21Surbhi : Ans pl. jaldi!!!
find the sum of all natural numbers n such that 1000 ≤ n ≤ 4000, that can be made with the digits 0, 1, 2, 3, 4 if repetition of digits in the same number is allowed.
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27 Answers
62@manipal.. yes it is very much in the syllabus.
I could even add that it is very standard question in a way..
1oh yes.... i got my mistake..[1].... thanx....... a lot.... main 4000 add karna bhool gayi thi!!!!
21Wat link re!
c this :
guyz n galzz i'll giv u da basic stuct. :
1) find tot no. of cases
2)Sum up the digits at each individual place (i.e. ones, tens, hundreds)
3) Rem only 1,2,3 appear at first dig. and for other digits : 0,1,2,3,4
4) last mein add 4000 to ur Ans. coz u didnt consider 4 initially
21WELL DONE SURBHI!!!!!!!!!!1
i did the same mistake..........
RE-READ pt4 in post #9
21Yah got da ans......
guyz n galzz i'll giv u da basic stuct. :
1) find tot no. of cases
2)Sum up the digits at each individual place (i.e. ones, tens, hundreds)
3) Rem only 1,2,3 appear at first dig. and for other digits : 0,1,2,3,4
4) last mein add 4000 to ur Ans.
11surbi
find the sum of all natural numbers n such that 1000 ≤ n ≤ 4000, that can be made with the digits 0, 1, 2, 3, 4 if repetition of digits in the same number is allowed.
this can be done by combination but u have asked 4 the sum of all the digits so u have to find each digit
1000
1001
1010
1100
1110
1101
1011
1111
1021
1023
1024
1042
1032
1012
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OR EITHER I AM GOING NUTS OR THERE IS A EASIER METHOD TO DO SO
PLEASE RECHECK THE QUES KINDLY[1]
1ohk... but what about tens, hundreds n thousands place... if we fix any no. of ten's place.. then there will be overcounting... so how to remove that??
1Look, on fixin the one's place with 1, the no. of nos. that can hv that is 3 x 5 x 5. so that means on adding those nos. the sum of the one's place wil be 75. Do the same while fixing for other digits....2, 3, 4. Not 0 since adding 0 +0+....= 0.
11surbhi check wat i said earlier
u didnt asked the number of digits but u asked their sum[2]
1hmmm...oh yeah...less than 4000. well, chek d method, although i dont think it's of much use.
1@ritika....
if u fix 1 as the last digit..
then the no. of natural no. formed will be 3 x 5 x 5..... not 4 x 5 x 5
1well, i'm no great fella in math, but what i usually do for p&c is use FPC. So, fix the last digit as, say, 1, coz adding zero infinitely will still leave you with just that. Now, the remaining digits that can be used in other places...4 x 5 x 5. These are the no. of nos. ending with 1. so 1 x 100. now do the same for 2, 3, 4. Then progress to ten's place.
Well, the time's u'll take for this, u might as well adopt manipal's method....
But i dunno...how WUD one do this?
1@Mani...
the question is correct.[1]... n there must be some easier method for dis....
