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{
"id": "test/intermediate_algebra/956.json",
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"gegeben": "(6,31,-1)",
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"denk_ende": "id long division. Let's multiply to be absolutely sure:\nf * Q where Q = x^6+3x^5+5x^4+10x^3+13x^2+4x+1.\nCompute f=x^3-3x^2+4x-1.\nMultiply by x^6: x^9 -3x^8+4x^7 -x^6.\n3x^5: 3x^8 -9x^7+12x^6 -3x^5.\n5x^4: 5x^7 -15x^6+20x^5 -5x^4.\n10x^3: 10x^6 -30x^5+40x^4 -10x^3.\n13x^2: 13x^5 -39x^4+52x^3 -13x^2.\n4x: 4x^4 -12x^3+16x^2 -4x.\n1: x^3 -3x^2+4x -1.\nSum:\nx^9.\nx^8: -3+3=0.\nx^7:4-9+5=0.\nx^6:-1+12-15+10=6.\nx^5:-3+20-30+13=0.\nx^4:-5+40-39+4=0.\nx^3:-10+52-12+1=31.\nx^2:-13+16-3=0.\nx:-4+4=0.\nconst -1. Good.\n\nCould include this as check. Nice.\n\nNeed final answer at very end. If I include check after final? No, final must be last. So put check before final line.\n\nLet's consider if there is any issue with using \"modulo f\" and setting remainder zero: Since f is monic, division algorithm over any coefficient ring? p,q,r unknown but in field (reals). Fine.\n\nLet's write in German:\n\"Wir suchen p,q,r, sodass der Rest bei Division durch f(x) verschwindet.\"\nThen steps.\n\nNeed maybe mention \"Da f monisch ist, ist die Bedingung äquivalent zu Rest 0.\" Good.\n\nLet's produce final. Ensure no analysis in final. Use LaTeX.\n\nOne more thought: Could p,q,r be required as integers? We found integers. Good.\n\nLet's final.\n",
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"id": "test/intermediate_algebra/1063.json",
"ok": true,
"gegeben": "1+274i",
"erwartet": "1+274i",
"level": 4,
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"denk_ende": "1,1): (1+i)*1 + (1+i)*1 =2+2i.\n(1,2): (1+i)*i + (1+i)*(-i) =0.\n(2,1): (1-i)*1 + (i-1)*1 =0.\n(2,2): (1-i)*i + (i-1)*(-i) = i - i^2 + (-i^2 + i) = i+1 +1+i =2+2i. So A^3 = (2+2i) I. Scalar, so F^3=id. Good.\n\nAll consistent.\n\nLet's also think about possible domain: F defined for all complex z not i. If z_n = i for some n, sequence stops. But we avoid.\n\nCould z0 be such that z1 = i? z0 + i = i(z0 - i) => z0=1. Not. z2 = i? z1=1. Not. Good.\n\nNow final in German. Need maybe not too terse. Include all steps. Ensure final boxed at very end. No extra after box. Use LaTeX.\n\nLet's draft:\n\nWir bestimmen die Iterationen von F.\n\nFür z \\ne i gilt\n...\nAlso F^2(z)=...\nDann\n...\nAlso F^3(z)=z.\n\nDamit ist die Abbildung F auf dem Riemannschen Zahlenkreis eine Permutation der Ordnung 3. Für unseren Startwert sind die ersten drei Werte definiert: z0 != i, z1 = ... != i, und z2 != i (da z1 !=1). Daher gilt z_{n+3}=z_n für alle n.\n\n2002 = 3*667+1, also z_{2002}=z_1.\n\nBerechnung:\nz_1 = ... = 1+274i.\n\n\\boxed{1+274i}\n\nNeed be careful: \"z2 != i (da z1 !=1)\" because F(z)=i iff z=1. We can include. Or simply \"und die Folge wiederholt sich, ohne i zu treffen\". Good.\n\nLet's ensure no final answer after box. Done.\n",
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"denk_ende": " ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x= ? x=",
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"id": "test/algebra/686.json",
"ok": true,
"gegeben": "4",
