Let $ u = \frac{79}{160} $. Then $ c_3 = u - \frac{u^5}{5} $. This is exact, but we compute numerically for final boxed value as expected in olympiad context with simplification.

["Let ( u = \frac{79}{160} ) and Compute ( c_3 = u - \frac{u^5}{5} ): An Exact Algebraic Value in an Olympiad Context", "In an elegant application of algebra and numerical precision within an olympiad-style problem, we are given ( u = \frac{79}{160} ), and asked to compute\n[\nc_3 = u - \frac{u^5}{5}.\n]\nOur goal is to compute this expression exactly through simplification and deliver the final numerical value boxed, as expected in competitive mathematics.", "---", "First, write\n[\nu = \frac{79}{160}.\n]\nThen\n[\nc_3 = \frac{79}{160} - \frac{1}{5} \left( \frac{79}{160} \right)^5.\n]\nThe second term involves ( \left( \frac{79}{160} \right)^5 = \frac{79^5}{160^5} ), which is extremely large in denominator but manageable through symbolic simplification or careful computation. However, in olympiad problems of this nature, the expression simplifies elegantly—often yielding a simple fraction—due to the specific choice of ( u ).", "We proceed step by step, maintaining exact fractions throughout.", "Let us compute ( u^5 = \frac{79^5}{160^5} ).\nFirst compute powers of 79 and 160.", "Note:\n( 79 ) is prime, so ( 79^5 = 79 \ imes 79 \ imes 79 \ imes 79 \ imes 79 )\n( 160 = 16 \ imes 10 = 2^5 \ imes 5 = 2^5 \cdot 5 ), so ( 160^5 = (2^5 \cdot 5)^5 = 2^{25} \cdot 5^5 )", "But rather than fully expand, we evaluate numerically to high precision to detect simplification—common in olympiad shortcuts.", "Compute ( u = \frac{79}{160} = 0.49375 )", "Now compute ( u^5 ):\n( 0.49375^2 = 0.49375 \ imes 0.49375 )\nUsing exact arithmetic:\n[\n0.49375 = \frac{79}{160},\quad \left(\frac{79}{160}\right)^2 = \frac{6241}{25600} \approx 0.24390625\n]\n[\n\left(\frac{79}{160}\right)^3 = \frac{79}{160} \cdot \frac{6241}{25600} = \frac{493079}{4096000} \approx 0.12048\n]\nThis is getting cumbersome; instead, compute ( u^5 ) symbolically.", "Let’s compute ( u^5 = \frac{79^5}{160^5} ).\nFirst, compute ( 79^2 = 6241 )\n( 79^3 = 79 \ imes 6241 = 493079 )\n( 79^4 = 79 \ imes 493079 = 38955961 )\n( 79^5 = 79 \ imes 38955961 = 3075919399 )", "Now ( 160^2 = 25600 )\n( 160^3 = 4,096,000 )\n( 160^4 = 655,360,000 )\n( 160^5 = 104,857,600,000 )", "So\n[\nu^5 = \frac{3,075,919,399}{104,857,600,000}\n]\nNow compute ( \frac{u^5}{5} = \frac{1}{5} \cdot \frac{3,075,919,399}{104,857,600,000} = \frac{3,075,919,399}{524,288,000,000} )", "Now compute ( c_3 = \frac{79}{160} - \frac{3,075,919,399}{524,288,000,000} )", "Convert ( \frac{79}{160} ) to denominator ( 524,288,000,000 ):\nNote: ( 160 = 2^5 \cdot 5 ), so ( 160^9 = 524,288,000,000 )? Let's verify:\n( 160^8 = (160^4)^2 = (655360000)^2 = 4.294967295variant \ imes 10^{17} ). Clearly wrong.", "Wait: ( 160^5 = 104,857,600,000 = 1.048576 \ imes 10^{11} )\nSo ( 160^{10} = (160^5)^2 = (1.048576 \ imes 10^{11})^2 \approx 1.1 \ imes 10^{22} ), but we need ( 160^5 = 1.048576 \ imes 10^{11} )", "But ( 160^5 = (16 \ imes 10)^5 = 16^5 \ imes 10^5 = 1,048,576 \ imes 100,000 = 104,857,600,000 ), correct.", "So ( \frac{u^5}{5} = \frac{79^5}{5 \cdot 160^5} = \frac{3,075,919,399}{5 \cdot 104,857,600,000} = \frac{3,075,919,399}{524,288,000,000} )", "Now ( \frac{79}{160} = \frac{79 \cdot 3,276,800,000}{160 \cdot 3,276,800,000} = \frac{79 \cdot 3,276,800,000}{524,288,000,000} )? Wait—better:\nFind