{
  "index": "1970-A-3",
  "type": "NT",
  "tag": [
    "NT"
  ],
  "difficulty": "",
  "question": "A-3. Find the length of the longest sequence of equal nonzero digits in which an integral square can terminate (in base 10) and find the smallest square which terminates in such a sequence.",
  "solution": "A-3 If \\( x \\) is an integer then \\( x^{2} \\equiv 0,1,4,6 \\) or \\( 9(\\bmod 10) \\). The case \\( x^{2} \\equiv 0 \\) \\( (\\bmod 10) \\) is eliminated by the statement of the problem. If \\( x^{2} \\equiv 11,55 \\) or 99 \\( (\\bmod 100) \\), then \\( x^{2} \\equiv 3(\\bmod 4) \\) which is impossible. Similarly, \\( x^{2} \\equiv 66(\\bmod 100) \\) implies \\( x^{2} \\equiv 2(\\bmod 4) \\) which is also impossible. Therefore \\( x^{2} \\equiv 44(\\bmod 100) \\). If \\( x^{2} \\equiv 4444(\\bmod 10,000) \\), then \\( x^{2} \\equiv 12(\\bmod 16) \\), but a simple check shows that this is impossible. Finally note that \\( (38)^{2}=1444 \\).",
  "vars": [
    "x"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "integerx"
      },
      "question": "A-3. Find the length of the longest sequence of equal nonzero digits in which an integral square can terminate (in base 10) and find the smallest square which terminates in such a sequence.",
      "solution": "A-3 If \\( integerx \\) is an integer then \\( integerx^{2} \\equiv 0,1,4,6 \\) or \\( 9(\\bmod 10) \\). The case \\( integerx^{2} \\equiv 0 \\) \\( (\\bmod 10) \\) is eliminated by the statement of the problem. If \\( integerx^{2} \\equiv 11,55 \\) or 99 \\( (\\bmod 100) \\), then \\( integerx^{2} \\equiv 3(\\bmod 4) \\) which is impossible. Similarly, \\( integerx^{2} \\equiv 66(\\bmod 100) \\) implies \\( integerx^{2} \\equiv 2(\\bmod 4) \\) which is also impossible. Therefore \\( integerx^{2} \\equiv 44(\\bmod 100) \\). If \\( integerx^{2} \\equiv 4444(\\bmod 10,000) \\), then \\( integerx^{2} \\equiv 12(\\bmod 16) \\), but a simple check shows that this is impossible. Finally note that \\( (38)^{2}=1444 \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "sunflower"
      },
      "question": "A-3. Find the length of the longest sequence of equal nonzero digits in which an integral square can terminate (in base 10) and find the smallest square which terminates in such a sequence.",
      "solution": "A-3 If \\( sunflower \\) is an integer then \\( sunflower^{2} \\equiv 0,1,4,6 \\) or \\( 9(\\bmod 10) \\). The case \\( sunflower^{2} \\equiv 0 \\) \\( (\\bmod 10) \\) is eliminated by the statement of the problem. If \\( sunflower^{2} \\equiv 11,55 \\) or 99 \\( (\\bmod 100) \\), then \\( sunflower^{2} \\equiv 3(\\bmod 4) \\) which is impossible. Similarly, \\( sunflower^{2} \\equiv 66(\\bmod 100) \\) implies \\( sunflower^{2} \\equiv 2(\\bmod 4) \\) which is also impossible. Therefore \\( sunflower^{2} \\equiv 44(\\bmod 100) \\). If \\( sunflower^{2} \\equiv 4444(\\bmod 10,000) \\), then \\( sunflower^{2} \\equiv 12(\\bmod 16) \\), but a simple check shows that this is impossible. Finally note that \\( (38)^{2}=1444 \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "noninteger"
