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| − | Problem: One of the following numbers is not divisible by any prime number less than 10. Which is it?
| + | #REDIRECT [[2022_AMC_10B_Problems/Problem_17]] |
| − | <math>\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}</math>
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| − | Solution 1 (Process of Elimination)
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| − | We examine option E first. <math>2^{607}</math> has a units digit of <math>8</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>) and <math>3^{607}</math> has a units digit of <math>7</math> (Taking the units digit of the first few powers of three gives a pattern of <math>3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1,\cdots</math>). Adding <math>7</math> and <math>8</math> together, we get <math>15</math>, which is a multiple of <math>5</math>, meaning that <math>2^{607}+3^{607}</math> is divisible by 5.
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| − | Next, we examine option D. We take the first few powers of <math>2</math> added with <math>1</math>:
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| − | <cmath>2^1+1=3</cmath>
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| − | <cmath>2^2+1=5</cmath>
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| − | <cmath>2^3+1=9</cmath>
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| − | <cmath>2^4+1=17</cmath>
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| − | <cmath>2^5+1=33</cmath>
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| − | <cmath>2^6+1=65</cmath>
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| − | <cmath>2^7+1=129</cmath>
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| − | We see that the odd powers of <math>2</math> added with 1 are multiples of three. If we continue this pattern, <math>2^{607}+1</math> will be divisible by <math>3</math>. (The reason why this pattern works: When you multiply <math>2 \equiv2\text{mod} 3</math> by <math>2</math>, you obtain <math>4 \equiv1 \text{mod} 3</math>. Multiplying by <math>2</math> again, we get <math>1\cdot2\equiv2 \text{mod} 3</math>. We see that in every cycle of two powers of <math>2</math>, it goes from <math>2 \text{mod}3</math> to <math>1 \text{mod}3</math> and back to <math>2 \text{mod}3</math>.)
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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>.
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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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| − | 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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