Given that n is a six digit palindrome, what is the probability that n/11 is also a palindrome?
Answers
Step-by-step explanation:
A Six digit palindrome is of the form
abccba.
where a is a digit form 1 or 9, while b and c
are both from 0 to 9. hence there are
9×10×10=900 possible to six- digit palindrome.
such a palindrome can be written as
100001a+10010b+1100c
which when divided by 11 , is equal to
9091a+910b+100c.
if a five digit number is a palindrome , then it must be of the form.
xyzyz.
or 10001x+1010y+100z, where x is a digit from 1 or 9 and y and z are digit form 0 to 9.
hence, when 9091a+910b+100c represents a palindrome, we must have
9091a+910b+100c=10001x+1010y++100z
but,
9091a+910b+100c=(10001-910)a +91+100c
=1001a+910(b-a) +100c
=10001a+(1010-100)(b-a) +100c
=1001a+1010(b-a) +100(c (b-a)).
hence, if 9091a+910b+100c is a palindrome,
then b must be chosen such that (b-a) is a digit from 0 to 9. and c must be chosen such that
c-(b-a) is a digit from 0 to 9.
let's consider these case -by-case. if a=1,then 0<b-a <9 means that 1<b<9.when a=2,0<b-a<9.
mean that 2<b<9.
continuing in this fraction, we obtain that
a<b<9.
simillary, since 0<c-(b-a) <9.we include that
(b-a) <c<9.
hence, if a=1, then there are
10+9+8+7+6+5+4-3+2 possible palindrome that result another palindrome when divided by 11.(10 is the case b=1, 9 Is a case b=2, and so on,)
simillary when a=2 there are 10+9+8+7+6+5+4-3 possible such palindrome and so on until a=9.
where there are just 10 possible such palindrome.
thus the number of six -digit palindrome that result in palindrome when divided by 11 is equal to
(10+9+8+7+6+5+4+3+2)
+(10+9+8+7+6+5+4-3)
+(10+9+8+7+6+5+4)
+...
+(10)
or
(9)(10)+(8)(9)+(7)(8)+...+(1)(2)
or 330.
hence, the required probability is
Answer:
Total favourable cases = 11*5!*4!
Heres the method:
As per the divisibility rule of 11 difference of sum of alternate digits should be divisible by 11.
let the digits be a,b,c... i
Now let the digits at odd places = a,b,c,d,e and the rest are at even places
now (a+b+c+d+e)-(f+g+h+i) has to be div by 11
max value that sum of 5 digits can take is 35 ( 5+6+7+8+9) and hence the sum of rest 4 digits would be 10 (45-35) , min value that (a+b+c+d+e) can take is 15, and hence sum of rest 4 digits wud be 30....
hence difference shall be like this 35-10 =25 ... 34-11 = 23 ... 33-12 = 21 ..... ... 15-30 = -15...
hence there can be only 2 cases when the difference is divisible by 11 ie 11 or -11
when difference = 11 ... a+b+c+d+e = 28 and f+g+h+i = 17 ....
theres shall be following cases for f+g+h+i = 17 --- > (9 1 2 5) , (9 1 3 4) , (8 1 2 6) , (8 1 3 5 ), (8 2 3 4) (7 1 3 6 ) ( 7 1 4 5 ) (7 2 3 5) ( 6 2 4 5) = 9 cases ... these can be arranged in 4!*5!ways.
Also when sum of a+b+c+d+e = 17 , then also no would be div by 11 .
(7 1 2 3 4 ) , (6 1 2 3 5 ) = 2 cases only.... hence total favourable cases = 11 * 4! * 5!
SO required probability =11×4!×5!9!