# Thread: Extremly hard question! Its a Challenge!

1. ## Extremly hard question! Its a Challenge!

aa

2. let the digits be represented by letters. Then the palindrome is
abcdefghhgfedcba

For multiplication this means $a^2 b^2 c^2 d^2 e^2 f^2 g^2 h^2 = (a b c d e f g h)^2 = 16$

then abcdefgh = 4

possible solutions:
one letter = 4, up to 7 of the others equal 1
two letters equal 2, up to 6 of the others equal 1

For addition this means $2a+2b+2c+2d+2e+2f+2g+2h = 2(a+b+c+d+e+f+g+h) = 16$

then a+b+c+d+e+f+g+h = 8

We know that they both equal 16 so possible solutions for addition from the set of possible solutions above (the intersection of their solution sets):
one letter equals 4, 4 letters equal 1, 3 letters equal zero
2 letters equal 2, 4 letters equal 1, 2 letters equal zero

Now we just need to find how many permutations there are in this set for the two solutions, then add them together. I'm not entirely sure how to do that.

My best attempt would look like this:
There are 8 possibilities for a 4, 7 for a 1, 6 for a 1, 5 for a 1, 4 for a 1, the rest are zero so 8*7*6*5*4 = 6720

There are 8 possibilities for a 2, 7 possibilities for a 2, 6 possibilities for a 1, 5 possibilities for a 1, 4 possibilities for a 1, 3 possibilities for a 1, the rest are zero so 8*7*6*5*4*3 = 20160

Their sum is 6720 + 20160 = 26880 different ways of making the palindrome.

I think it is correct up until the calculating of probability, but after that point I don't have much confidence.

3. Originally Posted by angel.white
We know that they both equal 16 so possible solutions for addition from the set of possible solutions above (the intersection of their solution sets)ne letter equals 4, 4 letters equal 1, 3 letters equal zero
2 letters equal 2, 4 letters equal 1, 2 letters equal zero
I think that the above answer is a gross over count. Although part of the analysis is right on.

Any 16-dight palindrome is completely determined by the first eight digits.
Because 16=(4)(4)=(8+8), it is clear that the sum must be 8 and the product of the non-zero digits must be 4.
Therefore as was found above, there are two possiblies: 41111000 or 22111100.
So the answer is the sum of the number of ways to rearrange each of those strings that do not begin with a zero.
Rearrangements of the string "41111000" can begin with 4 in $\frac{{7!}}{{\left( {4!} \right)\left( {3!} \right)}}$ ways and can begin with a 1 in $\frac{{7!}}{{\left( {3!} \right)^2 }}$.

Rearrangements of the string "22111100" can begin with a 2 in $\frac{{7!}}{{\left( {4!} \right)\left( {2!} \right)}}$ ways and can begin with a 1 in $\frac{{7!}}{{\left( {3!} \right)\left( {2!} \right)^2 }}$.

What is that sum?

4. Originally Posted by Plato
I think that the above answer is a gross over count. Although part of the analysis is right on.

Any 16-dight palindrome is completely determined by the first eight digits.
Because 16=(4)(4)=(8+8), it is clear that the sum must be 8 and the product of the non-zero digits must be 4.
Therefore as was found above, there are two possiblies: 41111000 or 22111100.
So the answer is the sum of the number of ways to rearrange each of those strings that do not begin with a zero.
Rearrangements of the string "41111000" can begin with 4 in $\frac{{7!}}{{\left( {4!} \right)\left( {3!} \right)}}$ ways and can begin with a 1 in $\frac{{7!}}{{\left( {3!} \right)^2 }}$.

Rearrangements of the string "22111100" can begin with a 2 in $\frac{{7!}}{{\left( {4!} \right)\left( {2!} \right)}}$ ways and can begin with a 1 in $\frac{{7!}}{{\left( {3!} \right)\left( {2!} \right)^2 }}$.

What is that sum?
35 + 140 + 105 + 210 = 490