Tangem Wallet Pack of 2 - Secure Crypto Wallet - Trusted Cold Storage for Bitcoin, Ethereum, NFT's &

Tangem Wallet Pack of 2 - Secure Crypto Wallet - Trusted Cold Storage for Bitcoin, Ethereum, NFT's &
Key item features Ultimate Security: Generates a private key that remains on the card, safeguarding crypto and NFTs from hackers with EAL6+ certification and audited firmware. Versatile Compatibility: Manages over 13,000 tokens across 70+ blockchains, supporting DeFi, NFTs, and DeEx without wires, Bluetooth, or USB. Effortless Operation: Utilizes NFC for secure transactions via a mobile device and the Tangem app, enabling buying and selling crypto with various payment methods. Smart Backup: Features a second Tangem Wallet as a backup, eliminating the need for paper, pictures, or seed phrases for recovery. Durable Design: Boasts IP68 protection against environmental conditions, ensuring longevity and robust physical security. Comprehensive Support: Compatible with Bitcoin, Ethereum, Solana, XRP, USDT, and over 6,000 cryptocurrencies, integrating with dApps and WalletConnect.

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Combinatorics of Bitcoin mining: *difficulty* vs number of trials of nonces and $32$ bit nonces

According to this question Mining Difficulty and Leading Zeros the Bitcoin mining problem is to find a string s = (hash of previous block + Merkle Tree Hash + nonce).

The current difficulty is roughly to find a small hash which has $19$ or $20$ leading zeros in Hexadecimal.

  1. $16^{20}=2^{80}$ which is a huge number and a random number hashed out has a probability $\leq\frac1{2^{80}}$ to have $80$ leading $0$ binary digits.

This number is much larger than trillions or quadrillions.

However the quoted difficulty of hashing to find leading $19$ or $20$ Hexadecimal digit $0$s is in order of $100$s of trillions.

a. Shouldn't the difficulty be roughly $2^{80}$ tries?

b. Why is the difficultynumber smaller?

c. How does the difficulty number related to number of random tries which is roughly $2^{80}$ in the case of $20$ leading zeros?

  1. How large in number of bits is the nonce? Isn't it $32$ bits? Is it bigger than $80$ bits? If not how can we try $2^{80}$ possible nonces given that hash of previous block + Merkle Tree Hash is a fixed value and only nonce change determines the changes in the computed hash?


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