Latest TCS Aptitude Question SOLUTION: The citizens of planet nigiet are 5 fingered and have thus developed their decimal system in base 8. A certain street in nigiet contains 1000 (in base 5) buildings numbered 1 to 1000. How many 3s are used in numbering these buildings? Express result in terms of base 10. a) 54 b) 64 c) 75 d) 100

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(#M40001523) TCS aptitude question latest Keep an EYE

The citizens of planet nigiet are 5 fingered and have thus developed their decimal system in base 8. A certain street in nigiet contains 1000 (in base 5) buildings numbered 1 to 1000. How many 3s are used in numbering these buildings? Express result in terms of base 10.
a) 54 b) 64 c) 75 d) 100

Asked In TCS Submited By-- Aditi Sharma (1 year ago)

Puzzle Status: Solved     By »» aratrika

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aratrika  (1 year ago)

no of 3s b/w 1-99 is 20,100-199 is 20 and so on. so leavn d range b/w 300-399 total no of 3s is(9*20=)180. no of 3s b/w 300-399 is(20+100=)120.
total no of 3s b/w 1-1000=120+180=300
in base 10 300=3*5^2=75.
ans:c

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Neha Ray  (1 year ago)

Keeping the fact in mind that the decimal system of citizens is in base 8,
we have,
unit place no of 3's are given as....3,13,23,.....73
103,113,123,.....173
203,213,223,.....273
303,313,333,.....373
likewise upto 703,713,723,.......773
hence total no of 3 in unit's place=8*8=64

again, no of 3's in decimal place are as follows
30,31,32,33.....37
130,131,132,.....137
.........
730,731,732,......733
i,e(8*8)=64

Hence total no of three between 1 to 1000=(64+64+64)=192
i,e (8*8)=64
and total no of 3s in 100's place will be
300,301.....307
310,311,....317
...............
370,371,....377

Is this solution Helpfull? Yes (4) | No (8)  

(#M40000886) TCS MATHS PUZZLE Base 8 Keep an EYE

The citizens of planet nigiet are 8 fingered and have thus developed their decimal system in base 8. A certain street in nigiet contains 1000 buildings numbered 1 to 1000 in base 8. How many 3s are used in numbering these buildings?Express your answer in base 10
a) 54 b) 64 c) 265 d) 192

Asked In TCS Submited By-- suchandra raavi (2 year ago)

Puzzle Status: Solved     By »» Arghadeep Chaudhury

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Arghadeep Chaudhury  (2 year ago)

ans is d)192
bcz there are 300 no of 3s in between 1 and 1000 (eg. 20 3s between 1 and 99 and 20 3sben 199 and so on. bt there are 120 3s between 300 and 399. so total no of 3s are 20+20+120+20+20+20+20+20+20+20=300). 300=3*8^2=3*64=192 in base 10 system.

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animesh paul  (2 year ago)

in between 1-100 there are 20 no of 3's.
as 03,13,23,30,31,32,33(here we use 2 3),34,35,36,37,38,39,43,53,63,73,83,93

similarly there are 20 3s in between each hundred no.
and there are 100 3's from 300-399.. so total n of 3's are.. 100+200=300

to convert it into base 10 system we use the same method as we use to convert from binary to decimel.
thus (300)base8=3*8^2+0+0=192 :)

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subhojyoti bose  (2 year ago)

In 8 base there can't be numbers more than 7. So 3s will be like 3,13,23,30,31,32,33,34,35,36,37,43,53,63,73 so that's make 16.So upto 1000 there will be 8x16=128 3s and from 300 to 377 there is 64 3s (counting all 3s ). so that's make the answer 192.

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VYSAKH SURESH  (2 year ago)

There are 1000 in base 8 buildings ie,512 buildings accg to dec no system.Now the houses are no:red in base 8.ie,for 512 housesnumbering will be 1-7,10-17,17-27,.........,70-77(from 1-100)similarly name 512 houses using base 8 ie, from 100-177,200-277,..........700-777,1000.
total no of 3 from (0-77.....200-277) is 16,300-377 is 16+64,rest also is 16.
so total no of 3 will be 16*8+64=192.

