Playing with Numbers
ICSE Class 8 · Mathematics · Chapter 5
What this chapter covers
- Numbers in generalised form
- Reversing digits: patterns that always work
- Letters for digits
- Making and solving number puzzles
- Divisibility tests for 10, 5 and 2
- Divisibility tests for 9 and 3
Key points
- The generalised form of a two-digit number is 10a + b, and of a three-digit number is 100a + 10b + c, where the first digit is from 1 to 9 and the others from 0 to 9.
- A two-digit number plus its reverse equals 11(a + b); the larger minus the smaller equals 9(a - b).
- A three-digit number minus its reverse equals 99(a - c); the middle digit cancels.
- In a letters-for-digits puzzle, each letter is one digit, different letters are different digits, and a leading digit is never 0.
- Solve letter puzzles column by column from the units, tracking carries and borrows, and check the answer at the end.
- A good puzzle is built backwards from a true calculation and checked for a single answer.
- A number is divisible by 10 if its units digit is 0, by 5 if it is 0 or 5, and by 2 if it is even.
- A number is divisible by 9 if its digit sum is divisible by 9, and by 3 if its digit sum is divisible by 3.
- These tests work because 100a + 10b + c = 10(10a + b) + c and 100a + 10b + c = 9(11a + b) + (a + b + c).
Questions answered
Write 4, 7 and 2 as the hundreds, tens and units digits of a number, and give its generalised form.
The number is 472, and 472 = 4 × 100 + 7 × 10 + 2.
Why can the tens digit a in 10a + b never be 0?
If a were 0, the number would be just b, a one-digit number, not a two-digit one.
Without adding, find 56 + 65.
It is 11 × (5 + 6) = 11 × 11 = 121.
Explain why the difference between a three-digit number and its reverse is always divisible by 99.
(100a + 10b + c) - (100c + 10b + a) = 99a - 99c = 99(a - c), which is a multiple of 99.
Find A and B: overline4A + overlineB6 = 93.
Units: A + 6 ends in 3, so A + 6 = 13, A = 7, carry 1. Tens: 4 + B + 1 = 9, so B = 4. Check: 47 + 46 = 93. Here A and B are different letters, and A = 7, B = 4 are different digits.
Find A: overlineA1 - overline1A = 45.
Using the reverse rule, 9(A - 1) = 45, so A - 1 = 5 and A = 6. Check: 61 - 16 = 45.
Is 2,580 divisible by 2, by 5 and by 10? Give reasons.
Yes to all three. Its units digit is 0, which is even, is 0 or 5, and is 0.
Without dividing, state whether 627 is divisible by 3 and by 9.
The digit sum is 6 + 2 + 7 = 15. This is divisible by 3 but not by 9, so 627 is divisible by 3 but not by 9.