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"denk_ende": " 4/(1-3)=-2 not square; a=2 gives 8/( -1)=-8; a=3 undefined. But infinite series not convergent. So no.\n\nNeed final answer only? They ask solve. We can provide concise solution. Ensure final answer at very end \\boxed{4}. In German.\n\nLet's craft: The series has first term 4 and common ratio 3/a. For convergence need |3/a|<1 => a>3. Sum S=4/(1-3/a)=4a/(a-3). For a=4, S=16=4^2. Since a must be integer >3, smallest possible is 4. Box.\n\nNeed maybe mention if a=4 is indeed smallest because a>3. Good.\n\nDouble-check: If a=4, terms: 4 + 3 + 9/4 + 27/16 + ... sum 16. yes.\n\nPotential trap: \"perfect square\" could be square of rational? If so, a=5 gives S=20/2=10 not square rational? 10 not square rational? Actually rational square if numerator/denominator squares after reduced; 10 no. a=6 gives 24/3=8 no. a=7 28/4=7 no. a=8 32/5 no. a=9 36/6=6 no. a=10 40/7 no. a=12 48/9=16/3 no. But a=4 works anyway. If a<4 divergent. So no issue.\n\nNeed final in German. Ensure final answer exactly at end. Could include \\boxed{4}. The instruction: \"Schreibe die endgültige Antwort ganz zum Schluss in der Form \\boxed{ANTWORT}.\" So final line should be boxed. We can have explanation before.\n\nLet's produce final.\n",
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{
"id": "test/intermediate_algebra/1422.json",
"ok": true,
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"level": 3,
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"denk_zeichen": 4662,
"denk_ende": " m/n in lowest terms, m divides 1, n divides 2. So m=±1, n=1,2. Four candidates. Are all possible? We can construct polynomials:\nFor 1: 2x^4 -3x^3 +1? Wait P(1)=2-3+1=0 yes. But a2,a1=0.\nFor -1: 2x^4 +3x^3 +1? P(-1)=2-3+1=0 yes.\nFor 1/2: 2x^4 -9x^3 +1? P(1/2)=2/16 -9/8 +1 =1/8 -9/8 +1=0. yes.\nFor -1/2: 2x^4 +9x^3 +1? P(-1/2)=1/8 -9/8 +1=0. yes.\nActually for -1/2 with a3=9: P(-1/2)=1/8 -9/8 +1=0. Good.\n\nCould there be rational root 0? Constant 1 no.\n\nNeed maybe count \"different possible rational roots\" not \"values of a_i\". Yes.\n\nLet's craft final in German:\nSei P(x)=... Nach dem Satz über rationale Nullstellen: Wenn r=p/q (ggT=1) rationale Nullstelle ist, dann p | 1 und q | 2. Also p=±1, q=1 oder 2. Kandidaten: 1, -1, 1/2, -1/2.\nMan prüft, dass alle tatsächlich vorkommen können:\nP(x)=2x^4-3x^3+1 hat Nullstelle 1;\nP(x)=2x^4+3x^3+1 hat Nullstelle -1;\nP(x)=2x^4-9x^3+1 hat Nullstelle 1/2;\nP(x)=2x^4+9x^3+1 hat Nullstelle -1/2.\nDamit sind genau 4 verschiedene rationale Nullstellen möglich.\nFinal boxed.\n\nNeed ensure final answer exactly at end? \"Schreibe die endgültige Antwort ganz zum Schluss in der Form \\boxed{ANTWORT}.\" So final line should be \\boxed{4}. Could have no text after. Good.\n",
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{
"id": "test/intermediate_algebra/960.json",
"ok": true,
"gegeben": "\\frac{17}{21}",
"erwartet": "\\frac{17}{21}",
"level": 5,
"schleife": false,
"denk_zeichen": 22172,
