JIPMAT 2024 (QA) - 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
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JIPMAT 2024

Number System
>
Divisibility Rules

Medium

If the seven digit number 94a29b694 a 29 b 6 is divisible by 72 , then what is the value of (2a+3b)(2 a+3 b) for a≠ba \neq b ?

Correct Option: 2
1) We have a seven digit number 94a29b694a29b6 which is divisible by 7272\newline - Since 72=8×972 = 8 × 9, the number must be divisible by both 88 and 99
2) For divisibility by 88:\newline - Last three digits 9b69b6 must be divisible by 88
- Try values for bb from 00-99 where 9b69b6 is divisible by 88
- b=3b = 3 (936÷8=117936 ÷ 8 = 117)\newline - b=7b = 7 (976÷8=122976 ÷ 8 = 122)
3) For divisibility by 99:\newline - Sum of digits must be divisible by 99\newline - 9+4+a+2+9+b+6=30+a+b9 + 4 + a + 2 + 9 + b + 6 = 30 + a + b must be divisible by 99
4) With b=3b = 3:\newline - 30+a+3=33+a30 + a + 3 = 33 + a must be divisible by 99\newline - When a=3a = 3, sum =36= 36 ✓
but the question states aa and bb can't be equal. So:
5) With b=7b = 7:\newline - 30+a+7=37+a30 + a + 7 = 37 + a must be divisible by 99\newline - When a=8a = 8, sum =45= 45 ✓
6) We need a≠ba ≠ b, so a=8a = 8, b=7b = 7 works.
7) Calculate 2a+3b2a + 3b:\newline = 2(8)+3(7)2(8) + 3(7)\newline = 16+2116 + 21\newline = 3737

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