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What is the sum of all 3 digit positive integers that can be formed us
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Updated on: 30 Dec 2009, 04:19
Question Stats:
66% (01:26) correct 34% (01:31) wrong based on 294 sessions
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What is the sum of all 3 digit positive integers that can be formed using the digits 1, 5, and 8, if the digits are allowed to repeat within a number? A. 126 B. 1386 C. 3108 D. 308 E. 13986 First off, I'm not even sure if this is worded coherently. Does this mean 111, 588, and 811 are included in this set?
I figured it couldn't be "A", "B", "C", or "D" because they would be too small:
A: 158 is already > 126 B: 855 + 588 + 585 is already > 126 C: 855 + 885 + 585 + 588 + 558 + 811 + 818 is already > 126 D: ... E: POE got me here.
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Originally posted by R2I4D on 30 Dec 2009, 03:29.
Last edited by R2I4D on 30 Dec 2009, 04:19, edited 1 time in total.




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Re: What is the sum of all 3 digit positive integers that can be formed us
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31 Oct 2015, 08:31
The Ans is E.
This can also be solved by using a formula
Sum of N numbers
With repetition = N ^ ( N  1 ) * Sum of the Numbers * 111..... N no. of times
With out repetition = ( N  1 ) ! * Sum of the Numbers * 111.... N no. of times.
Here repetition is allowed, therefore N is 3 ( No. of digits given ) and sum of No is 14.
3^2 * 14 * 111 = 13986.




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Re: What is the sum of all 3 digit positive integers that can be formed us
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30 Dec 2009, 03:47
there will 9 times 1 , 9 times 5 and nine times 8 at each place in a 3 digit no. with repetition sum will be =100*9*(1+58) + 10*9*(1+5+8) + 1*9*(1+5+8)=13986 in order to know how it is 9 times tot no. of ways =3*3*3=27 so in tot 27 words will be formed where each digit among 1,5,8 wil be repeated equal no. of times i.e 27/3=9 hence we have 9 , 1's ; 9 5's and 9 8's at each level
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Re: What is the sum of all 3 digit positive integers that can be formed us
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30 Dec 2009, 04:36
xcusemeplz2009 wrote: there will 9 times 1 , 9 times 5 and nine times 8 at each place in a 3 digit no. with repetition
sum will be =100*9*(1+58) + 10*9*(1+5+8) + 1*9*(1+5+8)=13986
in order to know how it is 9 times
tot no. of ways =3*3*3=27 so in tot 27 words will be formed where each digit among 1,5,8 wil be repeated equal no. of times i.e 27/3=9 hence we have 9 , 1's ; 9 5's and 9 8's at each level Wow. I have no idea what is happening here. Would you (or someone else) mind explaining a bit more...perhaps expounding on your method a little bit? Perhaps, the same deal, with a two digit number using 1 and 2, with digits allowed to repeat? [Answer: 11+12+21+22 = 66] How can I use xcusemeplz2009's method to arrive at the same answer?
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Re: What is the sum of all 3 digit positive integers that can be formed us
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30 Dec 2009, 06:34
R2I4D wrote: Wow. I have no idea what is happening here. Would you (or someone else) mind explaining a bit more...perhaps expounding on your method a little bit?
Perhaps, the same deal, with a two digit number using 1 and 2, with digits allowed to repeat?
[Answer: 11+12+21+22 = 66]
How can I use xcusemeplz2009's method to arrive at the same answer? let me try again let the no. be xyz Place value of no will be 100x+10y+Z.............eqn 1 TOTAL NO OF WAYS IN WHICH WE CAN MAKE A NO. WITH GIVEN CONDITION IS 3*3*3=27( as digits are getting repeated) now at unit place i.e Z if we fix 1 , then for y we have 3 options and for x we have 3 options , so total number where unit digit is one can be formed in 3*3=9 ways similarly for other digits at unit place can be done in 9 ways hence at unit place a digit is getting repeated 9 times and in the same manner in tenths place repetion for all digits will be 9 each now the task is to find the sum sum of digits at z place is 9*1+9*5+9*8 , WHICH IS SAME FOR Y AND X for that we need to put all the options in eqn 1 format i.e 100*9[1+5+8] + 10*9[1+5+8] + 9*[1+5+8]=13986 simple way is to remeber the quick formula to find out the repetition find out tot no. ways in which the no. can be formed and divide it by the no. of digit , then mutiply this factor with the summation of all the dig and the place value for second ex : 1,2 two dig no with repetn can be formewd in 2*2=4 ways , repetion factor=4/2 =2 ( tot no. of ways / tot no of dig) sum=10(place value) * 2 (rep factor) *(1+2 summation of dig)+2*3=60+6=66
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Re: What is the sum of all 3 digit positive integers that can be formed us
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01 Jan 2010, 11:06
Pretty impressive formula used by xcusemeplz2009. +1 for that! Nevertheless the question can be solved with POE as it is quite logical.... digits in scope = 1,5,8.. with this we can get highest 3 or 4 numbers  888, 885, 881, 858. Adding just these 4 numbers gives us = 3512 and we have 23 more numbers to be added... Hence the correct answer would be E of course! Cheers! JT
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Re: What is the sum of all 3 digit positive integers that can be formed us
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23 Aug 2016, 20:52
R2I4D wrote: What is the sum of all 3 digit positive integers that can be formed using the digits 1, 5, and 8, if the digits are allowed to repeat within a number?
A. 126 B. 1386 C. 3108 D. 308 E. 13986
I took a crude approach, the smallest number is 111 and largest is 888 and the number of integers are 3 * 3 * 3 = 27. the smallest is 111 and largest is 888, so the mean is around 500, so 500*27 = 13,500. E is the answer. +1 for kudos



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Re: What is the sum of all 3 digit positive integers that can be formed us
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31 Aug 2016, 22:16
What is the sum of all 3 digit positive integers that can be formed using the digits 1, 5, and 8, if the digits are allowed to repeat within a number?
A. 126 B. 1386 C. 3108 D. 308 E. 13986
Before solving, I knew that the number will be big but did not know how big. As I did not know a fancy way of solving this, I just tried to find the number if the digits are NOT allowed to repeat. Then it would be,
158 185 518 581 815 851
And just by rough estimation, it looked like it will be in 3000 range. As the question stem says digits ARE allowed to repeat, meaning, there should be more than 3000. The only answer more than 3000 was E.



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Re: What is the sum of all 3 digit positive integers that can be formed us
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05 Dec 2017, 15:28
Did a brute force one which took about 2 minutes 20, so I guess too long
158+185+851+581 = 1875==> 1875 + 555 +111+888 +155 +188 > exceeded option C. So picked E



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Re: What is the sum of all 3 digit positive integers that can be formed us
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13 Dec 2017, 22:00
[quote="R2I4D"]What is the sum of all 3 digit positive integers that can be formed using the digits 1, 5, and 8, if the digits are allowed to repeat within a number? A. 126 B. 1386 C. 3108 D. 308 E. 13986 Solution:It can be solved using below formula 111*(No. of ways 3 digit number can be formed with given numbers)*(1+5+8) No. of ways 3 digit number can be formed with repetition=3*3*3=27 This formula can also be extended for numbers formed w/o using repetition No. of 1 used in for multiplication = no of digits given=here it is 3 viz. 1,5,8
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