2022 AMC 12B Problems/Problem 15

Problem: One of the following numbers is not divisible by any prime number less than 10. Which is it? $\text{(A)} 2^{606}-1 \text{(B)} 2^{606}+1 \text{(C)} 2^{607}-1 \text{(D)} 2^{607}+1 \text{(E)} 2^{607}+3^{607}$

Solution 1 (Process of Elimination)

We examine option E first. $2^{607}$ has a units digit of $8$ (Taking the units digit of the first few powers of two gives a pattern of $2, 4, 8, 6, 2, 4, 8, 6, 2, 4, 8, 6,\cdots$ ) and $3^{607}$ has a units digit of $7$ (Taking the units digit of the first few powers of three gives a pattern of $3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1,\cdots$ ). Adding $7$ and $8$ together, we get $15$ , which is a multiple of $5$ , meaning that $2^{607}+3^{607}$ is divisible by 5.

Next, we examine option D. We take the first few powers of $2$ added with $1$ : $2^1+1=3$ $2^2+1=5$ $2^3+1=9$ $2^4+1=17$ $2^5+1=33$ $2^6+1=65$ $2^7+1=129$

We see that the odd powers of $2$ added with 1 are multiples of three. If we continue this pattern, $2^{607}+1$ will be divisible by $3$ .

Next, we examine option B. We see that $2^{606}$ has a units of digits of $4$ (Taking the units digit of the first few powers of two gives a pattern of $2, 4, 8, 6, 2, 4, 8, 6, 2, 4, 8, 6,\cdots$ ). Adding $1$ to $4$ , we get $5$ . Since $2^{606}+1$ has a units digit of $5$ , it is divisible by $5$ .

Lastly, we examine $2^{606}+1$ . Using the sum of cubes factorization $a^3+b^3=(a+b)(a^2-ab+b^2)$ , we have $2^{606}+1^3=(2^{202}+1)(2^{404}+2^{202}+1)$ . Since $2^{202}$ ends with a $4$ , and $4+1=5$ , $2^{606}+1$ is a multiple of $5$ , which means it is divisible by $5$ .

Since we have eliminated every option except one, $\boxed{\text{(C)}2^{607}-1}$ is not divisible by any prime less than $10$ .

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2022 AMC 12B Problems/Problem 15