Elitmus
Exam
Numerical Ability
Number System
A number 10,10,101 when multiplied with N gives product P, such that N is any no with only 1 as its digit (eg. N = 11,111,1111,.......) .How many different digit can be possible in the product P.
(a) 2
(b) 3
(c) 4
(D) not remembered
Read Solution (Total 20)
-
- N*10*10*101=P
P=N*100*(100+1)
P=10000*N+100*N
suppose if N=11 then P=111100
if N=111 then P=1121100
if N=1111 then P=11221100
and N=11111 then P=112221100
an so P only contains digits such as 0,1,2 with repeatations
so answer is 3 (option b) - 10 years agoHelpfull: Yes(72) No(10)
- ans is 4...coz after multiplying it many times there will come an ans having 4 different digits(1,2,3,4) and when u multiply it again with N having one more 1,then there will be repetation of the digits.
- 10 years agoHelpfull: Yes(18) No(12)
- N*10*10*101=P
P=N*100*(100+1)
P=10000*N+100*N
suppose if N=11 then P=111100
if N=111 then P=1121100
if N=1111 then P=11221100
an so P only contains digits such as 0,1,2 with repeatations
so answer is 3 - 10 years agoHelpfull: Yes(10) No(3)
- answer will be 9
- 10 years agoHelpfull: Yes(9) No(7)
- if number is 1010101 then ans is 4(1,2,3,4)
if number is 10,10,101 then ans is 3(0,1,2) - 10 years agoHelpfull: Yes(6) No(0)
- answer will be 4... 0,1,2,3...if u are in doubt...just multiply it with 11111 and see for yourself if u get 3 or not
- 10 years agoHelpfull: Yes(2) No(1)
- answer will be 3 and digits are 1,2,3.
- 10 years agoHelpfull: Yes(1) No(3)
- vijay could you please explain how it would be 9?
- 10 years agoHelpfull: Yes(1) No(1)
- take N= 11111111111
- 10 years agoHelpfull: Yes(1) No(2)
- @Ram gud one bro .. 0,1,2 is correct
- 10 years agoHelpfull: Yes(1) No(1)
- ans is 4 bcoz o multiplying with largest possible 1111111 we got com of 4 digits
- 10 years agoHelpfull: Yes(1) No(0)
- ans will be 2 because u can multipli with N to any no among the three u will always get either 1 or 2 differnt digit
ex:- 111111111*101=22222222211 - 10 years agoHelpfull: Yes(0) No(2)
- 2 diff no....i.e 0,1
- 10 years agoHelpfull: Yes(0) No(2)
- hh explain
- 10 years agoHelpfull: Yes(0) No(0)
- @rajat only 0,1,2 are repeated so how the answer is 4 instead of 3 pls explain.
- 10 years agoHelpfull: Yes(0) No(1)
- at most 3...start multiplying 1010101 with 1 as unit digit, then 11, 111 you will see the pattern that atmost 3 different digits are possible in P
- 10 years agoHelpfull: Yes(0) No(0)
- N*10*10*101=P
P=N*100*(100+1)
P=10000*N+100*N
suppose if N=11 then P=111100
if N=111 then P=1121100
if N=1111 then P=11221100
an so P only contains digits such as 0,1,2 with repeatations
so answer is 3
Read more at http://www.m4maths.com/placement-puzzles.php?ISSOLVED=&page=2&LPP=10&SOURCE=Elitmus&MYPUZZLE=&UID=&TOPIC=Numerical%20Ability&SUB_TOPIC=Number%20System#hFd5u7T02MPjt111.99 - 10 years agoHelpfull: Yes(0) No(1)
- 10 digits are possible 0 to 9.
Furthermore, the multiplication gives a pallindrome. - 10 years agoHelpfull: Yes(0) No(1)
- 10 digits are possible 0 to 9.
- 10 years agoHelpfull: Yes(0) No(1)
- 3 becoz digits are (0,1,2)
- 8 years agoHelpfull: Yes(0) No(0)
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