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Accenture exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer

Accenture exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer

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إظهار المزيد

📈 نظرة تحليلية على قناة تيليجرام Accenture exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer

تُعد قناة Accenture exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer (@coding_are) في القطاع اللغوي الإنكليزية لاعباً نشطاً. يضم المجتمع حالياً 13 221 مشتركاً، محتلاً المرتبة 15 316 في فئة التعليم والمرتبة 31 783 في منطقة الهند.

📊 مؤشرات الجمهور والحراك

منذ تأسيسه في невідомо، حقق المشروع نمواً سريعاً وجمع 13 221 مشتركاً.

بحسب آخر البيانات بتاريخ 23 يونيو, 2026، تحافظ القناة على نشاط مستقر. خلال آخر 30 يوماً تغيّر عدد الأعضاء بمقدار -162، وفي آخر 24 ساعة بمقدار 1، مع بقاء الوصول العام مرتفعاً.

  • حالة التحقق: غير موثّقة
  • معدل التفاعل (ER): يبلغ متوسط تفاعل الجمهور 2.68‎%. وخلال أول 24 ساعة من النشر يحصد المحتوى عادةً 0.93‎% من ردود الفعل نسبةً إلى إجمالي المشتركين.
  • وصول المنشورات: يحصل كل منشور على متوسط 354 مشاهدة. وخلال اليوم الأول يجمع عادةً 123 مشاهدة.
  • التفاعلات والاستجابة: يتفاعل الجمهور بانتظام؛ متوسط التفاعلات لكل منشور يبلغ 1.
  • الاهتمامات الموضوعية: يركز المحتوى على مواضيع رئيسية مثل placement, gaurntee, suree, capgemini, infosy.

📝 الوصف وسياسة المحتوى

يصف المؤلف القناة بأنها مساحة للتعبير عن الآراء الذاتية:
🔥Guys plz Stop fearing for daily exams 📝 👨‍💻 @srksvk is here to help you all at lowest cost possible.💪 🌀 ” Our Only Aim Is To Let Get Placed To You In A Reputed Company 🔥Effort from our side = 💯 📱Main Channel: @coding_are 📱Tel I'd : @srks...

بفضل وتيرة التحديث المرتفعة (أحدث البيانات بتاريخ 24 يونيو, 2026) تحافظ القناة على حداثتها ومستوى وصول مرتفع. وتُظهر التحليلات تفاعلاً نشطاً من الجمهور، ما يجعلها نقطة تأثير مهمة ضمن فئة التعليم.

13 221
المشتركون
+124 ساعات
-317 أيام
-16230 أيام
أرشيف المشاركات
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def is_modest(number): str_num = str(number) length = len(str_num) for i in range(1, length): left_part = int(str_num[:i]) right_part = int(str_num[i:]) if right_part != 0 and number % right_part == left_part: return True return False def find_modest_numbers(M, N): modest_numbers = [] for number in range(M, N + 1): if is_modest(number): modest_numbers.append(number) return modest_numbers M, N = map(int, input().split()) modest_numbers = find_modest_numbers(M, N) if modest_numbers: print(" ".join(map(str, modest_numbers))) else: print("No modest numbers found within the range")

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