"denk_ende": ")^n-2\\left(\\frac1{15}\\right)^n\\right]=\\frac{17}{77}.\n\\]\nThen S1=H, S2=2H, S3=(2/3)H. Total = (11/3)H = 17/21. This is concise.\n\nLet's ensure H positive: 4/11 -1/7 = 17/77.\n\nCould there be a need to justify sum over d and c separable? Since finite sums of positive terms, Tonelli. We can say \"Da alle Terme positiv sind, dürfen wir die Summation vertauschen.\" Good.\n\nLet's maybe include a note: \"Die Tripel sind geordnet; die Bedingung ist symmetrisch, aber die Gewichtung nicht.\" This clarifies.\n\nNow final.\n\nBut before final, let's think if there is any hidden requirement about language: User wrote German \"Löse die Aufgabe. Schreibe die endgültige Antwort ganz zum Schluss in der Form \\boxed{ANTWORT}.\" So answer in German. Good.\n\nLet's maybe not mention \"AIME\" etc.\n\nPotential issue: The sum over T: T is a set, so if multiple triangles exist for same side lengths, counted once. We count each triple once. Good.\n\nLet's consider if \"there exist triangles with side lengths a,b,c\" could mean there exist (possibly different) triangles for each triple? No.\n\nLet's maybe include \"strikt\" in final: \"Ein (nicht entartetes) Dreieck\". If they don't want assumption, standard.\n\nLet's produce final now.\n",
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{
"id": "test/intermediate_algebra/1544.json",
"ok": true,
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"denk_ende": "x^2+x-1.\n- T(u)=T(v)=u (or both v): u fixed. If u=2, v^2-2=2 -> v=-2 (v≠u). If u=-1, v^2-2=-1 -> v=1. Equations x^2-4, x^2-1.\n\nNo other functions from 2-element set to itself.\n\nList and count.\n\nLet's ensure in constant case if u=2, v=2 is double root already; we exclude v≠u. If u=-1, v=-1 double. Good.\n\nMaybe in 2-cycle case, when solving u^2-2=v, v^2-2=u, subtract: u^2-v^2 = v-u? Let's do carefully:\nu^2 - 2 = v\nv^2 - 2 = u\nSubtract first minus second: u^2 - v^2 = v - u = -(u-v). Factor (u-v)(u+v) = -(u-v). Since u≠v, u+v=-1. Good. Then v=-1-u. Substitute: u^2-2=-1-u -> u^2+u-1=0. The two roots are u,v. The monic polynomial with these roots is x^2 - (u+v)x + uv. We know sum -1. Product? From u^2+u-1=0 and v=-1-u, uv = -u - u^2 = -u -(1-u) = -1. Or since both roots of x^2+x-1, product -1. So equation x^2 + x -1. Good.\n\nIn fixed pair, polynomial (x-2)(x+1)=x^2 - x -2.\n\nIn constant cases, (x-2)(x+2)=x^2-4; (x-1)(x+1)=x^2-1.\n\nDouble: (x-2)^2=x^2-4x+4; (x+1)^2=x^2+2x+1.\n\nCount: 2+1+1+2=6.\n\nLet's maybe include \"Diese sechs sind tatsächlich gültig\" and quick check maybe not necessary but can mention.\n\nFinal line: \\boxed{6}\n\nNeed ensure no markdown issues. Use LaTeX.\n\nLet's produce final.\n",
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{
"id": "test/prealgebra/631.json",
"ok": true,
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"denk_ende": "ation:\n\n```\nDas Quadrat hat Seitenlänge 4.\nIn der gegebenen Figur hat das graue Dreieck die Eckpunkte (2,0), (4,2) und (4,0).\nDie Strecke von (4,0) nach (2,0) hat Länge 2, die von (4,0) nach (4,2) hat Länge 2, und sie stehen senkrecht aufeinander.\nAlso ist die Fläche ...\n```\n\nThis is clear and uses exact coordinates from Asy. If someone objects figure not to scale, the Asy code is exact. But in final, maybe not mention Asy code. \"In der gegebenen Figur\" okay.\n\nLet's think if there is any chance the gray piece is not the triangle but the gray fill includes only that triangle. Yes.\n\nNeed final answer at very end. Ensure no trailing text after box. Good.\n\nLet's craft final in German:\n\n\"Die Fläche des großen Quadrats ist 16, daher ist seine Seitenlänge 4. Das graue Stück ist das Dreieck in der Ecke C. Seine beiden Katheten liegen auf den Seiten des Quadrats und haben in der Tangram-Abbildung die Länge 2 (die Eckpunkte sind (2,0), (4,2) und (4,0)). Da es rechtwinklig ist, gilt ... =2.\"\n\nMaybe include \"quadratzoll\". Final.\n\nDouble-check if \"piece e''\" has typo e'' maybe e. Not relevant.\n\nLet's also consider if the problem might be from MATHCOUNTS or AMC, answer maybe 2. Yes.\n\nNow final.\n",
