Innings2
Create New Account

Sign in to Innings2

or

New Account     Reset Password     

Chapter 15: Playing with Numbers

    Introduction

    Divisibility Rules

    Some More Divisibility Rules

    Puzzles Based on Divisibility Rules

    Fun with 3-Digit Numbers

    Puzzles with Missing Digits

    Finding of Divisibility by taking Remainders of Place Values

    Some More Puzzles on Divisibility Rules

    Finding Sum of Consecutive Numbers

    Exercise 15.1

    Exercise 15.2

    Exercise 15.3

    Exercise 15.4

    Exercise 15.5

    Exercise 15.6

    What We Have Discussed?

    Easy Level Worksheet

    Moderate Level Worksheet

    Hard Level Worksheet

Powered by Innings 2

Reset Progress

This will delete your progress and chat data for all chapters in this course, and cannot be undone!

Glossary

Select one of the keywords on the left…

Chapter 15: Playing with Numbers > Fun with 3-Digit Numbers

Fun with 3-Digit Numbers

Here's another interesting number game:

  1. Choose any 3-digit number (keep it to yourself)
  2. Reverse all the digits to get a new number
  3. Subtract the smaller number from the larger number
  4. Divide your result by 13

I predict your answer will always be divisible by 13 with no remainder!

This is similar to the original puzzle:

Let's represent a 3-digit number as (100a + 10b + c), where: 'a' is the digit (1-9), 'b' is the digit (0-9) and 'c' is the digit (0-9).

When reversed, it becomes (100c + 10b + a)

Now there are two cases:

Case 1: If a > c:

Original - Reversed = (100a + 10b + c) - (100c + 10b + a) = 100a - 100c + c - a = a - c = (a - c) = × × (a - c)

Case 2: If c > a:

Reversed - Original = (100c + 10b + a) - (100a + 10b + c) = 100c - 100a + a - c = c - a = (c - a) = × × (c - a)

Verification: If someone picks 157:

Original:

Reversed:

Difference: - =

594 ÷ 9 =

594 ÷ 11 =

Both divide evenly!