Exercise 6.1 — Square Properties
Problems based on properties of square numbers.
Exercise 6.1 – Units Digit Patterns, Perfect Squares & Odd Number Sums
Exercise 6.1 of Class 8 Mathematics (CBSE, Telangana & Andhra Pradesh syllabus) puts the patterns from the chapter introduction into practice. Every question here is solved without performing the full multiplication — instead, you apply pattern-recognition rules about units digits, even/odd behaviour, and the relationships between consecutive squares and odd-number sums.
This exercise has six parts covering: finding the units digit of a square, identifying perfect squares, explaining why a number isn't a perfect square, determining whether a square is even or odd, counting integers between consecutive squares, and finding sums of consecutive odd numbers using the n² shortcut.
To find the units digit of a large number's square, you only need to look at the units digit of the original number and square that single digit. The units digit of the result becomes your answer — the rest of the number doesn't matter at all.
Units digit of n² depends only on the units digit of n
| Number | Units Digit of Number | Square of That Digit | Units Digit of n² |
|---|---|---|---|
| (i) 39 | 9 | 9² = 81 | 1 |
| (ii) 297 | 7 | 7² = 49 | 9 |
| (iii) 5125 | 5 | 5² = 25 | 5 |
| (iv) 7286 | 6 | 6² = 36 | 6 |
| (v) 8742 | 2 | 2² = 4 | 4 |
For each number, we first check the units digit. If it's 2, 3, 7, or 8, the number is immediately ruled out as a perfect square. If the units digit passes this test (0, 1, 4, 5, 6, or 9), we then check against the squares table from the chapter introduction (1² to 30²) to confirm.
| Number | Units Digit | Verdict | Reason |
|---|---|---|---|
| (i) 121 | 1 | Perfect square | 121 = 11² |
| (ii) 136 | 6 | Not a perfect square | Passes units-digit test, but 136 doesn't appear in the squares table (11² = 121, 12² = 144) |
| (iii) 256 | 6 | Perfect square | 256 = 16² |
| (iv) 321 | 1 | Not a perfect square | Passes units-digit test, but lies between 17² = 289 and 18² = 324 |
| (v) 600 | 0 | Not a perfect square | Ends in only one zero — squares of multiples of 10 must end in two zeroes |
This question asks for the reasoning, not just a yes/no answer. Every explanation here uses one of two rules: either the units digit doesn't belong to the set {0, 1, 4, 5, 6, 9}, or the number ends in an odd number of zeroes.
| Number | Units Digit / Ending | Reason It's Not a Perfect Square |
|---|---|---|
| (i) 257 | 7 | Units digit 7 — square numbers never end in 7 |
| (ii) 4592 | 2 | Units digit 2 — square numbers never end in 2 |
| (iii) 2433 | 3 | Units digit 3 — square numbers never end in 3 |
| (iv) 5050 | ends with one zero | Squares of multiples of 10 must end with two zeroes, not one |
| (v) 6098 | 8 | Units digit 8 — square numbers never end in 8 |
This is one of the simplest rules in the chapter: the square of an even number is always even, and the square of an odd number is always odd. You don't need to calculate the square at all — just check whether the original number is even or odd.
| Number | Even / Odd | Square Is |
|---|---|---|
| (i) 431 | Odd | Odd |
| (ii) 2826 | Even | Even |
| (iii) 8204 | Even | Even |
| (iv) 17779 | Odd | Odd |
| (v) 99998 | Even | Even |
Even number → Even square | Odd number → Odd square
This question applies the formula from the chapter introduction directly: the number of integers lying between n² and (n+1)² is always 2n. Simply take the smaller number, double it, and that's your answer — no need to calculate either square.
Number of integers between n² and (n + 1)² = 2n
| Between Squares Of | Calculation (2n) | Number of Integers |
|---|---|---|
| (i) 25 and 26 | 2 × 25 | 50 |
| (ii) 56 and 57 | 2 × 56 | 112 |
| (iii) 107 and 108 | 2 × 107 | 214 |
Instead of adding a long list of odd numbers one by one, simply count how many odd numbers are in the list (that's your value of n), and the sum is n². Since odd numbers always start from 1 in this pattern, the count directly gives the base number to be squared.
Sum of first n odd natural numbers = n²
| Sum | Count of Terms (n) | Result |
|---|---|---|
| (i) 1 + 3 + 5 + 7 + 9 | 5 | 5² = 25 |
| (ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 | 9 | 9² = 81 |
| (iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 | 13 | 13² = 169 |
Common Mistakes to Avoid in Exercise 6.1
- Squaring the entire number instead of just the units digit: For Q1, you only need to square the last digit of the number — squaring the whole number wastes time and increases error risk.
- Treating the units-digit rule as a guarantee: A number ending in 0, 1, 4, 5, 6, or 9 is not automatically a perfect square (see 136 and 321 in Q2) — it only passes the first filter. Always verify against known squares too.
- Confusing "ends in zero" with "ends in two zeroes": For Q3(iv), 5050 ends in just one zero. Perfect squares of multiples of 10 must end in exactly two zeroes (like 100, 400, 900).
- Miscounting terms in odd-number sums: For Q6, carefully count how many terms are in the sequence — an off-by-one error here directly changes the value of n and therefore the final answer.
- Forgetting the "strictly between" condition: In Q5, the 2n formula counts integers strictly between the two squares — don't include the squares themselves.
What Exercise 6.1 Prepares You For
The pattern-recognition skills built in this exercise — especially the units digit rule and perfect square identification — are essential for the next part of this chapter, where you'll learn to find the square root of a number using prime factorisation and the long division method. Being able to quickly judge whether a number is a perfect square saves significant time when checking your final answers.
These ideas also build directly on the foundational concepts covered in the Introduction to Square Roots and Cube Roots, and they connect with Exponents and Powers, where the same n² notation is explored with a wider range of exponents and bases.