− | 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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− | 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 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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