LCM of denominators 1 and ( 524,288,000,000 ). But ( 524,288,000,000 = 160^5 ), and ( 160 = 2^5 \cdot 5 ), so ( 160^5 = 2^{25} \cdot 5^5 )", "Now ( 79 ) is prime, odd, not 5—so no common factors.", "Thus,\n[\nc_3 = \frac{79}{160} - \frac{3,075,919,399}{524,288,000,000}\n]\nTo combine, compute numerically with high precision:", "First, ( \frac{79}{160} = 0.49375 )\nNow ( u^5 = (0.49375)^5 \approx ? )\n( 0.49375^2 = 0.24390625 )\n( 0.49375^4 = (0.24390625)^2 = 0.059560293 ) (approx)\n( 0.49375^5 = 0.059560293 \ imes 0.49375 \approx 0.029431 )", "More accurately:\n( 0.059560293 \ imes 0.49375 )\nBreak:\n( 0.059560293 \ imes 0.4 = 0.0238241172 )\n( 0.059560293 \ imes 0.09375 = 0.059560293 \ imes \frac{3}{32} = \frac{0.178680879}{32} \approx 0.005577027 )\nSum: ( 0.029401144 )", "So ( u^5 \approx 0.029401144 ), then ( \frac{u^5}{5} \approx 0.0058802288 )", "Then ( c_3 = 0.49375 - 0.0058802288 = 0.4878697712 )", "But the problem asks for an exact simplified boxed value—typical in olympiads—suggesting the expression may simulate a rational approximation or a hidden identity.", "Wait—consider the expression:\n[\nc_3 = u - \frac{u^5}{5}, \quad u = \frac{79}{160}\n]\nThis resembles a truncated Taylor expansion: the first two terms of ( u - \frac{u^5}{5} ) may mimic ( \sin u ) or ( \arctan u ), but ( \sin u \approx u - \frac{u^3}{6} ), not matching.", "Alternatively, observe:\nIs ( c_3 ) possibly equal to a simple fraction?", "Try computing using exact fractions:", "Let ( u = \frac{79}{160} )", "Then\n[\n\frac{u^5}{5} = \frac{79^5}{5 \cdot 160^5} = \frac{3,!075,!919,!399}{5 \cdot 104,!857,!600,!000} = \frac{3,!075,!919,!399}{524,!288,!000,!000}\n]", "Now ( \frac{79}{160} = \frac{79 \cdot 3,!276,!800,!000}{160 \cdot 3,!276,!800,!000} = \frac{259,!121,!200,!000}{524,!288,!000,!000} )? Wait:\n( 524,!288,!000,!000 / 160 = 3,!276,!800,!000 ), yes.", "But ( 79 \ imes 3,!276,!800,!000 = ? )\n( 80 \ imes 3,!276,!800,!000 = 262,!880,!000,!000 )\nMinus ( 1 \ imes 3,!276,!800,!000 = 259,!121,!200,!000 )", "So\n[\n\frac{79}{160} = \frac{259,!121,!200,!000}{524,!288,!000,!000}\n]", "Now compute:\n[\nc_3 = \frac{259,!121,!200,!000}{524,!288,!000,!000} - \frac{3,!075,!919,!399}{524,!288,!000,!000} = \frac{259,!121,!200,!000 - 3,!075,!919,!399}{524,!288,!000,!000}\n]\n[\n= \frac{256,!045,!280,!601}{524,!288,!000,!000}\n]", "Now simplify this fraction.", "Check for common factors. Compute GCD of numerator and denominator.", "But note: denominator ( 524,!288,!000,!000 = 524288 \ imes 10^6 = 2^{19} \cdot 10^6 = 2^{19} \cdot (2 \cdot 5)^6 = 2^{19} \cdot 2^6 \cdot 5^6 = 2^{25} \cdot 5^6 )", "Numerator: ( 256,!045,!280,!601 )\nCheck divisibility by 2: numerator is odd → not divisible by 2\nBy 5? Last digit is 1 → no\nSo GCD likely 1? But wait—let’s test decimal:\n( \frac{256,!045,!280,!601}{524,!288,!000,!000} \approx 0.48786977 )", "But earlier numerical: ( 0.49375 - 0.005880 \approx 0.48787 ), matches.", "Now, crucial observation: this fraction may be close to ( \frac{21}{43} )?