The number of times the digit 5 will appear while writing the integers from 1 to 1000 is

a)269

b)300

c)271

d)302

Correct answer is option 'C'. Can you explain this answer?

Related Test: UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT)

22 Answers

Raja Gopal
answered
Jan 17, 2020

By writing a Python Program or C Program

You can get the answer in short time .

You can get the answer in short time .

Manju Sood
answered
Mar 19, 2020

From 1 to 1000, the numbers in which 5 can occur could be of one digit, two digits or three digits.

(a) There is only one 5, this can happen in two ways _5 and 5_. In the first case (_5) the blank Place can be filled in 8 ways(as 0 and 5 cannot appear at that place), while in the second case (5_) the blank place can be filled in 9 ways (5 cannot appear there). Total 9 + 8 = 17 ways.

(b) There are two 5s. In this case only ONE possibility.

(a) Only one 5. Then, 5 can occupy three positions. 5 _ _ or _ 5 _ or _ _ 5. In the first case (5_ _), remaining two positions can be filled in 9 way each. So total 9 × 9 = 81 possibilities. In the second case (_ 5 _) first position can be filled in 8 ways and last position can be filled in 9 ways. So total 9 × 8 = 72 possibilities. Same will be true for the third (_ _ 5) case. So total 72 possibilities.

(b) Only two 5. This can be done in three ways 55_ or 5_5 or _55. In first (55_) and second (5_5) case it can be filled in 9 ways each. While in the third case (_55) it can be filled in 8 ways. So total 9 + 9 + 8 = 26 possibilities.

(c) All three digits are 5. This can be done in only ONE way. i.e, 555.

So, total = 1 + 17 + 1 + 81 + 72 + 72 + 26 + 1 = 271.

Bharath Reddy Gundala
answered
May 21, 2020

B) 300

Any number between 1 and 999 can be expressed in the form of xyz where 0 < x,y,z="" /> 9

Case 1. The numbers in which 5 occurs only once. This means that 5 is one of the digits and the remaining two digits will be any of the other 9 digits

You have 1*9*9 = 81 such numbers. However, 5 could appear as the first or the second or the third digit. Therefore, there will be 3*81 = 243 numbers (1-digit, 2-digits and 3- digits) in which 5 will appear only once.

Case 2. The numbers in which 5 will appear twice. In these numbers, one of the digits is not 5 and it can be any of the 9 digits.

There will be 9 such numbers. However, this digit which is not 5 can appear in the first or second or the third place. So there are 3 * 9 = 27 such numbers.

In each of these 27 numbers, the digit 5 is written twice. Therefore, 5 is written 27*2= 54 times.

Case 3. The number in which 5 appears thrice - 555 - 1 number. 5 is written thrice in it.

Therefore, the total number of times the digit 5 is written between 1 and 999 is 243 + 54 + 3 = 300

Any number between 1 and 999 can be expressed in the form of xyz where 0 < x,y,z="" /> 9

Case 1. The numbers in which 5 occurs only once. This means that 5 is one of the digits and the remaining two digits will be any of the other 9 digits

You have 1*9*9 = 81 such numbers. However, 5 could appear as the first or the second or the third digit. Therefore, there will be 3*81 = 243 numbers (1-digit, 2-digits and 3- digits) in which 5 will appear only once.

Case 2. The numbers in which 5 will appear twice. In these numbers, one of the digits is not 5 and it can be any of the 9 digits.

There will be 9 such numbers. However, this digit which is not 5 can appear in the first or second or the third place. So there are 3 * 9 = 27 such numbers.

In each of these 27 numbers, the digit 5 is written twice. Therefore, 5 is written 27*2= 54 times.

Case 3. The number in which 5 appears thrice - 555 - 1 number. 5 is written thrice in it.