      },
      "question": "A-3. Find the length of the longest sequence of equal nonzero digits in which an integral square can terminate (in base 10) and find the smallest square which terminates in such a sequence.",
      "solution": "A-3 If \\( noninteger \\) is an integer then \\( noninteger^{2} \\equiv 0,1,4,6 \\) or \\( 9(\\bmod 10) \\). The case \\( noninteger^{2} \\equiv 0 \\) \\( (\\bmod 10) \\) is eliminated by the statement of the problem. If \\( noninteger^{2} \\equiv 11,55 \\) or 99 \\( (\\bmod 100) \\), then \\( noninteger^{2} \\equiv 3(\\bmod 4) \\) which is impossible. Similarly, \\( noninteger^{2} \\equiv 66(\\bmod 100) \\) implies \\( noninteger^{2} \\equiv 2(\\bmod 4) \\) which is also impossible. Therefore \\( noninteger^{2} \\equiv 44(\\bmod 100) \\). If \\( noninteger^{2} \\equiv 4444(\\bmod 10,000) \\), then \\( noninteger^{2} \\equiv 12(\\bmod 16) \\), but a simple check shows that this is impossible. Finally note that \\( (38)^{2}=1444 \\)."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp"
      },
      "question": "A-3. Find the length of the longest sequence of equal nonzero digits in which an integral square can terminate (in base 10) and find the smallest square which terminates in such a sequence.",
      "solution": "A-3 If \\( qzxwvtnp \\) is an integer then \\( qzxwvtnp^{2} \\equiv 0,1,4,6 \\) or \\( 9(\\bmod 10) \\). The case \\( qzxwvtnp^{2} \\equiv 0 \\) \\( (\\bmod 10) \\) is eliminated by the statement of the problem. If \\( qzxwvtnp^{2} \\equiv 11,55 \\) or 99 \\( (\\bmod 100) \\), then \\( qzxwvtnp^{2} \\equiv 3(\\bmod 4) \\) which is impossible. Similarly, \\( qzxwvtnp^{2} \\equiv 66(\\bmod 100) \\) implies \\( qzxwvtnp^{2} \\equiv 2(\\bmod 4) \\) which is also impossible. Therefore \\( qzxwvtnp^{2} \\equiv 44(\\bmod 100) \\). If \\( qzxwvtnp^{2} \\equiv 4444(\\bmod 10,000) \\), then \\( qzxwvtnp^{2} \\equiv 12(\\bmod 16) \\), but a simple check shows that this is impossible. Finally note that \\( (38)^{2}=1444 \\)."
    },
    "kernel_variant": {
      "question": "(“Double–layered squares”)\n\nFor a non–zero digit $d\\in\\{1,2,\\dots ,9\\}$ and a positive integer $k$ put  \n\n\\[\nR_k(d)=d\\cdot\\frac{10^{\\,k}-1}{9},\n\\]\n\nso that $R_k(d)$ is the integer whose decimal expansion consists of $k$ consecutive copies of $d$.\n\nA positive integer $N$ is called a \\emph{double–layered square of thickness $k$} if  \n\n(i) $N$ is a perfect square,  \n\n(ii) the last $k$ decimal digits of $N$ equal $R_k(d)$ for some $d\\neq 0$,  \n\n(iii) deleting these $k$ digits once again leaves a perfect square, i.e. $\\bigl\\lfloor N/10^{\\,k}\\bigr\\rfloor$ is a square.\n\n(a) Determine the largest integer $k$ for which a double–layered square exists.  \n\n(b) For this maximal $k$ find the smallest double–layered square and state its trailing digit $d$.\n\n\n\n--------------------------------------------------------------------",
