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The sum of all the digits of the integers from 18 to 21 incl
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The sum of all the digits of the integers from 18 to 21 inclusive is 24 (1 + 8 + 1 + 9 + 2 + 0 + 2 + 1 = 24). What is the sum of all the digits of the integers from 0 to 99 inclusive? A. 450 B. 810 C. 900 D. 1000 E. 1100
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Originally posted by shrive555 on 28 Oct 2010, 17:05.
Last edited by Bunuel on 24 Jul 2014, 02:05, edited 1 time in total.
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Re: The sum of all the digits of the integers from 18 to 21 incl
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28 Oct 2010, 19:44
In the first 99 numbers, every digit appears 20 times. (Think 2  It appears 10 times in units digit and 10 times in tens digit) So all you need is (1 + 2 + 3 + ... + 9) x 20 = 45 x 20 = 900 Note: Remember 1 + 2 + 3 +...n = n(n+1)/2
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Re: The sum of all the digits of the integers from 18 to 21 incl
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28 Oct 2010, 17:34
What we want is 0+(1+2+3+....+9)+(1+1+1+2+1+3...1+9)+.....(9+1+9+2..+9+9)
1+2+..+9=(9*10)/2=45 Similarly 11+12+...+19 will have 10 1s and again a sequence of (1+2+3+..+9). 20 series will be (2+0+2+1..2+9)=20+(1+2+3..+9)
Extrapolating this, we will get 45+(10+45)+(20+45)+....(90+45) = 45+(9*45)+(10+20+30+...+90)=450+10(1+2+..+9)= 450+10(45)=900.
Not sure if there is a better way to do this.



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Re: sum of 099
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28 Dec 2010, 12:07
Simple counting is what needed in this question First of all we need to count how many 1 's are present from 099 i.e 1, 1119, 21, 31, 41, 51, 61, 71, 81 ,91.... thus total of 20 1's are present Thus total will be 1 * 20 = 20
Similarly there will be 20 2's and 20 3's ....... 20 9's Therefore of each number will be 2*20 = 40, 3*20 = 60 ....... 9*20 = 180
Thus total of 20+40+60+80+100+120+140+160+180 = 900
Answer is "C"



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Re: sum of 099
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28 Dec 2010, 19:40
This is my method
Sum of the digits from 1 to 9 = Sum of consecutive numbers = first plus last divided by two X number of terms = 5 X 9 = 45
Number of times the digits from one to nine appear in the sequence from 1 to 99 = 10
First part of the answer = 45 X 10 = 450
Number of times the number one (two..nine) appears in the sequence from 1 to 99 in the units digit = 10 1 X 10 = 10 2 X 10 = 20 ..... Sum of consecutive numbers = First Plus Last / 2 X Number of terms
90 + 10 = 100 / 2 = 50 X 9 = 450 Second part of the answer
Final answer 450 + 450 = 900
It worked for me...
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Re: sum of 099
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28 Dec 2010, 20:52
1+2+3+4+5+6+7+8+9 (1+0)+(1+1)+(1+2)+(1+3)+(1+4)+(1+5)+(1+6)+(1+7)+(1+8)+(1+9) (2+0)+(2+1)+(2+2)+(2+3)+(2+4)+(2+5)+(2+6)+(2+7)+(2+8)+(2+9) and so on.. You don't have to write the above in the test... just quickly visualize it. What do we see here? 1+2+... +9, which adds to 45 appears 10 times => 45*10 = 450  (1) also, we have 10 1's, 10 2's and so on => 10 (1 + 2 +...+ 9) => 10 * 45 = 450  (2) (1) + (2) = 900 Hope this helps!
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Re: sum of 099
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24 Jan 2011, 12:50
think of all numbers between 099 as doubledigit numbers(1=01, 2=02 and so on)
sum of unit digits digits 19 repeat 10 times (01 to 09, 11 to 19 .... 91 to 99)
sum of unit digits \(= 10*[\frac{9}{2}*(1+9)] = 450\)
sum of tens digits digits 19 repeat 10 times (10 to 19, 20 to 29 .... 90 to 99)
sum of unit digits \(= 10*[\frac{9}{2}*(1+9)] = 450\)
total = 450 + 450 = 900
Ans: C



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Re: sum of 099
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24 Jan 2011, 21:06
shrive555 wrote: The sum of all the digits of the integers from 18 to 21 inclusive is 24 (1+8 + 1+9 + 2+0 + 2+1 = 24). What is the sum of all the digits of the integers from 0 to 99 inclusive?
450 810 900 1000 1100 0 to 99 could be considered 100 two digit numbers(00, 01, 02....99) since when 0 is added, it doesn't change the sum. To write these 100 numbers, 200 digits will be used. Since there are 10 digits (0 to 9), each digit will appear 20 times. So total sum = 20*0 + 20*1 + 20*2 + 20*3 +.....20*9 = 20*9*10/2 = 900
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Re: sum of 099
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03 May 2011, 20:45
0+1+2+...+9 = x = 45
now we have, unit's digit sum from 099 = 10 * x ten's digit sum from 099 = 10* x
hence total is 20*x = 20 * (45) = 900. Hence C.



