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NCERT Solutions for Class 8th Maths Chapter 1 Patterns and Properties of Perfect Squares — In-text Questions

Book page 4 Updated on2026-09-19

Q1.
Find the squares of the first 30 natural numbers and fill in the table below.
1² = 111² = 12121² = 441
2² = 412² =22² =
3² = 913² =
4² = 1614² =
5² = 2515² =
6² =16² =
7² =17² =
8² =18² =
9² =19² =
10² =20² =
Answer

The completed table of the first 30 perfect squares:

12 =1112 =121212 =441
22 =4122 =144222 =484
32 =9132 =169232 =529
42 =16142 =196242 =576
52 =25152 =225252 =625
62 =36162 =256262 =676
72 =49172 =289272 =729
82 =64182 =324282 =784
92 =81192 =361292 =841
102 =100202 =400302 =900
Tip: You do not need to multiply each time. Since (n+1)2 = n2 + (2n + 1), each square is the previous one plus the next odd number: 400 + 41 = 441, 441 + 43 = 484, 484 + 45 = 529, and so on.
Q2.
What patterns do you notice? Share your observations and make conjectures.
Answer

Several patterns show up at once in that table.

  • Units digits. Every square ends in 0, 1, 4, 5, 6 or 9. None ends in 2, 3, 7 or 8.
  • Mirror pairs. Numbers whose units digits add to 10 have squares with the same units digit: 1 and 9 both give 1, 2 and 8 both give 4, 3 and 7 both give 9, 4 and 6 both give 6.
  • Differences. Consecutive squares differ by consecutive odd numbers: 4 – 1 = 3, 9 – 4 = 5, 16 – 9 = 7, …
  • Parity. The square of an even number is even, the square of an odd number is odd.
  • Zeros. 102 = 100, 202 = 400, 302 = 900 — one zero at the end of the number becomes two at the end of the square.
Why the units digit works: The last digit of a product depends only on the last digits of the factors. So the last digit of n2 depends only on the last digit of n, and there are only ten cases to check: 0→0, 1→1, 2→4, 3→9, 4→6, 5→5, 6→6, 7→9, 8→4, 9→1. The set of outcomes is exactly {0, 1, 4, 5, 6, 9}.
Q3.
If a number ends in 0, 1, 4, 5, 6 or 9, is it always a square?
Answer

No. The units digit is a one-way test.

16 = 42 ✓ and 36 = 62 ✓, but 26 ends in 6 and is not a square
Also 20, 21, 24, 30, 31 … all end in an allowed digit but are not squares
Why it happens: Ending in 0, 1, 4, 5, 6 or 9 is a necessary condition, not a sufficient one. A square must pass it, but passing it does not make a number a square. The useful direction is the contrapositive: if a number ends in 2, 3, 7 or 8, it is certainly not a square.
Tip: This is the fastest first filter you have. 1027 ends in 7 → not a square, and you never need to factorise it.
Q4.
Write 5 numbers such that you can determine by looking at their units digit that they are not squares.
Answer

Choose any numbers ending in 2, 3, 7 or 8. For example:

42, 153, 267, 638, 1082
NumberUnits digitConclusion
422not a square
1533not a square
2677not a square
6388not a square
10822not a square

No square of any natural number can end in these four digits, so a single glance settles all five.

Q5.
The squares, 1², 9², 11², 19², 21², and 29², all have 1 in their units place. Write the next two squares. Notice that if a number has 1 or 9 in the units place, then its square ends in 1.
Answer

The list 1, 9, 11, 19, 21, 29 continues with the next numbers ending in 1 or 9 — that is 31 and 39.

312 = 961
392 = 1521

Both end in 1, as expected.

Why it happens: Only the units digit of the number decides the units digit of the square. 1 × 1 = 1 ends in 1, and 9 × 9 = 81 also ends in 1. No other digit gives 1: the ten possibilities are 0→0, 1→1, 2→4, 3→9, 4→6, 5→5, 6→6, 7→9, 8→4, 9→1. So a square ends in 1 exactly when its root ends in 1 or 9.
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