      "solution": "Throughout we write  \n\n\\[\nN=x^{2},\\qquad \n\\Bigl\\lfloor\\frac{N}{10^{\\,k}}\\Bigr\\rfloor = y^{2},\n\\qquad x,y\\in\\mathbb{N}.\n\\tag{1}\n\\]\n\nWith $R_k(d)=d(10^{\\,k}-1)/9$ condition (ii) reads  \n\n\\[\nx^{2}=10^{\\,k}y^{2}+R_k(d).\n\\tag{2}\n\\]\n\nEquivalently  \n\n\\[\nx^{2}\\equiv R_k(d)\\pmod{10^{\\,k}},\\qquad 10^{\\,k}=2^{\\,k}\\cdot 5^{\\,k}.\n\\tag{3}\n\\]\n\nHence\n\n\\[\nx^{2}\\equiv R_k(d)\\pmod{2^{\\,k}},\\qquad \nx^{2}\\equiv R_k(d)\\pmod{5^{\\,k}}.\n\\tag{4}\n\\]\n\n\n--------------------------------------------------------------------\n1.  A joint $2$–adic / $5$–adic obstruction implies $k\\le 3$  \n\nAssume $k\\ge 4$ and put  \n\n\\[\nS:=R_k(d)=d\\cdot\\frac{10^{\\,k}-1}{9}.\n\\]\n\n\\emph{(a) The $2$–adic side.}  \nBecause $k\\ge 4$, $10^{\\,k}\\equiv 0\\pmod{16}$, whence  \n\n\\[\nS\\equiv -d\\cdot 9^{-1}\\equiv -d\\cdot 9\\equiv 7d\\pmod{16}\n\\qquad\\bigl(9^{-1}\\equiv 9\\bigr).\n\\tag{5}\n\\]\n\nQuadratic residues modulo $16$ form the set $\\{0,1,4,9\\}$.  \nThus $7d\\pmod{16}$ must belong to this set; checking $d=1,\\dots ,9$ shows\n\n\\[\n7d\\pmod{16}\\in\\{0,1,4,9\\}\\Longleftrightarrow d=7.\n\\tag{6}\n\\]\n\nHence $d$ would have to be $7$.\n\n\\emph{(b) The $5$–adic side.}  \nSince $10^{\\,k}\\equiv 0\\pmod{5}$ for every $k\\ge 1$,\n\n\\[\nS\\equiv -d\\cdot 9^{-1}\\equiv -d\\cdot(-1)\\equiv d\\pmod{5},\n\\tag{7}\n\\]\nas $9\\equiv -1\\pmod5$.  \nQuadratic residues modulo $5$ are $0,1,4$, so $d\\equiv 0,1,4\\pmod5$.  \nBut $d=7$ found in (6) satisfies $d\\equiv 2\\pmod5$, contradicting (7).\n\nConsequently no double–layered square can exist for $k\\ge 4$, and therefore  \n\n\\[\nk_{\\max}\\le 3.\n\\tag{8}\n\\]\n\n--------------------------------------------------------------------\n2.  The admissible digit when $k=3$\n\nPut $k=3$.  Then $10^{3}=8\\cdot125$ and  \n\n\\[\nR_3(d)=111d.\n\\tag{9}\n\\]\n\nApplying (4):\n\n(i)  Modulo $8$: $111\\equiv 7$.  \nSince the quadratic residues modulo $8$ are $\\{0,1,4\\}$ we need  \n\n\\[\n7d\\equiv 0,1,4\\pmod8\\Longrightarrow d\\in\\{4,7,8\\}.\n\\tag{10}\n\\]\n\n(ii) Modulo $5$: $111d\\equiv d\\pmod5$.  \nBecause a square is congruent to $0,1,4\\pmod5$, we must have $d\\equiv 0,1,4\\pmod5$.  \nCombining with (10) leaves the unique possibility  \n\n\\[\nd=4.\n\\tag{11}\n\\]\n\nThus every double–layered square of thickness $3$ ends in the block $444$.\n\n--------------------------------------------------------------------\n3.  Solving $x^{2}\\equiv 444\\pmod{1000}$\n\nBecause $1000=8\\cdot125$ we work separately.\n\n•  Modulo $8$: $444\\equiv 4$, so $x\\equiv 2\\text{ or }6\\pmod8$.  \n\n•  Modulo $125$: Hensel lifting gives the two solutions  \n\n\\[\nx\\equiv 38 \\text{ or } 87\\pmod{125}.\n\\]\n\nChinese remaindering yields  \n\n\\[\nx\\equiv  38,\\;462,\\;538,\\;962\\pmod{1000}.\n\\tag{12}\n\\]\n\n--------------------------------------------------------------------\n4.  Parametrisation of all candidates\n\nWrite  \n\n\\[\nx=r+1000m,\\qquad r\\in\\{38,462,538,962\\},\\;m\\ge 0.\n\\tag{13}\n\\]\n\nThen  \n\n\\[\nx^{2}=r^{2}+2000rm+10^{6}m^{2}\n     \\;=\\; 1000\\bigl(1000m^{2}+2rm+\\tfrac{r^{2}-444}{1000}\\bigr)+444,\n\\tag{14}\n\\]\nso deleting the last three digits produces  \n\n\\[\nS_r(m):=1000m^{2}+2rm+c_r,\\quad \nc_r:=\\frac{r^{2}-444}{1000}.