JIPMAT 2023 (QA) - If the eight-digit number 5a32465b is divisible by 88, then 2a+ 5b = | PYQs + Solutions | AfterBoards
Skip to main contentSkip to question navigationSkip to solution
IPMAT Indore Free Mocks Topic Tests

JIPMAT 2023 (QA) PYQs

JIPMAT 2023

Number System
>
Divisibility Rules

Conceptual

If the eight-digit number 5a32465b5a32465b is divisible by 8888, then 2a+5b=2a+ 5b =

Correct Option: 1
Let me solve this carefully:
For divisibility by 8888 (=8×11)(= 8 × 11): \newline Number must be divisible by both 88 and 1111
1)1) For divisibility by 88: \newline - Last three digits (65b)(65b) must be divisible by 88 \newline - Testing values: b=6b = 6 makes 656656 divisible by 88
2)2) For divisibility by 1111: \newline - Difference of sums of alternate digits must be divisible by 1111 \newline - (5+3+4+5)(a+2+6+6)=1714a(5+3+4+5)-(a+2+6+6)=17-14-a \newline - (3a)(3 - a) must be divisible by 1111 \newline - Therefore a=3a = 3
So the number is 5332465b5332465b where b=6b = 6
2a+5b=2(3)+5(6)2a + 5b = 2(3) + 5(6)
=6+30= 6 + 30
=36= 36

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question