Number N is obtained by writing all the digits from 1 to 100 in an

Question

Number N is obtained by writing all the digits from 1 to 100 in an ascending order. What is remainder when sum of first 31 digits of N is divided by 9?

A. 0
B. 1
C. 2
D. 3
E. 4

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Charli08 · Accepted Answer

The sum of the first 31 digits:
 First 9 digits  will be: 1, 2, 3.., 9

Leaving us 31-9 =  22 digits remaining  which will be  2 digit numbers  (starting from 10)
==> 22/2 =  11 digits starting from 10  
i.e.: 10, 11, 12... 20

Total set: {1, 2, .. 20)

Lets think how many times each digits occurs in this set:
1 : 1*12 = 12 
2: 2* 3 = 6
3: 2*3 = 6
4: 2*4 = 8
5: 2*5 = 10
...
9: 9*2 = 18
We know that each of the other digits aside from 1 and 2 will occur only twice since the range is 1 to 20

We then take the  sum of individual digits :
12 + 6 + 6 + 8 + 10 + 12 + 14 + 16 + 18 = 102
 
Now all is left is to  find the remainder :
 102/9 = 11  R3

==>  D

Not sure if there is a faster way!

aaronagyemang · Answer

median of 31 is 16.

Sum of first 31 digits of N is = 16x31 = 496

496 / 9 = 55.11

Remainder 
= 496 - (9x55) 
= 1

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mraparthy · Answer

This is the way i solved it:
so, the numbers when in order are these:
123456789
10111213141516171819
20212223242526272829
3031

Since it asks the sum of all of the indivudual numbers, it would be
1+2+3+...+9+1+0+1+1+1+2+1+3+...+2+0+2+1+2+2+...+3+0+3+1

Since the order does not matter in a additive process, we can rewrite above expression as:

1+2+3+...+9+1+1+1+1+...+1+2+3+...+9+2+2+2+...+1+2+3+...+9+3+0+3+1

So, in words, this is a summation of sums of smaller sections

Summation of x from 1 to 9 = 9(10)/2 = 45, and this repeats thrice.
summation of number 1 10 times = 1(10) = 10
summation of number 2 10 times = 2(10) = 20

So, the sum is 45(3)+10+20+3+1+3+0 = 172

and add individual numbers of 172 = 1+7+2 and divide by 9 to get remainder. 1+7+2 = 10/9 = R1

theunicornbuzz · Answer

At first I also made the mistake of calculating the sum from 1-31 giving me the remainder 1 when the sum is divided by 9. But the question asks for the sum of first 31 digits, which are
1,2,3...10 = 11 digits
11,12,...20 = 20 digits

Sum = 1+2+3+....+20 = 210 (Sum of A.P. from 1 to 20)

210 = 9*23 + 3

Hence remainder = 3

tirthshah2013 · Answer

here the first 31 digits of N will be 1234567891011121314151617181920, So the sum of all this digits can be divided into parts - sum of 1 to 10 = 55 and sum of 11 to 20 =155 so the sum of 1 to 20 = 210. therefore remainder when 210 is divided by 9 is 3. Hence answer is D.

shaurya_gmat · Answer

N = 1 2 3 4 5..9 10 11..20
(31 digits - 1 to 9 = 9 digits.. 31- 9 = 22 => 22/2 to 11 2 digit numbers, hence range is from 10 to 20)

Sum of the digits = 20/2   = 210
(210/9)R = 3

Therefore answer = 3

anshikag22 · Answer

the question asks to sum the digits and not the numbers so shouldn't we take the sum as 1+2+3+4+5+6+7+8+9+1+0+1+1+1+2......+1+9+2+0?

shaurya_gmat · Answer

Yeah.. I just got this message and re-solved the question without looking at my previous answer and the approach i followed was to add 1-9 + 10 1s + 0-9 + 2 (same as you've mentioned) and then divide by 9
So 102/9 -> 3
I think I misread the question in my first attempt and took a wrong approach. Lucky that I got the right answer in the first attempt :p

tten234 · Answer

The question is asking for sum of the first 31 digits, then divided by 9, then tell us the remainder. The way I did it:

Single digits: First 9 digits = 1, 2, 3, 4, ..., 9 = (1+9)/2 * 9 = 10/2 * 9 = 4*5 = 45
Double digits: 31 digits - 9 = 22 digits. 22/2 = 11 numbers
10, 11, 12, ..., 19, 20 (10-20 is 11 numbers, inclusively counting 10)
(10+20)/2 * 11 = (30/2) *11 = 15*11 = 165
45+165 = 210
210 divided by 9 = 207 R3

Answer, remainder is 3.

Vibhatu · Answer

question asked for first 31 digits of N
N=1234567891011121314151617181920
we should add this not upto 31.
What do you think  Bunuel ?

Prathammola · Answer

I divided them into 2 groups
1st 9 digits - 1 to 9
Sum of N terms = 9/2 * (1+9) =  45

Remaining Digits = 31-9 = 22 (so 11 numbers) (not digits - remember this)

In terms of numbers i.e., 10-20 (11 numbers and 22 digits)
but we can't just add the terms as they are, we need to add the digits so we have 10, 11, 12, 13, ... 20
- There are 9 (1s) in tens place =  9 
- Units place has 1-9. Sum =  45 
- Remaining 1+2 =  3  (From 10 and 20)

Now -  45+9+45+3

45 and 9 are multiples of 9 and 3 is not. So remainder = 3

Alternatively, 102/9 = Remainder = 3

DTT9 · Answer

So first 31 digits we have:

((9-1)/1) +1 = 9 digits from 1 to 9
Their sum will 9 * 5 = 45 (#of terms * avg)

We then need 22 more digits to arrive at 31, so 11 numbers (11 numbers with 2 digits each = 22 digits)
and from 20 to 10 we have 11 numbers ((20-10)/1) +1 =11 
Their sum will be, 11 * 15 =165 (#of terms * avg)

Thus, total sum is 45+165=210, which leaves a remainder of 3 when divided by 9.

Answer D

Number N is obtained by writing all the digits from 1 to 100 in an

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Number N is obtained by writing all the digits from 1 to 100 in an

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