Dhaka Secondary 2011/8

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Dhaka Secondary 2011/8

Unread post by BdMO » Fri Jan 28, 2011 10:08 pm

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?

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leonardo shawon
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Re: Dhaka Secondary 2011/8

Unread post by leonardo shawon » Sat Jan 29, 2011 1:22 pm

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 $N=5!*5!*5!*5!*5!*10$

Ibtehaz Shawon
BRAC University.

long way to go .....

MUjtahid Akon
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Re: Dhaka Secondary 2011/8

Unread post by MUjtahid Akon » Mon Jan 31, 2011 7:56 pm

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

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Re: Dhaka Secondary 2011/8

Unread post by Shifat » Sat Aug 27, 2011 12:11 am

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... :?

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