- 7 years ago

The same as mine

Ramasubramaniyan Gunasridharan
- 7 years, 2 months ago

i solved it almost the same way 1-91=55 2-92=65 3-93=75 . . 9-99=135 10-100=46

```
total=901
```

Vasudev Kp
- 7 years, 2 months ago

same as mine

Nihar Mahajan
- 6 years, 8 months ago

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 $450 + 450+1 = \boxed{901}$ .

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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)

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

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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.

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×

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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 = $\boxed{901}$