Dhaka Secondary 2011/8
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Please don't post problems (by starting a topic) in the "Secondary: Solved" forum. This forum is only for showcasing the problems for the convenience of the users. You can post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
A palindrome, such as $83438$, is a number that remains the same when its digits are reversed. $N$ is a six digit palindrome which is divisible by $6$. The number obtained by eliminating its leftmost and rightmost digits is divisible by $4$. How many possible values of $N$ are possible?
- leonardo shawon
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Re: Dhaka Secondary 2011/8
suppose ABCDEF is the six digit Polindrom. So A=F and B=F. C and D must be the same number.
ABCDEF=N is divisible by 6. So the last digit should be 2,4,8,6,0. And BCDF is divisible by 4. So the last digit is 2,4,8,6,0..
And for CD we have 10 numbers. 0-9.
so We have 5 digits for F and 5 digits for E.
Possible values of $N=5!*5!*5!*5!*5!*10$
?????!!!
ABCDEF=N is divisible by 6. So the last digit should be 2,4,8,6,0. And BCDF is divisible by 4. So the last digit is 2,4,8,6,0..
And for CD we have 10 numbers. 0-9.
so We have 5 digits for F and 5 digits for E.
Possible values of $N=5!*5!*5!*5!*5!*10$
?????!!!
Ibtehaz Shawon
BRAC University.
long way to go .....
BRAC University.
long way to go .....
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Re: Dhaka Secondary 2011/8
Let the number be ABCCBA
As BCCB=0 (mod 4)
therefore, CB=0 (mod 4)
Again, ABCCBA=0 (mod 6)
so A+B+C+C+B+A=0 (mod 3) and A is even
Or, C+B+A=0 (mod 3)
we get, CBA=0 (mod 3)
As A is even, it may take values as 2,4,6,8.
(we can't consider 0 here.if it does,the numbers of digit will become 4)
Now we'll consider 3 cases:
(1) CB=0 (mod 4) and CB=0 (mod 3)
(2) CB=0 (mod 4) and CB=1 (mod 3)
(3) CB=0 (mod 4) and CB=2 (mod 3)
from case (1) we obtain,CB=0(mod 12)
it is possible in 8 ways.this time A can only be 6 (not 0).
so in case (1) the possibility is 8.1=8
accordingly, in case (2),
CB=4 (mod 12)
it is also possible in 8 ways (4,16,28,....,88)
this time A takes 2 and 8.
so in case (2) the possibility is 8.2=16
in case (3),we get CB=8 (mod 12)
it is possible in 8 ways too.(8,20,32,....,92)
A will be only 4
so in case (3) the possibility is 8.1=8
notice, we avoid the value 600006.
so the total probability will be 8+16+8+1=33
As BCCB=0 (mod 4)
therefore, CB=0 (mod 4)
Again, ABCCBA=0 (mod 6)
so A+B+C+C+B+A=0 (mod 3) and A is even
Or, C+B+A=0 (mod 3)
we get, CBA=0 (mod 3)
As A is even, it may take values as 2,4,6,8.
(we can't consider 0 here.if it does,the numbers of digit will become 4)
Now we'll consider 3 cases:
(1) CB=0 (mod 4) and CB=0 (mod 3)
(2) CB=0 (mod 4) and CB=1 (mod 3)
(3) CB=0 (mod 4) and CB=2 (mod 3)
from case (1) we obtain,CB=0(mod 12)
it is possible in 8 ways.this time A can only be 6 (not 0).
so in case (1) the possibility is 8.1=8
accordingly, in case (2),
CB=4 (mod 12)
it is also possible in 8 ways (4,16,28,....,88)
this time A takes 2 and 8.
so in case (2) the possibility is 8.2=16
in case (3),we get CB=8 (mod 12)
it is possible in 8 ways too.(8,20,32,....,92)
A will be only 4
so in case (3) the possibility is 8.1=8
notice, we avoid the value 600006.
so the total probability will be 8+16+8+1=33
Re: Dhaka Secondary 2011/8
The answer is 33, to my reckon, here is how I got it:
Let the Number is $ABCCBA$
$BCCB$ is dividable by 4
which means the probable value of BC is 04 to 96= 24 ways
and the value of A is 2,4,6,8
among the 24 values of BC, 24/3= 8 values are dividable by 6(also 00 is dividable by 6, but will count it next line)
So if the value of A is 6 then the numbers can be written in 9 ways(600006 counting as well)
now if the value of A= 2, then
A+A= 4(mod 6), which means BCCB is not dividable by 6,
amongst the other 16 values of BC, we can now write 8 of them now to make sure that 2*(A+B+C) is dividable by 6.
For A=4, or 2A= 8\equiv 2 (mod6) the other 8 numbers become applicable
For A=8 or 2A= 16\equiv 4(mod 6) the values used for 2 becomes applicable,
so the total ways of mentioning N
is 9+8+8+8=33
(wrote in short form, if necessary, try writing down all 4 dividend values from 04 to 96 )
I am not sure though, i still think I missed something here...
Let the Number is $ABCCBA$
$BCCB$ is dividable by 4
which means the probable value of BC is 04 to 96= 24 ways
and the value of A is 2,4,6,8
among the 24 values of BC, 24/3= 8 values are dividable by 6(also 00 is dividable by 6, but will count it next line)
So if the value of A is 6 then the numbers can be written in 9 ways(600006 counting as well)
now if the value of A= 2, then
A+A= 4(mod 6), which means BCCB is not dividable by 6,
amongst the other 16 values of BC, we can now write 8 of them now to make sure that 2*(A+B+C) is dividable by 6.
For A=4, or 2A= 8\equiv 2 (mod6) the other 8 numbers become applicable
For A=8 or 2A= 16\equiv 4(mod 6) the values used for 2 becomes applicable,
so the total ways of mentioning N
is 9+8+8+8=33
(wrote in short form, if necessary, try writing down all 4 dividend values from 04 to 96 )
I am not sure though, i still think I missed something here...