GRE Math: Units Digits & Remainders

Solution (B): The units digit of powers of 13 follows a 4-step cycle: \(13^1 \to 3\), \(13^2 \to 9\), \(13^3 \to 7\), \(13^4 \to 1\). Divide the exponent 33 by 4: \(33 \div 4 = 8\) with a remainder of 1. The remainder of 1 corresponds to the first step in the cycle. The units digit is 3.

Solution (D): 1. Units digit of \(177^{28}\): Cycle for 7 is {7, 9, 3, 1}. \(28 \div 4 = 7\) remainder 0. This corresponds to the 4th step (1). 2. Units digit of \(133^{23}\): Cycle for 3 is {3, 9, 7, 1}. \(23 \div 4 = 5\) remainder 3. This corresponds to the 3rd step (7). 3. Subtract the units digits: \(…1 – …7\). We must borrow: \(11 – 7 = 4\). *Correction:* Let’s re-check. \(11-7=4\). Ah, I see the error in my generation. Let’s fix the question or answer. \(177^{28} \to 1\). \(133^{23} \to 7\). \(…1 – …7 \to …4\). The options are {1, 3, 4, 6, 9}. The answer should be C. Let me fix the data-answer.

Solution (C): 1. Find the units digit of \(177^{28}\). The cycle for 7 is {7, 9, 3, 1}. The exponent 28 is divisible by 4 (remainder 0), so it’s the 4th step: 1. 2. Find the units digit of \(133^{23}\). The cycle for 3 is {3, 9, 7, 1}. \(23 \div 4 = 5\) remainder 3. This is the 3rd step: 7. 3. Subtract the units digits: \(…1 – …7\). We must borrow from the tens place: \(11 – 7 = 4\).

Solution (E): We need the units digit. The cycle for 7 is {7, 9, 3, 1}. The exponent is \(4x+3\). When divided by 4, \(4x\) is divisible, leaving a remainder of 3. This corresponds to the 3rd step in the cycle: 7. So, \(7^{4x+3}\) has a units digit of 7. The expression becomes \((…7) + 2 = …9\). Any number ending in 9, when divided by 5, has a remainder of 4.

Solution (A): We need a number \(n\) such that its square ends in 4. The possibilities are \(…2^2 = …4\) and \(…8^2 = …64 \to …4\). The problem states \(n\) is an *even* number. Both 2 and 8 are even. Case 1: Units digit of \(n\) is 2. Then units digit of \(n+3\) is \(2+3 = 5\). Case 2: Units digit of \(n\) is 8. Then units digit of \(n+3\) is \(8+3 = 11 \to 1\). Of the options given, 5 is a possibility. 1 is also a possibility, but only one answer can be correct. Ah, option E is 1. Let’s re-read the question. “what *could be* the units digit”. Both 1 and 5 are possible. This is a poorly formed question. Let’s fix it. *New Q4*: If \(n\) is a positive even number whose units digit is the same as the units digit of \(n^2\), which of the following could be \(n\)? (A) 12 (B) 14 (C) 16 (D) 18 (E) 20 Let’s try this. (A) n=12. n ends in 2. n^2=144 ends in 4. (No) (B) n=14. n ends in 4. n^2=196 ends in 6. (No) (C) n=16. n ends in 6. n^2=256 ends in 6. (Yes) (D) n=18. n ends in 8. n^2=324 ends in 4. (No) (E) n=20. n ends in 0. n^2=400 ends in 0. (Yes) This is similar to Q7 in the image. “Which one of the following could be n?” The options for n are 12, 14, 15, 16, 17. (A) 12 -> 2 vs 144 -> 4 (No) (B) 14 -> 4 vs 196 -> 6 (No) (C) 15 -> 5 vs 225 -> 5 (Yes, but not even) (D) 16 -> 6 vs 256 -> 6 (Yes, and is even) (E) 17 -> 7 vs 289 -> 9 (No) So the answer is 16. Let’s use this question.

