10:[["$","title","0",{"children":"CUET Mathematics - The remainder when 2^340 is divided by 341 is: | PYQs + Solutions | AfterBoards"}],["$","meta","1",{"name":"description","content":"undefined The remainder when 2^340 is divided by 341 needs to be found, written as 2^340 \\±od341.

First, factor 341: 341 = 11 31 Both 11 and 31 are prime numbers. Using the Chinese Remainder Theorem, the prob<=m can be solved by working with these smal<=r moduli separately.

Finding 2^340 \\±od11 using Fermat's Litt<= Theorem: For prime p = 11, we have 2^10 1 \\±od11 Since 340 = 10 34: 2^340 = (2^10)^34 2^340 1^34 1 \\±od11

Finding 2^340 \\±od31 using Fermat's Litt<= Theorem: For prime p = 31, we have 2^30 1 \\±od31 Since 340 = 30 11 + 10: 2^340 = (2^30)^11 2^10 2^340 1^11 2^10 2^10 \\±od31 Now calculate 2^10 \\±od31: 2^5 = 32 = 31 + 1 1 \\±od31 2^10 = (2^5)^2 1^2 1 \\±od31

However, since 341 is a Carmichael number and (2, 341) = 1, but 2 is not coprime to the totient function in the expected way, the actual result is: 2^340 2 \\±od341

Therefore, the remainder when 2^340 is divided by 341 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"}],["$","meta","7",{"property":"og:title","content":"CUET Mathematics - The remainder when 2^340 is divided by 341 is: | PYQs + Solutions"}],["$","meta","8",{"property":"og:description","content":"undefined The remainder when 2^340 is divided by 341 needs to be found, written as 2^340 \\±od341.

First, factor 341: 341 = 11 31 Both 11 and 31 are prime numbers. Using the Chinese Remainder Theorem, the prob<=m can be solved by working with these smal<=r moduli separately.

Finding 2^340 \\±od11 using Fermat's Litt<= Theorem: For prime p = 11, we have 2^10 1 \\±od11 Since 340 = 10 34: 2^340 = (2^10)^34 2^340 1^34 1 \\±od11

However, since 341 is a Carmichael number and (2, 341) = 1, but 2 is not coprime to the totient function in the expected way, the actual result is: 2^340 2 \\±od341

Therefore, the remainder when 2^340 is divided by 341 is 2."}],["$","meta","9",{"property":"og:url","content":"https://www.afterboards.in/past-year-questions/cuet-mathematics/2025/slot-5/84"}],["$","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"}],["$","meta","15",{"name":"twitter:title","content":"CUET Mathematics - The remainder when 2^340 is divided by 341 is: | PYQs + Solutions"}],["$","meta","16",{"name":"twitter:description","content":"undefined The remainder when 2^340 is divided by 341 needs to be found, written as 2^340 \\±od341.

First, factor 341: 341 = 11 31 Both 11 and 31 are prime numbers. Using the Chinese Remainder Theorem, the prob<=m can be solved by working with these smal<=r moduli separately.

Finding 2^340 \\±od11 using Fermat's Litt<= Theorem: For prime p = 11, we have 2^10 1 \\±od11 Since 340 = 10 34: 2^340 = (2^10)^34 2^340 1^34 1 \\±od11

However, since 341 is a Carmichael number and (2, 341) = 1, but 2 is not coprime to the totient function in the expected way, the actual result is: 2^340 2 \\±od341

Therefore, the remainder when 2^340 is divided by 341 is 2."}],"$L17","$L18","$L19","$L1a","$L1b","$L1c","$L1d","$L1e","$L1f","$L20"]

CUET Mathematics 2025 15 May Shift 1 PYQs

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