10:[["$","title","0",{"children":"CUET Mathematics - The least non-negative remainder when 3^128 is divided by 7 is: | PYQs + Solutions | AfterBoards"}],["$","meta","1",{"name":"description","content":"undefined To find the remainder when 3^128 is divided by 7, calculate the first few powers of 3 modulo 7: 3^1 = 3 → Remainder = 3 3^2 = 9 → Remainder = 2 3^3 = 27 → Remainder = 6 3^4 = 81 → Remainder = 4 3^5 = 243 → Remainder = 5 3^6 = 729 → Remainder = 1 When 3^6 gives remainder 1, the pattern repeats every 6 powers.

To determine where 128 falls in this cyc<=, divide 128 by 6: 128 6 = 21 remainder 2 Therefore: 128 = 6 21 + 2

Express 3^128 using this decomposition: 3^128 = 3^6 21 + 2 = 3^6 21 3^2 = (3^6)^21 3^2

Since 3^6 gives remainder 1 when divided by 7: (3^6)^21 gives remainder 1^21 = 1 3^2 = 9 gives remainder 2 Therefore, the remainder of 3^128 when divided by 7: 1 2 = 2 The <=ast non-negative remainder when 3^128 is divided by 7 is 2."}],["$","link","2",{"rel":"manifest","href":"https://www.afterboards.in/manifest.json","crossOrigin":"$undefined"}],["$","meta","3",{"name":"keywords","content":"CUETMAT,MAT,Arithmetic,Numericals"}],["$","meta","4",{"name":"robots","content":"index, follow"}],["$","meta","5",{"name":"category","content":"education"}],["$","meta","6",{"name":"google-site-verification","content":"google"}],["$","meta","7",{"property":"og:title","content":"CUET Mathematics - The least non-negative remainder when 3^128 is divided by 7 is: | PYQs + Solutions"}],["$","meta","8",{"property":"og:description","content":"undefined To find the remainder when 3^128 is divided by 7, calculate the first few powers of 3 modulo 7: 3^1 = 3 → Remainder = 3 3^2 = 9 → Remainder = 2 3^3 = 27 → Remainder = 6 3^4 = 81 → Remainder = 4 3^5 = 243 → Remainder = 5 3^6 = 729 → Remainder = 1 When 3^6 gives remainder 1, the pattern repeats every 6 powers.

To determine where 128 falls in this cyc<=, divide 128 by 6: 128 6 = 21 remainder 2 Therefore: 128 = 6 21 + 2

Express 3^128 using this decomposition: 3^128 = 3^6 21 + 2 = 3^6 21 3^2 = (3^6)^21 3^2

Since 3^6 gives remainder 1 when divided by 7: (3^6)^21 gives remainder 1^21 = 1 3^2 = 9 gives remainder 2 Therefore, the remainder of 3^128 when divided by 7: 1 2 = 2 The <=ast non-negative remainder when 3^128 is divided by 7 is 2."}],["$","meta","9",{"property":"og:url","content":"https://www.afterboards.in/past-year-questions/cuet-mathematics/2025/slot-3/63"}],["$","meta","10",{"property":"og:site_name","content":"AfterBoards | CUET Mathematics Preparation"}],["$","meta","11",{"property":"og:image","content":"https://www.afterboards.in/pyp/AfterBoards%20CUET%20Past%20Year%20Papers%20with%20Solutions.webp"}],["$","meta","12",{"property":"og:type","content":"website"}],["$","meta","13",{"name":"twitter:card","content":"summary_large_image"}],["$","meta","14",{"name":"twitter:site","content":"afterboards.in"}],["$","meta","15",{"name":"twitter:title","content":"CUET Mathematics - The least non-negative remainder when 3^128 is divided by 7 is: | PYQs + Solutions"}],["$","meta","16",{"name":"twitter:description","content":"undefined To find the remainder when 3^128 is divided by 7, calculate the first few powers of 3 modulo 7: 3^1 = 3 → Remainder = 3 3^2 = 9 → Remainder = 2 3^3 = 27 → Remainder = 6 3^4 = 81 → Remainder = 4 3^5 = 243 → Remainder = 5 3^6 = 729 → Remainder = 1 When 3^6 gives remainder 1, the pattern repeats every 6 powers.

To determine where 128 falls in this cyc<=, divide 128 by 6: 128 6 = 21 remainder 2 Therefore: 128 = 6 21 + 2

Express 3^128 using this decomposition: 3^128 = 3^6 21 + 2 = 3^6 21 3^2 = (3^6)^21 3^2

CUET Mathematics 2025 14 May Shift 1 PYQs

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