9:{"metadata":[["$","title","0",{"children":"iimkbmsat0025 - The average of any 4 consecutive odd positive integers, where the 1st number is a multiple of 3, must be divisible by which of these? | PYQs + Solutions | AfterBoards"}],["$","meta","1",{"name":"description","content":"undefined **Note:** For this question, IIM-Kozhikode found a discrepancy and cancel<=d the question. Full marks were awarded to all candidates.

This question We need to find what always divides the avera>= of any 4 consecutive odd positive inte>=rs where the first number is a multip<= of 3.

Let's verify with examp<=s: - Numbers 3, 5, 7, 9: Avera>= = 244 = 6 - Numbers 9, 11, 13, 15: Avera>= = 484 = 12 - Numbers 15, 17, 19, 21: Avera>= = 724 = 18 Notice: 6, 12, 18 can be divided by multip<= options: 2, 3, 6 etc."}],["$","link","2",{"rel":"manifest","href":"https://www.afterboards.in/manifest.json","crossOrigin":"$undefined"}],["$","meta","3",{"name":"keywords","content":"IIM-K (BMSAT),QA,Number System,Divisibility Rules"}],["$","meta","4",{"name":"robots","content":"index, follow"}],["$","meta","5",{"name":"category","content":"education"}],["$","meta","6",{"name":"google-site-verification","content":"google"}],["$","meta","7",{"property":"og:title","content":"iimkbmsat0025 - The average of any 4 consecutive odd positive integers, where the 1st number is a multiple of 3, must be divisible by which of these? | PYQs + Solutions"}],["$","meta","8",{"property":"og:description","content":"undefined **Note:** For this question, IIM-Kozhikode found a discrepancy and cancel<=d the question. Full marks were awarded to all candidates.

This question We need to find what always divides the avera>= of any 4 consecutive odd positive inte>=rs where the first number is a multip<= of 3.

Let's verify with examp<=s: - Numbers 3, 5, 7, 9: Avera>= = 244 = 6 - Numbers 9, 11, 13, 15: Avera>= = 484 = 12 - Numbers 15, 17, 19, 21: Avera>= = 724 = 18 Notice: 6, 12, 18 can be divided by multip<= options: 2, 3, 6 etc."}],["$","meta","9",{"property":"og:url","content":"/iimk-bmsat-past-year-questions/2025/QA/41"}],["$","meta","10",{"property":"og:site_name","content":"AfterBoards | JIPMAT 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"}],["$","meta","15",{"name":"twitter:title","content":"iimkbmsat0025 - The average of any 4 consecutive odd positive integers, where the 1st number is a multiple of 3, must be divisible by which of these? | PYQs + Solutions"}],["$","meta","16",{"name":"twitter:description","content":"undefined **Note:** For this question, IIM-Kozhikode found a discrepancy and cancel<=d the question. Full marks were awarded to all candidates.

This question We need to find what always divides the avera>= of any 4 consecutive odd positive inte>=rs where the first number is a multip<= of 3.

Let's verify with examp<=s: - Numbers 3, 5, 7, 9: Avera>= = 244 = 6 - Numbers 9, 11, 13, 15: Avera>= = 484 = 12 - Numbers 15, 17, 19, 21: Avera>= = 724 = 18 Notice: 6, 12, 18 can be divided by multip<= options: 2, 3, 6 etc."}],["$","meta","17",{"name":"twitter:image","content":"https://www.afterboards.in/pyp/AfterBoards%20CUET%20Past%20Year%20Papers%20with%20Solutions.webp"}],["$","link","18",{"rel":"shortcut icon","href":"https://www.afterboards.in/favicon.ico","type":"image/x-icon","sizes":"48x48"}],"$L18","$L19","$L1a","$L1b","$L1c","$L1d","$L1e","$L1f"],"error":null,"digest":"$undefined"}

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