\n\\tag{15}\n\\]\n\nFor the four residues one finds  \n\n\\[\nc_{38}=1,\\qquad c_{462}=213,\\qquad c_{538}=289,\\qquad c_{962}=925.\n\\tag{16}\n\\]\n\n--------------------------------------------------------------------\n5.  Which $S_r(m)$ can be a square?  \n\n(a)  $r=38$.  Already $S_{38}(0)=1^{2}$, hence $x=38,m=0$ gives a valid\ndouble–layered square $N=38^{2}=1\\,444$.\n\n(b)  $r=538$.  $S_{538}(0)=17^{2}$, so $x=538$ yields the further example\n$N=538^{2}=289\\,444$.\n\n(c)  $r=462,962$.  \nFor $m=0$ neither $S_{462}(0)=213$ nor $S_{962}(0)=925$ is a square.\nIf $m\\ge 1$ then $x\\ge r+1000\\ge 1\\,462$, whence  \n\n\\[\nx^{2}>1\\,462^{2}=2\\,137\\,444>289\\,444.\n\\]\n\nEven if some $m\\ge 1$ made $S_r(m)$ a square, the resulting\n$N=x^{2}$ would exceed the already obtained value $289\\,444$ and\ntherefore could not be minimal.  Thus the classes $r=462,962$ are irrelevant\nfor part (b).\n\n(The question whether $S_{462}(m)$ or $S_{962}(m)$ is ever a square can be\ndecided negatively by a deeper modular analysis, but this is not needed\nhere.)\n\n--------------------------------------------------------------------\n6.  Maximal thickness and minimal example  \n\nSections 1–2 proved that double–layered squares exist for $k=3$ but for no\nlarger $k$, hence  \n\n\\[\nk_{\\max}=3.\n\\]\n\nAmong those, Section 5 showed that the smallest one is obtained for\n$x=38$, giving  \n\n\\[\nN_{\\min}=38^{2}=1\\,444,\\qquad d=4.\n\\]\n\n--------------------------------------------------------------------\n7.  Final answer  \n\n(a) The largest possible thickness is  \n\n\\[\n\\boxed{k_{\\max}=3}.\n\\]\n\n(b) The least double–layered square of that thickness is  \n\n\\[\n\\boxed{N_{\\min}=1\\,444=38^{2}},\\qquad\\text{with trailing digit }d=4.\n\\]\n\n(For completeness: the next example is $538^{2}=289\\,444$, and both residue\nclasses $r=38$ and $r=538$ generate infinitely many further double–layered\nsquares via the Pell–type equation implicit in (14).)\n\n\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.589263",
        "was_fixed": false,
        "difficulty_analysis": "The original task only asked whether a perfect square can end with a long block of identical non–zero digits.  The enhanced variant introduces two substantial layers of extra difficulty:\n\n1. A second, interlocking square condition — after trimming the trailing block the remaining prefix must again be a perfect square.  This forces one to track two unknown squares simultaneously, not just one.\n\n2. A full 2–adic/5–adic analysis up to any desired precision is required:  \n   • For k≥4 the argument must hold modulo 16 and modulo 5 simultaneously;  \n   • For k=3 existence must be settled by a non-trivial CRT calculation modulo 1000, which itself demands solving a quadratic congruence modulo 125 and lifting it from modulo 5.  \n\nThus the solver must blend quadratic–residue theory, Hensel lifting, and the Chinese Remainder Theorem, far beyond the quick “look-at-the-last-two-digits” observation that is enough for the original problem."
