Question: What is the remainder when $ 1001 + 1003 + 1005 + \cdots + 1099 $ is divided by $ 16 $?

["Title: Find the Remainder of the Sum $ 1001 + 1003 + 1005 + \cdots + 1099 $ When Divided by 16", "Introduction\nUnderstanding modular arithmetic is essential in number theory and everyday computation, but sometimes the numbers involved feel large and unwieldy. Today, we explore a compelling sum: $ 1001 + 1003 + 1005 + \cdots + 1099 $, and specifically determine the remainder when this sum is divided by 16. Whether you're preparing for a math competition or just sharpening your modular arithmetic skills, this step-by-step breakdown will clarify how modular properties simplify complex sums.", "---", "Step 1: Identify the Structure of the Sum\nThe sequence $ 1001, 1003, 1005, \ldots, 1099 $ is an arithmetic progression of odd numbers.", "- First term $ a = 1001 $\n- Common difference $ d = 2 $\n- Last term $ l = 1099 $", "To find the number of terms $ n $:\n$$\nl = a + (n - 1)d \Rightarrow 1099 = 1001 + (n - 1) \cdot 2\n$$\n$$\n98 = (n - 1) \cdot 2 \Rightarrow n - 1 = 49 \Rightarrow n = 50\n$$\nThere are 50 terms in the sequence.", "---", "Step 2: Compute the Sum of the Sequence\nThe sum $ S $ of an arithmetic series is:\n$$\nS = \frac{n}{2}(a + l) = \frac{50}{2}(1001 + 1099) = 25 \cdot 2100 = 52,500\n$$", "We now seek:\n$$\n52,500 \mod 16\n$$", "---", "Step 3: Compute $ 52,500 \mod 16 $\nInstead of dividing directly, use modular arithmetic simplifications. Note that:\n$$\n16 = 2^4, \ ext{ and we can reduce in parts using powers of 2.}\n$$", "Alternatively, reduce the numerator and divisor modulo 16:", "- First, compute $ 52,500 \div 16 $:\nBut better: simplify $ 52,500 \mod 16 $ using repeated division by 2.", "Or observe:\n$$\n52,500 \div 16 = 3281.25 \Rightarrow \ ext{not helpful directly}\n$$\nInstead, reduce $ 52,500 \mod 16 $ using known powers.", "A more efficient method:\nNote that:\n$$\n52,500 = 52,500 \ imes 1 = ?\n$$\nWe can reduce modulo 16 by breaking up:", "Since $ 1001 \mod 16 $ determines the pattern, compute each term mod 16 and sum.", "---", "Step 4: Use Modulo 16 to Simplify the Sum\nInstead of summing 50 terms, note:\nAll terms are odd, so each $ \equiv 1, 3, 5, \ldots \mod 16 $.\nBut they follow an arithmetic sequence mod 16.", "Let’s compute $ 1001 \mod 16 $:\n$$\n16 \ imes 62 = 992 \Rightarrow 1001 - 992 = 9 \Rightarrow 1001 \equiv 9 \mod 16\n$$\nSince the sequence increases by 2 each time, modulo 16:\n$$\n1001 \equiv 9, \quad 1003 \equiv 11, \quad 1005 \equiv 13, \quad 1007 \equiv 15, \quad 1009 \equiv 1, \quad \ ext{then } 11, \ldots\n$$\nThe sequence mod 16 cycles every 8 terms:\n9, 11, 13, 15, 1, 3, 5, 7, then repeats (since $ 1001 + 14 \cdot 2 = 1029 $, but we go to 1099).\nTotal: 50 terms, and $ 50 \div 8 = 6 $ full cycles (48 terms) + 2 extra.", "Cycle (8 terms):\n$$\n9 + 11 + 13 + 15 + 1 + 3 + 5 + 7 = 64\n$$\n$$\n64 \mod 16 = 0\n$$\nSo each full cycle contributes $ 0 \mod 16 $.\nThus, sum of 6 full cycles $ \equiv 0 \mod 16 $.", "Remaining two terms:\nAfter 48 terms: last term is $ 1001 + 47 \cdot 2 = 1001 + 94 = 1095 $\nNext is $ 1097 $, then $ 1099 $ — but only two extra: terms $ n = 49 $ and $ n = 50 $:\n- $ a_{49} = 1001 + 48 \cdot 2 = 1001 + 96 = 1097 \equiv 9 + 1 = 10 \mod 16 $?\nWait — better: since step is 2 mod 16, and $ a_1 = 9 $:\n$$\na_k \equiv 9 + 2(k-1) \mod 16\n$$\nSo for $ k = 49 $: $ a_{49} \equiv 9 + 2(48) = 9 + 96 = 105 \mod 16 $\n$ 105 \div 16 = 6 \cdot 16 = 96 $, remainder $ 9 $ → $ \equiv 9 $\n$ k = 50 $: $ a_{50} = 9 + 2(49) = 9 + 98 = 107 \Rightarrow 107 - 96 = 11 \Rightarrow \equiv 11 \mod 16 $", "Wait — but earlier manual check got 1097 and 1099. Let's correct indexing.", "Actually, the sequence:\n$ a_k = 1001 + 2(k-1) $, $ k = 1 $ to $ 50 $", "So:\n- $ a_{49} = 1001 + 2(48) = 1001 + 96 = 1097 $\n- $ a_{50} = 1001 + 2(49) = 1001 + 98 = 1099 $", "Now compute $ 1097 \mod 16 $:\n$ 16 \ imes 68 = 1088 $, so $ 1097 - 1088 = 9 \Rightarrow 1097 \equiv 9 \mod 16 $\n$ 1099 - 1088 = 11 \Rightarrow 1099 \equiv 11 \mod 16 $", "So the last two terms mod 16 are 9 and 11.", "But earlier full cycle sum: 9+11+13+15+1+3+5+7 =\nLet’s compute: $ 9+11=20 $, $ 20+13=33 $, $ 33+15=48 $, $ 48+1=49 $, $ +3=52 $, $ +5=57 $, $ +7=64 $. Yes, $ 64 \equiv 0 \mod 16 $.", "So 6 cycles × 8 terms = 48 terms → sum $ \equiv 0 \mod 16 $.", "Add two terms: $ 1097 \equiv 9 $, $ 1099 \equiv 11 $\nTotal remainder: $ 9 + 11 = 20 \equiv 4 \mod 16 $", "Thus:\n$$\n52,500 \equiv 0 + 4 = 4 \mod 16\n$$", "Wait — but earlier direct sum gave $ 52,500 $. Let’s verify $ 52,500 \div 16 $:\n$ 16 \ imes 3281 = 16 \ imes 3000 = 48,000 $, $ 16 \ imes 281 = 4496 $, total $ 48,000 + 4,496 = 52,496 $\nThen $ 52,500 - 52,496 = 4 $\nSo $ 52,500 \equiv 4 \mod 16 $", "Conclusion:\nThe remainder when $ 1001 + 1003 + 1005 + \cdots + 1099 $ is divided by 16 is $ \boxed{4} $.", "---", "Why This Matters\nUnderstanding how modular arithmetic applies to arithmetic sequences helps solve large sums efficiently—critical in cryptography, algorithm design, and competitive math. Recognizing cycles and simplifying with modulo properties avoids tedious full computation.", "Key Takeaway:\nAlways use properties of modular arithmetic—especially periodicity—to reduce large sums with minimal computation.", "---", "Final Answer:\nThe remainder is $ \boxed{4} $."]