Bakayev E.V.

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 backward as forward.)
Exists. Consider a palindrome consisting of \$n \$ ones \$ 1 \_n=1 \ldots 1 \$.

Method 1. If \$n\$ is divisible by \$k\$, then \$1_k\$ divides \$1_n\$. For example, \$111111\$ is divisible by \$111\$ and \$11\$. It remains to choose a number \$n\$ that has more than 100 proper divisors. For example, \$ 2 \wedge{101} \$.

Remark. The number \$6 \$ divides not only \$1_k\$, but also \$2_k\$, \$3_k\$, and \$ n \$. To reduce the number of divisors, we can choose a number \$n\$ that has more than 25 proper divisors, for example, \$ n=720=2 \wedge 4 \mid c d o t 3 \wedge 2 \cdot 5\$.

Idea of Method 2. Notice that if there are no carries 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 when multiplying any number of these factors, no carries occur, and the result is a palindrome consisting of zeros and ones. Since 8 factors can be divided into two groups in \$ 2 \wedge 7 = 128 \$ ways, we can obtain 128 different (since the factors are pairwise coprime) representations of the number \$1_{256}\$ as a product of two palindromes.

Remark 2. The considerations from the remark to Method 1 show that the number \$6_{64}\$ is suitable. It can be shown,

![](https://cdn.mathpix.com/cropped/2024_05_06_5dbee4d95f83bceb3891g-46.jpg?height=55&width=1973&top_left_y=2105&top_left_x=14)
|cdot 1 1 \underbrace{0 \ldots 0}_{5} 1 \cdot 1 \underbrace{0 \ldots 0}_{6} 1 \cdot 1 1 \underbrace{0 \ldots 0}_{7} 1 . \$

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