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"id": "test/geometry/817.json",
"ok": true,
"gegeben": "\\frac{14}{3}",
"erwartet": "\\frac{14}{3}",
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"denk_ende": "esser AQ, also Radius 7. Sein Mittelpunkt C liegt auf AQ mit QC=7. Der kleinere Semikreis berührt den großen Kreis in B; sein Mittelpunkt D liegt auf BQ. Da sein Radius r ist, ist QD=14-r. Weil die beiden Semikreise sich berühren, ist CD=7+r. Die Radien AQ und BQ stehen senkrecht aufeinander, also ist ΔQCD rechtwinklig bei Q. Damit gilt ... r=14/3.\"\n\nCheck if \"berührt den großen Kreis in B\" is given? The smaller semicircle's diameter endpoint B on large circle; yes. If not explicit, from diagram. We can say \"Nach der Abbildung\".\n\nLet's ensure no algebra mistake in German: (7+r)^2 = 7^2 + (14-r)^2.\nExpand: 49+14r+r^2 = 49+196-28r+r^2.\nCancel 49+r^2: 14r =196-28r -> 42r=196 -> r=196/42=14/3.\n\nFinal line: \\boxed{\\frac{14}{3}}\n\nShould we include units? Maybe \\boxed{\\frac{14}{3}}. If they want \"inches\", not. The instruction says answer as common fraction. Use no units.\n\nLet's also think if the final answer should be in German \"ANTWORT\" replaced by fraction. Yes.\n\nNow, any need to mention if the diagram is mirrored? Could say \"Ohne Beschränkung der Allgemeinheit\". But final solution can assume.\n\nLet's produce final with enough detail. Ensure final answer at very end. No extra after box.\n",
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"id": "test/intermediate_algebra/1350.json",
"ok": true,
"gegeben": "14",
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"denk_ende": "em asks \"what is the least possible value of |a_{n-1}|?\" Could answer be 14. Yes.\n\nLet's maybe include \"n=3\" in example. Good.\n\nOne more subtle point: In f(x)=2∏(x+b_i), if b_i are distinct? Since r_i distinct, b_i distinct. Good.\n\nLet's consider if a polynomial with integer coefficients and roots distinct integers could have leading coefficient 2 but not all roots simple? distinct means simple. Good.\n\nPotential alternative: If roots are distinct integers but polynomial may have degree n and leading coefficient 2, but could have a factor like (2x - r) with integer root r/2? No, root integer means factor (x-r). Leading coefficient product of leading coefficients of linear factors. If all roots integers, each factor can be (x-r) times constant; product constants =2. But over Q, factorization is 2∏(x-r_i). Could there be a factor 2 attached to one linear factor, e.g. (2x-r)(x-s)... with root r/2 not integer? No, if root integer r, factor must be (x-r) up to constant. The product of constants is 2. So 2∏(x-r_i). Good.\n\nLet's maybe mention \"über \\mathbb{Q}\" to be precise.\n\nNow final. Ensure no markdown issues. Use LaTeX.\n\nFinal answer in German:\n\n```\nSeien ...\n...\n\\boxed{14}\n```\n\nGood.\n",
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