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Re: sum of 099
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07 May 2011, 09:44
we can see the pattern here
units digit : 1,2,3,4,5,6,7,8,9 repeated 10 times
=> unit digit sum = 10*(1+2+3+4+5+6+7+8+9)
tens digit : 1 repeated 10 times, 2 repeated 10 times...9 repeated 10 times
=> tens digit sum = 10*(1+2+3+4+5+6+7+8+9)
so their sum = 20*(1+2+3+4+5+6+7+8+9) = 900
Answer is C.



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Re: The sum of all the digits of the integers from 18 to 21 incl
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03 Jan 2014, 05:57
shrive555 wrote: The sum of all the digits of the integers from 18 to 21 inclusive is 24 (1+8+1+9+2+0+ 2+1=24). What is the sum of all the digits of the integers from 0 to 99 inclusive?
A. 450 B. 810 C. 900 D. 1000 E. 1100 1+2+3+4...+9 sum of all digits is 45 Each digit will appear ten times in the units digit = 10*45= 450 Each digit will appear once in the tens digit = 1*45*10 = 450 (The ten is cause its in the tens digit) Total= 900 Answer is C Hope it helps Gimme Kuddos Cheers J



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Re: The sum of all the digits of the integers from 18 to 21 incl
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03 Jan 2014, 06:12
Sum from 0  9 = 45 Sum from 10  19 = (1x10 + 45) = 55 Sum from 20  29 = (2x10 + 45) = 65 . . . Sum from 90  99 = (9x10 + 45) = 135 Total sum = 55 + 65 ..... 135 = 10(45+135)/2 = 900
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Re: The sum of all the digits of the integers from 18 to 21 incl
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24 Feb 2014, 22:24
0 to 9 in units place appear 10 times, so 10x0 + 10x1 + ............ + 10x9 = 450 0 to 9 in tens place appear 10 times, so 10x0 + 10x1 + ............ + 10x9 = 450 450+450=900 = Answer = C
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The sum of all the digits of the integers from 18 to 21 incl
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17 Jun 2016, 09:08
shrive555 wrote: The sum of all the digits of the integers from 18 to 21 inclusive is 24 (1 + 8 + 1 + 9 + 2 + 0 + 2 + 1 = 24). What is the sum of all the digits of the integers from 0 to 99 inclusive?
A. 450 B. 810 C. 900 D. 1000 E. 1100 Many of the overmentioned require a good number sense, which I do not posses, so I'll show you my " layman" approachWe want the sum of the digits from 0 to 99, so I approximated: 09 > 45 > (9+0)*10/2 4049 > 85 (13+4)*10/2 9099 > 135 (18+9)*10/2 We can see at a glance that the "weight" goes up as the numbers go up (meaning the difference between 85 and 45 is 40, while 13585 is 50, this means that the second part of this sequence carries more weight for our result), so we know that the final answer has to be more than 850 (85*10) but close to it, and that's just 900: the answer is C.Kudos if you like.Den.



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Re: The sum of all the digits of the integers from 18 to 21 incl
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25 Aug 2018, 04:21
VeritasKarishma wrote: In the first 99 numbers, every digit appears 20 times. (Think 2  It appears 10 times in units digit and 10 times in tens digit)
So all you need is (1 + 2 + 3 + ... + 9) x 20 = 45 x 20 = 900
Note: Remember 1 + 2 + 3 +...n = n(n+1)/2 VeritasKarishma, your solution to this is great, but do you not think that there must be reason to give 18 to 21 = 24. Is there a possibility to use this and reach to the answer earlier? Usually Questions mentions " such as" or " for example" if they add info just to make us understand the question better!!! Since this question is from Magoosh, mikemcgarry, can you provide the official solution?



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Re: The sum of all the digits of the integers from 18 to 21 incl
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Re: The sum of all the digits of the integers from 18 to 21 incl
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