      }
    },
    "original_kernel_variant": {
      "question": "(“Double–layered squares”)\n\nFor a non–zero digit $d\\in\\{1,2,\\dots ,9\\}$ and a positive integer $k$ put  \n\n\\[\nR_k(d)=d\\cdot\\frac{10^{\\,k}-1}{9},\n\\]\n\nso that $R_k(d)$ is the integer whose decimal expansion consists of $k$ consecutive copies of $d$.\n\nA positive integer $N$ is called a \\emph{double–layered square of thickness $k$} if  \n\n(i) $N$ is a perfect square,  \n\n(ii) the last $k$ decimal digits of $N$ equal $R_k(d)$ for some $d\\neq 0$,  \n\n(iii) deleting these $k$ digits once again leaves a perfect square, i.e. $\\bigl\\lfloor N/10^{\\,k}\\bigr\\rfloor$ is a square.\n\n(a) Determine the largest integer $k$ for which a double–layered square exists.  \n\n(b) For this maximal $k$ find the smallest double–layered square and state its trailing digit $d$.\n\n\n\n--------------------------------------------------------------------",
      "solution": "Throughout we write  \n\n\\[\nN=x^{2},\\qquad \n\\Bigl\\lfloor\\frac{N}{10^{\\,k}}\\Bigr\\rfloor = y^{2},\n\\qquad x,y\\in\\mathbb{N}.\n\\tag{1}\n\\]\n\nWith $R_k(d)=d(10^{\\,k}-1)/9$ condition (ii) reads  \n\n\\[\nx^{2}=10^{\\,k}y^{2}+R_k(d).\n\\tag{2}\n\\]\n\nEquivalently  \n\n\\[\nx^{2}\\equiv R_k(d)\\pmod{10^{\\,k}},\\qquad 10^{\\,k}=2^{\\,k}\\cdot 5^{\\,k}.\n\\tag{3}\n\\]\n\nHence\n\n\\[\nx^{2}\\equiv R_k(d)\\pmod{2^{\\,k}},\\qquad \nx^{2}\\equiv R_k(d)\\pmod{5^{\\,k}}.\n\\tag{4}\n\\]\n\n\n--------------------------------------------------------------------\n1.  A joint $2$–adic / $5$–adic obstruction implies $k\\le 3$  \n\nAssume $k\\ge 4$ and put  \n\n\\[\nS:=R_k(d)=d\\cdot\\frac{10^{\\,k}-1}{9}.\n\\]\n\n\\emph{(a) The $2$–adic side.}  \nBecause $k\\ge 4$, $10^{\\,k}\\equiv 0\\pmod{16}$, whence  \n\n\\[\nS\\equiv -d\\cdot 9^{-1}\\equiv -d\\cdot 9\\equiv 7d\\pmod{16}\n\\qquad\\bigl(9^{-1}\\equiv 9\\bigr).\n\\tag{5}\n\\]\n\nQuadratic residues modulo $16$ form the set $\\{0,1,4,9\\}$.  \nThus $7d\\pmod{16}$ must belong to this set; checking $d=1,\\dots ,9$ shows\n\n\\[\n7d\\pmod{16}\\in\\{0,1,4,9\\}\\Longleftrightarrow d=7.\n\\tag{6}\n\\]\n\nHence $d$ would have to be $7$.\n\n\\emph{(b) The $5$–adic side.}  \nSince $10^{\\,k}\\equiv 0\\pmod{5}$ for every $k\\ge 1$,\n\n\\[\nS\\equiv -d\\cdot 9^{-1}\\equiv -d\\cdot(-1)\\equiv d\\pmod{5},\n\\tag{7}\n\\]\nas $9\\equiv -1\\pmod5$.  \nQuadratic residues modulo $5$ are $0,1,4$, so $d\\equiv 0,1,4\\pmod5$.  \nBut $d=7$ found in (6) satisfies $d\\equiv 2\\pmod5$, contradicting (7).\n\nConsequently no double–layered square can exist for $k\\ge 4$, and therefore  \n\n\\[\nk_{\\max}\\le 3.\n\\tag{8}\n\\]\n\n--------------------------------------------------------------------\n2.  The admissible digit when $k=3$\n\nPut $k=3$.  Then $10^{3}=8\\cdot125$ and  \n\n\\[\nR_3(d)=111d.\n\\tag{9}\n\\]\n\nApplying (4):\n\n(i)  Modulo $8$: $111\\equiv 7$.  \nSince the quadratic residues modulo $8$ are $\\{0,1,4\\}$ we need  \n\n\\[\n7d\\equiv 0,1,4\\pmod8\\Longrightarrow d\\in\\{4,7,8\\}.\n\\tag{10}\n\\]\n\n(ii) Modulo $5$: $111d\\equiv d\\pmod5$.  \nBecause a square is congruent to $0,1,4\\pmod5$, we must have $d\\equiv 0,1,4\\pmod5$.  \nCombining with (10) leaves the unique possibility  \n\n\\[\nd=4.\n\\tag{11}\n\\]\n\nThus every double–layered square of thickness $3$ ends in the block $444$.\n\n--------------------------------------------------------------------\n3.  Solving $x^{2}\\equiv 444\\pmod{1000}$\n\nBecause $1000=8\\cdot125$ we work separately.\n\n•  Modulo $8$: $444\\equiv 4$, so $x\\equiv 2\\text{ or }6\\pmod8$.  \n\n•  Modulo $125$: Hensel lifting gives the two solutions  \n\n\\[\nx\\equiv 38 \\text{ or } 87\\pmod{125}.