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Difference between revisions of "2022 AMC 12B Problems/Problem 15"
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Next, we examine option B. We see that <math>2^{606}</math> has a units of digits of <math>4</math> (Taking the units digit of the first few powers of two gives a pattern of <math>2, 4, 8, 6, 2, 4, 8, 6, 2, 4, 8, 6,\cdots</math>). Adding <math>1</math> to <math>4</math>, we get <math>5</math>. Since <math>2^{606}+1</math> has a units digit of <math>5</math>, it is divisible by <math>5</math>.
 
Next, we examine option B. We see that <math>2^{606}</math> has a units of digits of <math>4</math> (Taking the units digit of the first few powers of two gives a pattern of <math>2, 4, 8, 6, 2, 4, 8, 6, 2, 4, 8, 6,\cdots</math>). Adding <math>1</math> to <math>4</math>, we get <math>5</math>. Since <math>2^{606}+1</math> has a units digit of <math>5</math>, it is divisible by <math>5</math>.
  
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Lastly, we examine <math>2^{606}+1</math>. Using the sum of cubes factorization <math>a^3+b^3=(a+b)(a^2-ab+b^2)</math>, we have <math>2^{606}+1^3=(2^{202}+1)(2^{404}+2^{202}+1)</math>. Since <math>2^{202}</math> ends with a <math>4</math>, and <math>4+1=5</math>, <math>2^{606}+1</math> is a multiple of <math>5</math>, which means it is divisible by <math>5</math>.
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Lastly, we examine option A. Using the difference of cubes factorization <math>a^3-b^3=(a-b)(a^2+ab+b^2)</math>, we have <math>2^{606}-1^3=(2^{202}-1)(2^{404}+2^{202}+1)</math>. Since <math>2^{404}+2^{202}+1\equiv0\text{mod}3</math> (Every term in the sequence is equivalent to <math>1\text{mod}3</math>), <math>2^{606-1}</math> is divisible by <math>3</math>.
  
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Since we have eliminated every option except one, <math>\boxed{\text{(C)}2^{607}-1}</math> is not divisible by any prime less than <math>10</math>.
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Since we have eliminated every option except C, <math>\boxed{\text{(C)}2^{607}-1}</math> is not divisible by any prime less than <math>10</math>.

Revision as of 01:16, 18 November 2022

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 $3$. (The reason why this pattern works: When you multiply $2 \equiv2\text{mod} 3$ by $2$, you obtain $4 \equiv1 \text{mod} 3$. Multiplying by $2$ again, we get $1\cdot2\equiv2 \text{mod} 3$. We see that in every cycle of two powers of $2$, it goes from $2 \text{mod}3$ to $1 \text{mod}3$ and back to $2 \text{mod}3$.)

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 option A. Using the difference 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 $2^{404}+2^{202}+1\equiv0\text{mod}3$ (Every term in the sequence is equivalent to $1\text{mod}3$), $2^{606-1}$ is divisible by $3$.

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

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