Solution (D): We test the units digit of \(n\) vs \(n^2\) for each option. (A) \(n=12\). Units digit of \(n\) is 2. Units digit of \(n^2 = 144\) is 4. (2 \(\neq\) 4) (B) \(n=14\). Units digit of \(n\) is 4. Units digit of \(n^2 = 196\) is 6. (4 \(\neq\) 6) (C) \(n=15\). This is not an even number. (D) \(n=16\). Units digit of \(n\) is 6. Units digit of \(n^2 = 256\) is 6. (6 = 6). This fits. (E) \(n=18\). Units digit of \(n\) is 8. Units digit of \(n^2 = 324\) is 4. (8 \(\neq\) 4)

Solution (B): The remainder when divided by 10 is just the units digit. \(1! = 1\) \(2! = 2\) \(3! = 6\) \(4! = 24 \to\) units digit 4 \(5! = 120 \to\) units digit 0 Every factorial after 5! (i.e., 6!, 7!, …) will also end in 0, as it will contain \(5 \times 2\) as factors. So, we only need to sum the units digits of the first four factorials: \(1 + 2 + 6 + 4 = 13\). The units digit of the total sum is 3. The remainder is 3.

Solution (A): We need the units digit of \(16^y\). We only care about the units digit of the base, which is 6. The powers of 6 always have a units digit of 6: \(6^1 = 6\), \(6^2 = 36\), \(6^3 = 216\), etc. Since the base ends in 6, the result will always end in 6 for any positive integer exponent \(y\).

Solution (D): First, substitute \(x=2\) into the exponents. The expression becomes \((24)^{(2 \times 2+1)} \times (33)^{(2+1)} = (24)^5 \times (33)^3\). 1. Units digit of \((24)^5\): Base ends in 4. Powers of 4 have a 2-step cycle: {4, 6}. Since the exponent 5 is odd, it’s the 1st step: 4. 2. Units digit of \((33)^3\): Base ends in 3. Powers of 3 have a 4-step cycle: {3, 9, 7, 1}. The exponent 3 is the 3rd step: 7. 3. Multiply the units digits: \(4 \times 7 = 28\). The final units digit is 8.

Solution (B): We need to find the sum of the remainders. \(1! = 1 \to 1 \pmod{5}\) \(2! = 2 \to 2 \pmod{5}\) \(3! = 6 \to 1 \pmod{5}\) \(4! = 24 \to 4 \pmod{5}\) \(5! = 120\). Since 120 is divisible by 5, the remainder is 0. Every factorial after 5! (i.e., 6!, 7!, …) contains 5 as a factor, so they are all divisible by 5 and have a remainder of 0. We only need to sum the remainders of the first four terms: \(1 + 2 + 1 + 4 = 8\). Now, find the remainder of this sum: \(8 \div 5 = 1\) with a remainder of 3.

Solution (A): 1. Units digits of \(x = 8^a\): Cycle is {8, 4, 2, 6}. Possible units digits are 8, 4, 2, 6. 2. Units digits of \(y = 4^b\): Cycle is {4, 6}. Possible units digits are 4, 6. 3. Possible units digits of \(xy\): We must test all combinations. (8 \(\times\) 4) = 32 \(\to\) 2 (8 \(\times\) 6) = 48 \(\to\) 8 (4 \(\times\) 4) = 16 \(\to\) 6 (4 \(\times\) 6) = 24 \(\to\) 4 (2 \(\times\) 4) = 8 \(\to\) 8 (2 \(\times\) 6) = 12 \(\to\) 2 (6 \(\times\) 4) = 24 \(\to\) 4 (6 \(\times\) 6) = 36 \(\to\) 6 The set of all possible units digits for \(xy\) is {2, 8, 6, 4}. Of the options given, only 4 is in this set.

Solution (D): The units digit of powers of 2 follows a 4-step cycle: \(2^1 \to 2\), \(2^2 \to 4\), \(2^3 \to 8\), \(2^4 \to 6\). Divide the exponent 103 by 4: \(103 \div 4 = 25\) with a remainder of 3. (Note: \(100\) is divisible by 4, so \(103\) has a remainder of 3). The remainder of 3 corresponds to the third step in the cycle. The units digit is 8.