\n\\]\n\nChinese remaindering yields  \n\n\\[\nx\\equiv  38,\\;462,\\;538,\\;962\\pmod{1000}.\n\\tag{12}\n\\]\n\n--------------------------------------------------------------------\n4.  Parametrisation of all candidates\n\nWrite  \n\n\\[\nx=r+1000m,\\qquad r\\in\\{38,462,538,962\\},\\;m\\ge 0.\n\\tag{13}\n\\]\n\nThen  \n\n\\[\nx^{2}=r^{2}+2000rm+10^{6}m^{2}\n     \\;=\\; 1000\\bigl(1000m^{2}+2rm+\\tfrac{r^{2}-444}{1000}\\bigr)+444,\n\\tag{14}\n\\]\nso deleting the last three digits produces  \n\n\\[\nS_r(m):=1000m^{2}+2rm+c_r,\\quad \nc_r:=\\frac{r^{2}-444}{1000}.\n\\tag{15}\n\\]\n\nFor the four residues one finds  \n\n\\[\nc_{38}=1,\\qquad c_{462}=213,\\qquad c_{538}=289,\\qquad c_{962}=925.\n\\tag{16}\n\\]\n\n--------------------------------------------------------------------\n5.  Which $S_r(m)$ can be a square?  \n\n(a)  $r=38$.  Already $S_{38}(0)=1^{2}$, hence $x=38,m=0$ gives a valid\ndouble–layered square $N=38^{2}=1\\,444$.\n\n(b)  $r=538$.  $S_{538}(0)=17^{2}$, so $x=538$ yields the further example\n$N=538^{2}=289\\,444$.\n\n(c)  $r=462,962$.  \nFor $m=0$ neither $S_{462}(0)=213$ nor $S_{962}(0)=925$ is a square.\nIf $m\\ge 1$ then $x\\ge r+1000\\ge 1\\,462$, whence  \n\n\\[\nx^{2}>1\\,462^{2}=2\\,137\\,444>289\\,444.\n\\]\n\nEven if some $m\\ge 1$ made $S_r(m)$ a square, the resulting\n$N=x^{2}$ would exceed the already obtained value $289\\,444$ and\ntherefore could not be minimal.  Thus the classes $r=462,962$ are irrelevant\nfor part (b).\n\n(The question whether $S_{462}(m)$ or $S_{962}(m)$ is ever a square can be\ndecided negatively by a deeper modular analysis, but this is not needed\nhere.)\n\n--------------------------------------------------------------------\n6.  Maximal thickness and minimal example  \n\nSections 1–2 proved that double–layered squares exist for $k=3$ but for no\nlarger $k$, hence  \n\n\\[\nk_{\\max}=3.\n\\]\n\nAmong those, Section 5 showed that the smallest one is obtained for\n$x=38$, giving  \n\n\\[\nN_{\\min}=38^{2}=1\\,444,\\qquad d=4.\n\\]\n\n--------------------------------------------------------------------\n7.  Final answer  \n\n(a) The largest possible thickness is  \n\n\\[\n\\boxed{k_{\\max}=3}.\n\\]\n\n(b) The least double–layered square of that thickness is  \n\n\\[\n\\boxed{N_{\\min}=1\\,444=38^{2}},\\qquad\\text{with trailing digit }d=4.\n\\]\n\n(For completeness: the next example is $538^{2}=289\\,444$, and both residue\nclasses $r=38$ and $r=538$ generate infinitely many further double–layered\nsquares via the Pell–type equation implicit in (14).)\n\n\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.473124",
        "was_fixed": false,
        "difficulty_analysis": "The original task only asked whether a perfect square can end with a long block of identical non–zero digits.  The enhanced variant introduces two substantial layers of extra difficulty:\n\n1. A second, interlocking square condition — after trimming the trailing block the remaining prefix must again be a perfect square.  This forces one to track two unknown squares simultaneously, not just one.\n\n2. A full 2–adic/5–adic analysis up to any desired precision is required:  \n   • For k≥4 the argument must hold modulo 16 and modulo 5 simultaneously;  \n   • For k=3 existence must be settled by a non-trivial CRT calculation modulo 1000, which itself demands solving a quadratic congruence modulo 125 and lifting it from modulo 5.  \n\nThus the solver must blend quadratic–residue theory, Hensel lifting, and the Chinese Remainder Theorem, far beyond the quick “look-at-the-last-two-digits” observation that is enough for the original problem."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}