Palindromes Formed by Multiplying by 55
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Theorem
$55$ multiplied by any of the odd integers between $91$ and $109$ inclusive produces a palindromic number.
Proof
We have that:
- $55 \times 91 = 5005$
Then we have that:
- $55 \times 2 = 110$
Thus:
- $\forall k: 1 \le k \le 9: 55 \times 2 k = 110k = 110, 220, 330, \ldots, 990$
Thus for $1 \le k \le 9$:
- $55 \times \paren {91 + 2 k} = \sqbrk {5kk5}_{10}$
That is:
\(\ds 55 \times 91\) | \(=\) | \(\ds 5005\) | ||||||||||||
\(\ds 55 \times 93\) | \(=\) | \(\ds 5115\) | ||||||||||||
\(\ds 55 \times 95\) | \(=\) | \(\ds 5225\) | ||||||||||||
\(\ds 55 \times 97\) | \(=\) | \(\ds 5335\) | ||||||||||||
\(\ds 55 \times 99\) | \(=\) | \(\ds 5445\) | ||||||||||||
\(\ds 55 \times 101\) | \(=\) | \(\ds 5555\) | ||||||||||||
\(\ds 55 \times 103\) | \(=\) | \(\ds 5665\) | ||||||||||||
\(\ds 55 \times 105\) | \(=\) | \(\ds 5775\) | ||||||||||||
\(\ds 55 \times 107\) | \(=\) | \(\ds 5885\) | ||||||||||||
\(\ds 55 \times 109\) | \(=\) | \(\ds 5995\) |
$\blacksquare$
Sources
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $55$