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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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Ko'proq ko'rsatish

📈 Telegram kanali Accenture exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer analitikasi

Accenture exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer (@coding_are) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 13 221 obunachidan iborat bo'lib, Taʼlim toifasida 15 316-o'rinni va Hindiston mintaqasida 31 783-o'rinni egallagan.

📊 Auditoriya ko‘rsatkichlari va dinamika

невідомо sanasidan buyon loyiha tez o‘sib, 13 221 obunachiga ega bo‘ldi.

23 Iyun, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -162 ga, so‘nggi 24 soatda esa 1 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.

  • Tasdiqlash holati: Tasdiqlanmagan
  • Jalb etish (ER): Auditoriya o‘rtacha 2.68% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 0.93% ini tashkil etuvchi reaksiyalarni to‘playdi.
  • Post qamrovi: Har bir post o‘rtacha 354 marta ko‘riladi; birinchi sutkada odatda 123 ta ko‘rish yig‘iladi.
  • Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 1 ta reaksiya keladi.
  • Tematik yo‘nalishlar: Kontent placement, gaurntee, suree, capgemini, infosy kabi asosiy mavzularga jamlangan.

📝 Tavsif va kontent siyosati

Muallif resursni shaxsiy fikrni ifoda etish maydoni sifatida ta’riflaydi:
🔥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...

Yuqori yangilanish chastotasi (oxirgi ma’lumot 24 Iyun, 2026 da olingan) sababli kanal doimo dolzarb va katta qamrovli bo‘lib qoladi. Analitika auditoriya kontent bilan faol hamkorlik qilishini, uni Taʼlim toifasidagi muhim ta’sir nuqtasiga aylantirishini ko‘rsatadi.

13 221
Obunachilar
+124 soatlar
-317 kunlar
-16230 kunlar
Postlar arxiv
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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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