*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?

## Dhaka Secondary 2011/8

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- leonardo shawon
**Posts:**169**Joined:**Sat Jan 01, 2011 4:59 pm**Location:**Dhaka

### 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.

BRAC University.

*long way to go .....*-
**Posts:**2**Joined:**Mon Jan 31, 2011 3:09 pm**Location:**Dhaka

### 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...