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Mohammad Anwar  (9 Month ago)

Ans: 192

Number 3 in 1-100 comes 20 times.
101-200-> 20 times
201-300-> 20 times
301-400->120 times
401-500-> 20 times
501-600-> 20 times
601-700-> 20 times
701-800-> 20 times
801-900-> 20 times
901-1000-> 20 times
Total times: 300 times (any no. from 1 to 9 except 0 will occur 300 times)
8 based planet.. then
3*8^2 + 0*8^1 + 0*8^0 = 192

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tk  (2 year ago)

(512*3)/8=192

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animesh paul  (2 year ago)

in 1 to 1000 there are 300 3s used.. in between evry 100.. there are 20 no 3s,, counting 2 3s in 33.. so 20*10=200 and as is 300-399 there are 100 3's at d begining. so total 3=300.
in base 10.. no(3)=3*6^2+0+0=108 :)

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sourav baidya  (2 year ago)

192

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subhojyoti bose  (2 year ago)

192 is the answer.

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RAVI  (2 year ago)

192

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(#M40001614) TCS MATHS PUZZLE latest puzzle Keep an EYE

The citizens of planet nigiet are 8 fingered and have thus developed their decimal system in base 8. A certain street in nigiet contains 1000 (in base 8) buildings numbered 1 to 1000. How many 3s are used in numbering these buildings?
a) 54 b) 64 c) 265 d) 192

Asked In tcs Submited By-- Aditi Sharma (1 year ago)

Puzzle Status: Solved     By »» Aritra Das

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Aritra Das  (1 year ago)

(N-1)*(BASE)^(N-2)
N=NUMBER OF LAST DIGIT=4(100 contains 4 digits)
base=8
(4-1)*8^(4-2)
=192

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(#M40001714) TCS MATHS PUZZLE Keep an EYE

The citizens of planet nigiet are 8 fingered and have thus developed their decimal system in base 8. A certain street in nigiet contains 1000 (in base 8) buildings numbered 1 to 1000. How many 3s are used in numbering these buildings?
a) 54 b) 64 c) 265 d) 192

Asked In tcs Submited By-- UN (1 year ago)

Puzzle Status: Solved     By »» vamsi krishna reddy

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vamsi krishna reddy  (1 year ago)

in unit's place, the 3's are:
3, 13, 23 ... 73
103, 113, 123 ... 173
...
703, 713, .... 773
=> total number of 3's in unit's place is 8*8 = 64
in 10's place (that's 8 in decimal), the 3's are:
30, 31, 32... 37
130, 131, ... 137
...
730, 731, .... 737
=> total number of 3's in 10's place is 8*8 = 64
in 100's place (that's 64 in decimal), the 3's are:
300, 301, ... 307
310, 311 ... 317
...
370, 371 .... 377
=> total number of 3's in 100's place is 8*8 = 64
thus, the total number of 3's in total is 64 + 64 + 64 = 192

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Mohammad Anwar  (9 Month ago)

Ans: 192

Number 3 in 1-100 comes 20 times.
101-200-> 20 times
201-300-> 20 times
301-400->120 times
401-500-> 20 times
501-600-> 20 times
601-700-> 20 times
701-800-> 20 times
801-900-> 20 times
901-1000-> 20 times
Total times: 300 times (any no. from 1 to 9 except 0 will occur 300 times)
8 based planet.. then
3*8^2 + 0*8^1 + 0*8^0 = 192

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DIPIN  (1 year ago)

192

64+64+64 =192

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(#M40002438) TCS MATHS PUZZLE Keep an EYE

The citizens of planet nigiet are 6 fingered and have thus developed their decimal system in base 8. A certain street in nigiet contains 1000 (in base 8) buildings numbered 1 to 1000. How many 3s are used in numbering these buildings?
192
54
64
256

Asked In tcs Submited By-- satyajit kumar singh (1 year ago)

Puzzle Status: Solved     By »» Arpit Agarwal

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Arpit Agarwal  (1 year ago)

Consider 3 in one's place. The possible numbers are
3,13,23....,73
103,113...,173....
703,713...773
i.e 8*8=64 times.
Similarly consider 10's place
30,31,31...37
130,131,...137....
730,731...737
Again 8*8=64
Now to hundred's place
300,301...307....
370,371,...377
Again 8*8=64
so total 64+64+64=192

so a????

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divya  (1 year ago)

ans:192
3*8*8

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vivek chauhan  (8 Month ago)

for this type of questions answer will be always 3n^2
so 3(8)^2=192

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aashima  (1 year ago)

first calculate no.of 3 in 1 to 1000(on base 10)...
i.e. 300
now convert 300 into base 8... 8*2+0+0=64 answer

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prashant kumar  (1 year ago)

convert 1000 into base 8 form
i.e 512
No of 3s=512/8=64

Is this solution Helpfull? Yes | No (31)