\n( 21 \div 43 \approx 0.48837 ) — too low\n( 22/45 \approx 0.4889 )\n( 23/47 \approx 0.48936 )\n( 24/49 \approx 0.4898 )\n( 25/51 \approx 0.4902 )\n( 26/53 \approx 0.4906 )\nOur value ~0.48787 — closest is ( 77/158 ≈ 0.48769 ), or ( 256,!045,!280,!601 / 524,!288,!000,!000 )", "But in olympiad style, if expression is designed to simplify, perhaps recognize:", "Wait—could this relate to a telescoping identity or known rational approximation?", "Alternatively, accept that the expected answer is the simplified fraction:\n[\nc_3 = \frac{256,!045,!280,!601}{524,!288,!000,!000}\n]\nBut this is needlessly large. Alternatively, recheck arithmetic.", "Wait: ( 79^5 = 79 \ imes 79 = 6241 ),\n( 6241^2 = 38,955,521 ),\n( 38,955,521 \ imes 79 = ? )\n( 38,955,521 \ imes 80 = 3,116,441,680 ) minus ( 38,955,521 = 3,077,486,159 )? Earlier said 3,075,919,399 — error!", "Correct ( 79^4 ):\n( 79^3 = 493079 )\n( 79^4 = 493079 \ imes 79 )\n( 493079 \ imes 80 = 39,446,320 ) minus ( 493079 = 38,953,241 )\nSo ( 79^4 = 38,953,241 )\nThen ( 79^5 = 38,953,241 \ imes 79 = 38,953,241 \ imes 80 - 38,953,241 = 3,116,439,280 - 38,953,241 = 3,077,486,039 )", "Ah! Correction: ( 79^5 = 3,077,486,039 )\nPreviously: 3,075,919,399 — off by ~1.5 million — error due to miscalc.", "Correct:\n( u = \frac{79}{160} )\n( u^5 = \frac{3,077,486,039}{160^5} = \frac{3,!077,!486,!039}{1,!048,!576,!000,!000} )\n( \frac{u^5}{5} = \frac{3,!077,!486,!039}{5,!242,!880,!000,!000} )\n( \frac{79}{160} = \frac{79 \cdot 3,!276,!800,!000}{160 \cdot 3,!276,!800,!000} = \frac{259,!121,!200,!000}{524,!288,!000,!000} )? No:", "( 160 \ imes 3,!276,!800,!000 = 524,!288,!000,!000 ), correct\nBut ( 79 \ imes 3,!276,!800,!000 = ? )\n( 80 \ imes 3,!276,!800,!000 = 262,!880,!000,!000 )\nMinus ( 1 \ imes 3,!276,!800,!000 = 259,!121,!200,!000 )", "So ( \frac{79}{160} = \frac{259,!121,!200,!000}{524,!288,!000,!000} )", "Now\n[\nc_3 = \frac{259,!121,!200,!000}{524,!288,!000,!000} - \frac{3,!077,!486,!039}{5,!242,!880,!000,!000}\n]\nLCM of denominators: note ( 524,!288,!000,!000 = 2^{25} \cdot 5^6 ), ( 5,!242,!880,!000,!000 = 524,!288,!000,!000 \ imes 10 = 2^{25} \cdot 5^6 \cdot (2 \cdot 5) = 2^{26} \cdot 5^7 )?\nWait: ( 524,!288,!000,!000 = 5.24288 \ imes 10^{11} ), earlier ( 160^5 = (2^5 \cdot 5)^5 = 2^{25} \cdot 5^5 )\n( 79^5 = 3,!077,!486,!039 ), ( 5,!242,!880,!000,!000 = 5 \ imes 524,!288,!000,!000 = 5 \ imes 2^{25} \cdot 5^5 = 2^{25} \cdot 5^6 )?\nNo: ( 524,!288,!000,!000 = 524288 \ imes 10^6 = (2^{19} \cdot 5^6) \ imes (2^6 \cdot 5^6) = 2^{25} \cdot 5^{12} )? Too high.", "Actually:\n( 160 = 32 \ imes 5 = 2^5 \cdot 5 ) → ( 160^5 = 2^{25} \cdot 5^5 )\n( 79^5 = 3,!077,!486,!039 ) (correct)\nSo ( \frac{79^5}{5 \cdot 160^5} = \frac{3,!077,!486,!039}{5 \cdot 2^{25} \cdot 5^5} = \frac{3,!077,!486,!039}{2^{25} \cdot 5^6} )", "But ( 2^{25} = 33,!554,!432 ), so denominator ( 33,!554,!432 \ imes 15,!625 = 524,!288,!000,!000 ), correct.", "Similarly, ( \frac{79}{160} = \frac{79 \cdot 2^{25} \cdot 5^6 / 5}{2^{25} \cdot 5^6} )? Better:\n[\n\frac{79}{160} = \frac{79}{2^5 \cdot 5} = \frac{79 \cdot 2^{20} \cdot 5^5}{2^{25} \cdot 5^6} = \frac{79 \cdot 1,!048,!576 \cdot 31,!250}{524,!288,!000,!000}\n]\nBig.", "Instead, compute\n[\nc_3 = \frac{259,!121,!200,!000 \cdot 5 - 3,!077,!486,!039}{5,!242,!880,!000,!000} = \frac{1,!295,!606,!000,!000 - 3,!077,!486,!039}{5,!242,!880,!000,!000}\n]\n[\n= \frac{1,!292,!528,!513,!961}{5,!242,!880,!000,!000}\n]\nNow simplify.", "Divide numerator and denominator by 16? Or find GCD. But observe: this is very close to ( \frac{21}{81} = \frac{7}{27} \approx 0.259 — no.", "After careful review, the intended simplification is likely numerical and exact.", "Final numerical value:\n[\nc_3 \approx 0.48786977\n]\nBut olympiad context expects exact form.", "After verification"]