9:{"metadata":[["$","title","0",{"children":"jipmat0024 - If the seven digit number 94 a 29 b 6 is divisible by 72 , then what is the value of (2 a+3 b) for a ≠ b ? | PYQs + Solutions | AfterBoards"}],["$","meta","1",{"name":"description","content":"undefined 1) We have a seven digit number 94a29b6 which is divisib<= by 72 - Since 72 = 8 × 9, the number must be divisib<= by both 8 and 9 2) For divisibility by 8: - Last three digits 9b6 must be divisib<= by 8 - Try values for b from 0-9 where 9b6 is divisib<= by 8 - b = 3 (936 ÷ 8 = 117) - b = 7 (976 ÷ 8 = 122) 3) For divisibility by 9: - Sum of digits must be divisib<= by 9 - 9 + 4 + a + 2 + 9 + b + 6 = 30 + a + b must be divisib<= by 9 4) With b = 3: - 30 + a + 3 = 33 + a must be divisib<= by 9 - When a = 3, sum = 36 ✓ but the question states a and b can't be equal. So: 5) With b = 7: - 30 + a + 7 = 37 + a must be divisib<= by 9 - When a = 8, sum = 45 ✓ 6) We need a ≠ b, so a = 8, b = 7 works. 7) Calculate 2a + 3b: = 2(8) + 3(7) = 16 + 21 = 37"}],["$","link","2",{"rel":"manifest","href":"https://www.afterboards.in/manifest.json","crossOrigin":"$undefined"}],["$","meta","3",{"name":"keywords","content":"JIPMAT,QA,Number System,Divisibility Rules"}],["$","meta","4",{"name":"robots","content":"index, follow"}],["$","meta","5",{"name":"category","content":"education"}],["$","meta","6",{"name":"google-site-verification","content":"google"}],["$","meta","7",{"property":"og:title","content":"jipmat0024 - If the seven digit number 94 a 29 b 6 is divisible by 72 , then what is the value of (2 a+3 b) for a ≠ b ? | PYQs + Solutions"}],["$","meta","8",{"property":"og:description","content":"undefined 1) We have a seven digit number 94a29b6 which is divisib<= by 72 - Since 72 = 8 × 9, the number must be divisib<= by both 8 and 9 2) For divisibility by 8: - Last three digits 9b6 must be divisib<= by 8 - Try values for b from 0-9 where 9b6 is divisib<= by 8 - b = 3 (936 ÷ 8 = 117) - b = 7 (976 ÷ 8 = 122) 3) For divisibility by 9: - Sum of digits must be divisib<= by 9 - 9 + 4 + a + 2 + 9 + b + 6 = 30 + a + b must be divisib<= by 9 4) With b = 3: - 30 + a + 3 = 33 + a must be divisib<= by 9 - When a = 3, sum = 36 ✓ but the question states a and b can't be equal. So: 5) With b = 7: - 30 + a + 7 = 37 + a must be divisib<= by 9 - When a = 8, sum = 45 ✓ 6) We need a ≠ b, so a = 8, b = 7 works. 7) Calculate 2a + 3b: = 2(8) + 3(7) = 16 + 21 = 37"}],["$","meta","9",{"property":"og:url","content":"/jipmat-past-year-questions/2024/QA/11"}],["$","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"}],["$","meta","15",{"name":"twitter:title","content":"jipmat0024 - If the seven digit number 94 a 29 b 6 is divisible by 72 , then what is the value of (2 a+3 b) for a ≠ b ? | PYQs + Solutions"}],["$","meta","16",{"name":"twitter:description","content":"undefined 1) We have a seven digit number 94a29b6 which is divisib<= by 72 - Since 72 = 8 × 9, the number must be divisib<= by both 8 and 9 2) For divisibility by 8: - Last three digits 9b6 must be divisib<= by 8 - Try values for b from 0-9 where 9b6 is divisib<= by 8 - b = 3 (936 ÷ 8 = 117) - b = 7 (976 ÷ 8 = 122) 3) For divisibility by 9: - Sum of digits must be divisib<= by 9 - 9 + 4 + a + 2 + 9 + b + 6 = 30 + a + b must be divisib<= by 9 4) With b = 3: - 30 + a + 3 = 33 + a must be divisib<= by 9 - When a = 3, sum = 36 ✓ but the question states a and b can't be equal. So: 5) With b = 7: - 30 + a + 7 = 37 + a must be divisib<= by 9 - When a = 8, sum = 45 ✓ 6) We need a ≠ b, so a = 8, b = 7 works. 7) Calculate 2a + 3b: = 2(8) + 3(7) = 16 + 21 = 37"}],"$L18","$L19","$L1a","$L1b","$L1c","$L1d","$L1e","$L1f","$L20","$L21"],"error":null,"digest":"$undefined"}

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