Does there exist a natural number that can be represented as the product of two palindromes in more than

100 ways? (A palindrome is a natural number that reads the same from left to right as from right to left.)

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Consider a palindrome consisting of \$n\$ ones \$1_n = 1...1\$.

Method 1. If \$n\$ is divisible by \$k\$, then \$1_n\$ is divisible by the palindrome \$1_k\$, and the quotient is also a palindrome, consisting of ones separated by groups of \$k-1\$ zeros. It remains to choose a number \$n\$ that has more than 100 proper divisors. For example, \$2^{101}\$.

Remark. The number \$6^n\$ is divisible not only by \$1_k\$ but also by \$2_k\$, \$3_k\$, and \$6_k\$. This allows reducing \$n\$ to a number with more than 25 proper divisors, for example, \$n=720=2^4 \cdot 3^2 \cdot 5\$.

Idea of Method 2. Notice that if no carries occur between digits when multiplying palindromes, the product is also a palindrome. Consider the product \$\$11 \cdot 101 \cdot

1 \underbrace{0 \ldots 0}_{3} 1 \cdot 1 \underbrace{0 \ldots 0}_{7} 1 \cdot 1 \underbrace{0 \ldots 0}_{15} 1 \cdot 1 \underbrace{0 \ldots 0}_{31} 1 \cdot 1 \underbrace{0 \ldots 0}_{63} 1 \cdot 1 \underbrace{0 \ldots 0}_{127} 1 = 1_{256}.\$\$ It can be proven that no carries occur when multiplying any number of these factors, and the result is a palindrome consisting of zeros and ones. Since 8 factors can be divided into two groups in \$2^7=128\$ ways, 128 different (since the factors are pairwise coprime) representations of the number \$1_{256}\$ as a product of two palindromes can be obtained.

Remark 2. The considerations from the remark to Method 1 show that the number \$6_{64}\$ is suitable. It can be shown that even the number \$11 \cdot 101 \cdot 1001 \cdot 1 \underbrace{0 \ldots 0}_{3} 1 \cdot 1 \underbrace{0 \ldots 0}_{4} 1 \cdot 1 \underbrace{0 \ldots 0}_{5} 1 \cdot 1 \underbrace{0 \ldots 0}_{6} 1 \cdot 1 \underbrace{0 \ldots 0}_{7} 1\$ fits.