Elitmus
Exam
Numerical Ability
Permutation and Combination
Q. A number 'Z' contains all the digits from 1 to 9 exactly once. Z is divisible by 99. what will b the number on its hundredth place.
Option
a) 1
b) 2
c) 3
d) 4
Read Solution (Total 13)
-
- Let d digits be abcdefghi
if it is divisible by 99 , it shud be divisible by both '9' nd '11'
divisible by 11 means ----> (a+c+e+g+i)-(b+d+f+h) / 11
--------------------------> (7+5+6+2+8)-(1+3+4+9) = 28-17 which is divisible by 11
so the number is 715364298
nd the 100th digit is "2" - 10 years agoHelpfull: Yes(16) No(22)
- more than one answers are possible like
142738596
542738196
541738296
541728396
- 10 years agoHelpfull: Yes(7) No(1)
- can i know the right answer if its wrong...
- 10 years agoHelpfull: Yes(5) No(3)
- with all due respect wt abt 917365482 i got this ans
- 10 years agoHelpfull: Yes(2) No(4)
- suppose digits be abcdefghi
if it is divisible by 99 ,
it shud be divisible by both '9' (7+5+6+2+8)+(1+3+4+9) = 45 which is divisible by 9
and
divisible by 11 means is (a+c+e+g+i)-(b+d+f+h)/11
(7+5+6+2+8)-(1+3+4+9) = 28-17 which is divisible by 11
so the number is 715364298
so the 100th digit is 2 - 10 years agoHelpfull: Yes(2) No(4)
- approach is right but in the question not mentioned that number is greatest or smallest .so question is incomplete because if u find any no which divisible by 99 if u exchange the no from odd place sum of the digits of odds place always be 28 and the even place 17 see ur self in above examples
like 7+5+6+2+8 or if exchang the no 5+7+6+2+8 or 2+8+7+5+6 and similar for even place. - 7 years agoHelpfull: Yes(2) No(0)
- this can be answer 214365789 then 100 digit is 7
- 10 years agoHelpfull: Yes(1) No(2)
- but there is no such option...
- 10 years agoHelpfull: Yes(1) No(1)
- At 100th place we can put 9,8,6,4,1 .Among the options 4 can be the answer.
- 10 years agoHelpfull: Yes(1) No(0)
- 99 = 9×11 where 9 and 11 are coprime. Therefore, if a number is divisible by 9 and 11, it is divisible by 99 also
Here all the digits from 1 to 9 are used exactly once. 1+2+...+9 = 45. 45 is divisible by 9.
=> any number formed by using the digits 1 to 9 exactly one is divisive by 9
Hence, any number formed by using 1 to 9 exactly once is divisible by 99 if the same is divisible by 11
For number to be divisible by 11, [Sum of odd numbered digits] - [sum of even numbered digits] = 0 or divisible by 11
for example, (5+6+7+2+8) - (1+3+4+9) = 11
Hence, 516374298 is a number which is divisible by 99
In this case, the digit at hundredth place is 2
But this is not the only answer. We can get other numbers also in a similar way such as 245837196 and accordingly different answers for the hundredth place digit(like 1 in this case)
- 8 years agoHelpfull: Yes(1) No(1)
- 99 = 9×11 where 9 and 11 are coprime. Therefore, if a number is divisible by 9 and 11, it is divisible by 99 also
Here all the digits from 1 to 9 are used exactly once. 1+2+...+9 = 45. 45 is divisible by 9.
=> any number formed by using the digits 1 to 9 exactly one is divisive by 9
Hence, any number formed by using 1 to 9 exactly once is divisible by 99 if the same is divisible by 11
For number to be divisible by 11, [Sum of odd numbered digits] - [sum of even numbered digits] = 0 or divisible by 11
for example, (5+6+7+2+8) - (1+3+4+9) = 11
Hence, 516374298 is a number which is divisible by 99
In this case, the digit at hundredth place is 2
AND
But this is not the only answer. We can get other numbers also in a similar way such as 245837196 and accordingly different answers for the hundredth place digit like 1 - 7 years agoHelpfull: Yes(1) No(0)
- number is divisible by 99 means divisible by 9 and 11 both
number formed by digit 1 to 9 i.e (1+2+3+4+5+6+7+8+9)=45 is always divisible by 9
but not divisible by 11
condition for divisibility by 11 is-(sum of odd places number-sum of even places number)
so here, hit and trial method
digit taken from 1 to 9
for e.g (5+6+7+8+2)-(3+4+1+9)=11
is divisible by 11
so number is 536471298
hundredth place digit is 2 - 7 years agoHelpfull: Yes(1) No(0)
- wt abt 928675341
- 10 years agoHelpfull: Yes(0) No(1)
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