Therefore, the total number of times the digit 5 is written between 1 and 999 is 243 + 54 + 3 = 300

Parimi Gayathri
answered
May 03, 2020

From 1 to 100

5,15,25,35,45,55,65,75,85,95 Among these ten 5's are in units place

50,51,52,53,54,55,56,57,58,59 Among these ten 5's are in tens place

so ten 5's in units place + ten 5's in tens place = 20 times

=> from 1 to 1000 the number of times the digit 5 appears =

20 * 10 = 200 times

=> from 500 to 600 the number of times the digit 5 appears= 100 times

then total number of times 5 appears from 1 to 1000=

200+ 100 = 300 times

Hence 300 is the right answer

5,15,25,35,45,55,65,75,85,95 Among these ten 5's are in units place

50,51,52,53,54,55,56,57,58,59 Among these ten 5's are in tens place

so ten 5's in units place + ten 5's in tens place = 20 times

=> from 1 to 1000 the number of times the digit 5 appears =

20 * 10 = 200 times

=> from 500 to 600 the number of times the digit 5 appears= 100 times

then total number of times 5 appears from 1 to 1000=

200+ 100 = 300 times

Hence 300 is the right answer

Bidyapati Biswaranjan Sa
answered
19 hours ago

5 will come 100 times in ones place (1-1000).

5 will come 100 times in tens place (1-1000).

5 will come 100 times in hundreds place (1-1000).

So in total 5 will come 300(100+100+100) times from 1 to 1000. So I think 300 is the right answer.

5 will come 100 times in tens place (1-1000).

5 will come 100 times in hundreds place (1-1000).

So in total 5 will come 300(100+100+100) times from 1 to 1000. So I think 300 is the right answer.

Krishnendu S Nair
answered
Mar 16, 2020

How can the answer be 271 ...it's 300.

I am not able to get 271 as it's answer....

I am not able to get 271 as it's answer....

Vikash Anand
answered
2 weeks ago

Correct answer is 300,

in 1-100 (1-49 -- 5-5s exist,51-59-- 11-5s,60-100--4-5s)

total=11+5+4=20

so,there are 10 such groups,1-100,101-200,201-300-901-1000,

so we will do -- 20*10=200,

but in 500-599-there is additional 100- 5s in 100th place ,

so we will add 200+100=300,

simple

in 1-100 (1-49 -- 5-5s exist,51-59-- 11-5s,60-100--4-5s)

total=11+5+4=20

so,there are 10 such groups,1-100,101-200,201-300-901-1000,

so we will do -- 20*10=200,

but in 500-599-there is additional 100- 5s in 100th place ,

so we will add 200+100=300,

simple

Vinod Patil
answered
Apr 25, 2020

(i:.e; 1-100 normally counts 5,15,25...95 =10

50=1

51,52,53...59=9

So Total 20)

105 to 195= Counts again 20

205 to 295= Counts again 20

305 to 395= Counts again 20

But 405 to 500 = Counts 21 (because last number 500 will be added)

So total 1 to 500= Counts 20+20+20+20+21=101

[ Next, 501 to 510 here remember 505 =Total Counts 11

up to 550 =11*5=55+1 =56 (+1 to include 550)

551 to 560 = Counts 20

561 to 599 = Counts 44

Total 56+20+44= 120 ]

remaining 605 to 695 (same counts as 1 to 100) 20

705 to 795 = 20

805 to 895 = 20

905 to 995 = 20

So total 20+20+20+20= 80

1 to 1000

{Number 5 counts 101+120+80= 301 times}

50=1

51,52,53...59=9

So Total 20)

105 to 195= Counts again 20

205 to 295= Counts again 20

305 to 395= Counts again 20

But 405 to 500 = Counts 21 (because last number 500 will be added)

So total 1 to 500= Counts 20+20+20+20+21=101

[ Next, 501 to 510 here remember 505 =Total Counts 11

up to 550 =11*5=55+1 =56 (+1 to include 550)

551 to 560 = Counts 20

561 to 599 = Counts 44

Total 56+20+44= 120 ]

remaining 605 to 695 (same counts as 1 to 100) 20

705 to 795 = 20

805 to 895 = 20

905 to 995 = 20

So total 20+20+20+20= 80

1 to 1000

{Number 5 counts 101+120+80= 301 times}

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