5	a	b	b	a	5

For divisible by 55 it shall be divisible by 11 and 5 both, for divisibility by 5 unit digit shall be 0 or 5 but as the number is six digit palindrome unit digit is 5.

A number is divisible by 11 if the difference between sum of the digits in the odd places and the sum of the digits in the even places is a multiple of 11 or zero.

Sum of the digits in the even place = a + b + 5

Sum of the digits in the odd places = a + b + 5

Difference between the two sums = (a + b + 5 ) - (a + b + 5) = 0

0 is divisible by 11.

Hence, 5abba5 is divisible by 11.

So, required number = 10 $$\times$$ 10 = 100

(i) When '0' is at unit place



Number of numbers = 20

(ii) When 4 or 6 are at unit place



Number of numbers = 32

Total three digit even number = 20 + 32 = 52

$${}^1{P_1} + 2.{}^2{P_2} + 3.{}^3{P_3} + .... + 15.{}^{15}{P_{15}}$$

= 1! + 2 . 2! + 3 . 3! + ..... 15 $$\times$$ 15!

$$ = \sum\limits_{r = 1}^{15} {(r + 1)! - (r)!} $$

= 16! $$-$$ 1

= $${}^{16}{P_{16}}$$ $$-$$ 1

$$\Rightarrow$$ q = r = 16, s = 1

$${}^{q + s}{C_{r - s}} = {}^{17}{C_{15}}$$ = 136

N = 2¹⁰ $$\times$$ 5¹⁰ $$\times$$ 11¹¹ $$\times$$ 13¹³

Now, power of 2 must be zero,

power of 5 can be anything,

power of 13 can be anything

But, power of 11 should be even.

So, required number of divisors is

1 $$\times$$ 11 $$\times$$ 14 $$\times$$ 6 = 924