GRE Math: Units & Remainders (Level 3)

Solution (D): 1. For \(17^{17}\): Cycle for 7 is {7, 9, 3, 1}. \(17 \div 4 = 4\) remainder 1. This is the 1st step: 7. 2. For \(23^{23}\): Cycle for 3 is {3, 9, 7, 1}. \(23 \div 4 = 5\) remainder 3. This is the 3rd step: 7. 3. Product of units digits: \(7 \times 7 = 49\). The final units digit is 9.

Solution (A): We must find a pattern for remainders of powers of 3 when divided by 7. \(3^1 = 3 \equiv 3 \pmod{7}\) \(3^2 = 9 \equiv 2 \pmod{7}\) \(3^3 = 27 \equiv 6 \pmod{7}\) \(3^4 = 81 \equiv 4 \pmod{7}\) \(3^5 = 243 \equiv 5 \pmod{7}\) \(3^6 = 729 \equiv 1 \pmod{7}\) The cycle of remainders has a length of 6: {3, 2, 6, 4, 5, 1}. We divide the exponent 101 by 6: \(101 \div 6 = 16\) with a remainder of 5. The 5th step in the cycle corresponds to a remainder of 5.

Solution (C): We can factor out the smaller term, \(4^{10n-1}\). \(4^{10n-1} \times (4^1 – 1) = 4^{10n-1} \times 3\). The cycle for 4 is {4, 6}. The exponent \(10n-1\) is always odd (e.g., if n=2, 19). An odd exponent for 4 always results in a units digit of 4. So, we have \((…4) \times 3 = …12\). The final units digit is 2. *Correction:* Let me re-check. \(10n-1\) is odd. \(4^{\text{odd}} \to 4\). \(4 \times 3 = 12 \to 2\). The answer should be B.

Solution (B): Alternatively, find the units digit of each part. 1. \(4^{10n}\): The exponent \(10n\) is always even. For 4, an even exponent gives a units digit of 6. 2. \(4^{10n-1}\): The exponent \(10n-1\) is always odd. For 4, an odd exponent gives a units digit of 4. 3. Subtract: \((…6) – (…4) = …2\). The final units digit is 2.

Solution (B): First, evaluate the exponent: \(2^4 = 16\). The problem is to find the units digit of \(3^{16}\). The cycle for 3 is {3, 9, 7, 1}. We need the remainder of the exponent 16 when divided by 4. \(16 \div 4 = 4\) with a remainder of 0. A remainder of 0 corresponds to the 4th step in the cycle, which is 1. *Correction:* \(3^{16} \to 1\). The answer should be A.

Solution (A): First, evaluate the exponent: \(2^4 = 16\). The problem is to find the units digit of \(3^{16}\). The cycle for 3 is {3, 9, 7, 1}. We need the remainder of the exponent 16 when divided by 4. \(16 \div 4 = 4\) with a remainder of 0. A remainder of 0 corresponds to the 4th step in the cycle, which is 1.

Solution (D): We find the remainder of each term when divided by 13. \(14 \equiv 1 \pmod{13}\) \(15 \equiv 2 \pmod{13}\) \(16 \equiv 3 \pmod{13}\) \(17 \equiv 4 \pmod{13}\) The remainder of the product is the product of the remainders: \(1 \times 2 \times 3 \times 4 = 24\). Now, find the remainder of 24 when divided by 13. \(24 \div 13 = 1\) with a remainder of 11. *Correction:* \(24 – 13 = 11\). The answer should be 11. None of the options is 11. Let me fix the options.

Solution (E): We find the remainder of each term when divided by 13. \(14 \equiv 1 \pmod{13}\) \(15 \equiv 2 \pmod{13}\) \(16 \equiv 3 \pmod{13}\) \(17 \equiv 4 \pmod{13}\) The remainder of the product is the product of the remainders: \(1 \times 2 \times 3 \times 4 = 24\). Now, find the remainder of 24 when divided by 13. \(24 \div 13 = 1\) with a remainder of 11.

Solution (A): 1. For \(2^{100}\): Cycle {2, 4, 8, 6}. Exponent 100 is divisible by 4 (rem 0). 4th step: 6. 2. For \(7^{100}\): Cycle {7, 9, 3, 1}. Exponent 100 is divisible by 4 (rem 0). 4th step: 1. 3. Subtract: \((…6) – (…1) = …5\). The final units digit is 5.

Solution (E): We find the remainder of each part when divided by 5. 1. For \(2^{50}\): Cycle of remainders {2, 4, 3, 1}. \(50 \div 4 = 12\) rem 2. 2nd step: 4. 2. For \(3^{50}\): Cycle of remainders {3, 4, 2, 1}. \(50 \div 4 = 12\) rem 2. 2nd step: 4. 3. Sum of remainders: \(4 + 4 = 8\). 4. Remainder of the sum: \(8 \div 5\) has a remainder of 3. *Correction:* \(2^2=4\). \(3^2=9 \equiv 4\). \(4+4=8 \equiv 3\). The answer should be C.

Solution (C): We find the remainder of each part when divided by 5. 1. For \(2^{50}\): The cycle of remainders is {2, 4, 3, 1}. The cycle length is 4. \(50 \div 4 = 12\) remainder 2. This corresponds to the 2nd step: 4. 2. For \(3^{50}\): The cycle of remainders is {3, 4, 2, 1}. The cycle length is 4. \(50 \div 4 = 12\) remainder 2. This corresponds to the 2nd step: 4. 3. Sum of remainders: \(4 + 4 = 8\). 4. Find the remainder of the sum: \(8 \div 5\) has a remainder of 3.

Solution (B): If \(x^4\) ends in 6, the units digit of \(x\) must be 2, 4, 6, or 8. (e.g., \(2^4=16\), \(4^4=256\), \(6^4=…6\), \(8^4=…6\)). Case 1: \(x\) ends in 2. \((x-1)\) ends in 1. \((x-1)^2\) ends in \(1^2=1\). Case 2: \(x\) ends in 4. \((x-1)\) ends in 3. \((x-1)^2\) ends in \(3^2=9\). Case 3: \(x\) ends in 6. \((x-1)\) ends in 5. \((x-1)^2\) ends in \(5^2=…5\). Case 4: \(x\) ends in 8. \((x-1)\) ends in 7. \((x-1)^2\) ends in \(7^2=…9\). The possible units digits are {1, 9, 5}. Of the options, 1 is possible.

Solution (D): We check the remainder of each factorial. \(1! = 1 \equiv 1 \pmod{12}\) \(2! = 2 \equiv 2 \pmod{12}\) \(3! = 6 \equiv 6 \pmod{12}\) \(4! = 24\). 24 is divisible by 12, so \(4! \equiv 0 \pmod{12}\). Every factorial after 4! (e.g., 5!, 6!, …) contains \(4 \times 3\) as factors, so they are all divisible by 12. Their remainders are all 0. We only need to sum the remainders of the first three terms: \(1 + 2 + 6 = 9\). The remainder of the sum is 9.

Solution (A): We only need the units digit of each term. \(1! = 1\) \(3! = 6\) \(5! = 120 \to\) units digit 0. Every odd factorial after 5! (7!, 9!, etc.) will also contain \(5 \times 2\) as factors, so they will all end in 0. We only need to sum the units digits of the first two terms in the series: \(1 + 6 + 0 + 0 + … + 0 = 7\). The final units digit is 7.