IIT JEE , AIEEE Forum - Mathematics , Algebra

AYUSHITANDON

how many times u write 5 while listing all integers from

1- 1000

giv solutn too

its of p and c

rvidya93

If im not wrong i think its this way:

1000/5 =200.

but out of these 200 , 100 end with zero but not 5.

so the remaining 100 is where u write 5 while listing the integers from 0 to 1000

if u consider 1 to hundread numbers,

then {50, 51, 52,...., 59} -{55}

will give you 9*10=90( multiplied with ten becoz u have to follow the same method for {150,...159},..........{950,......959})

and

{500, 501, ........599} - {505,515,....595} -{550, 551,........559}

will give you 100-20= 80numbers

so i think the final answer is 100+90+80=270

if i am wrong plz correct me

prathyu

is the answer 190 times if s the procedure is as follows:

we write 5; 19 times from 1-100 then from 1-1000 we write it 1000*19/100 =190

i think this is the answer

cheers!!!!!!!!!!!

if it is correct please do rate me.....

nadeemoidu

The question says "how many times u write 5"
So u can count 55 two times as u write 5 two times.
so i think the answer is 300

thedumbheadwithnobrain

is the answer 270 rate if its correct

dannyfoo

i also get 300 as the answer,

here is how i get it:
let n b number of 5 written:
we seperate the numbers in 1-1000 in 1,2,3,n 4 digit,

for 1 digit- n(1)=1
2 digit- n(2)=10+9
3 digits- n(3)=200+90+90

totL=300!

I GET IT BY COUNTING THE NUMBER OF 5 WRITTEN SEPERATELY...
example:
for 2 digit,
_ _ the possible number with 5 in 1st colum is 50-59,so n=10
2nd column is 15,25,...95=9

avinash.sharma

Dear AYUSHITANDON

The correct answer is 271 times.

Numbers	Description	Number Formats	Total outcomes
1-9	Only unit place digit is 5	5	1
10-99	(i) Unit place digit is 5 = 81 = 8 (Tenth place cannot contains 0 and 5) (ii) Tenth place digit is 5 = 19 = 9 (iii) Both digits are 5 = 1*1 = 1	$5 5* 55	18
100-999	(i) Unit place digit is 5 = 891 = 72 (Hundred place must not 0 & 5) (ii) Tenth place digit is 5 = 819 = 72 (Hundred place must not 0 & 5) (iii) Hundred place has 5 = 199 =81 (iv) Any two digits are 5 = (2* 119 + 811) = 26 (v) All three digits are 5 = 1	$5 $5 5** 55,55,$55 555	252
	Total outcomes		271

Note: * Mean any digit from 9 {0,1,2,3,4,6,7,8,9} digits.

$ Mean any digit from 8 {1,2,3,4,6,7,8,9} digits.

Raghudevan

5,15,25,35,45,55,65,75,85,95, (11)
105,115,125,135,145,155,165,175,185,195, (11)
205-----(11)
305--------(11)
405--------(11)
505,515,525,535,545,555,565,575,585,595(21)
605----------(11)
705---------(11)
805------------(11)

905--------------(11)

so the ans is 9*11 + 21 = 120

i think thats correct

nadeemoidu

I dont know if there's any thing wrong in the question but it clearly says " how many times u write 5 ".
So that means 55 is counted as 2 times , 555 as 3 times and so on.
So the answer should be 300 times ( 100 times in units position , 100 times in tens position ,100 times in hundreds position) .

If the question was how many integers in 1-1000 contain the digit 5, then 271 is correct.

avinash.sharma

Hi nadeemoidu,

you are right, thanks.

Numbers	Methods Description	Number Formats	Total outcomes
1-9	Only unit place digit is 5	5	1
10-99	(i) Unit place digit is 5 = 81 = 8 (Tenth place cannot contains 0 and 5) (ii) Tenth place digit is 5 = 19 = 9 (iii) Both digits are 5 = 1*1 = 1	$5 5* 55	18
100-999	(i) Unit place digit is 5 = 891 = 72 (Hundred place must not 0 & 5) (ii) Tenth place digit is 5 = 819 = 72 (Hundred place must not 0 & 5) (iii)

Ask & Discuss Questions with Community & Experts