− | 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 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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