Emordnilap palindromE 2

there are 77, 717,7117 and 71117 so thats 4/5 = 80%

The 4 digit palindromes must have 1st and last digit same $=> \text{ 9 ways possible in unit's place and thousand's place excluding 0 }$

And 2nd and 3rd digit must be same $\text{=> 10 ways possible including 0}$ $$ $\text{Total palindromes possible = 90 ways }$

All palindromes are divisible by 11.The least palindrome is 1001 which is divisible by 7.Addition of 770 and 1001 givies palindromes divisible by both 7 and 11.The palindromes divisible by both 7 and 11 are $1001,1771,2002,2772,3003,3773,4004,4774,5005,$ $5775,6006,6776,7007,7777,8008,8888,9009,9999$

$=> (90-18)/90 × 100=80\text{\% palindromes not divisible by 77 }$

We know that all 4-digit palindromes are divisible by 11. So we just need to count how many palindromes aren't divisible by 7. We'll use complementary counting. Let's count how many 4-digit palindromes ARE divisible by 7. We start off with 1001 and check to see if it is divisible by 7. The prime factorization of $1001$ is $7\times11\times13$ . So it is divisible by 7. Now, all multiples of 1001 are divisible by 7, so $1001,2002,3003,...,9009$ are divisible by 7. Also, when we add a multiple of 7 to another multiple of 7, the result is also a multiple of 7. So we add $770$ to $1001$ to get $1771$ another palindrome divisible by 7. We add $770$ to all the other palindromes we came up with, so we have altogether 18 palindromes that are divisible by 77. There are no more ways to come up with more palindromes divisible by 7. Now we count how many palindromes there are. We use constructive counting. We have 9 choices for the thousands digit (because of 0) and 10 choices for the hundreds. The tens digit has to be equal to the hundreds digit and the ones needs to be equal to the thousands so we are only left with 1 option for the tens and ones digit. This gives us $9\times10=90$ 4-digit palindromes. Of these palindromes, $18$ are multiples of $77$ . So we have $90-18=72$ that are not multiples of $77$ Now we are asked for the percent of the 4-digit palindromes that aren't divisible by $77$ Our answer is $\frac{72}{90}=\frac{\boxed{80}}{100}$

EDIT: oops sniped.