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NCERT Solutions Ganita Prakash (Part 1) Chapter 1 Patterns and Properties of Perfect Squares — In-text Questions

Book page 5 Updated on2026-09-05

Q1.
Which of the following numbers have the digit 6 in the units place? (i) 38² (ii) 34² (iii) 46² (iv) 56² (v) 74² (vi) 82²
Answer

Square only the units digit of each number and look at the last digit of the result.

NumberUnits digitIts squareUnits digit of the square
(i)388644
(ii)344166 ✓
(iii)466366 ✓
(iv)566366 ✓
(v)744166 ✓
(vi)82244

So (ii) 342, (iii) 462, (iv) 562 and (v) 742 have 6 in the units place.

Why it happens: A square ends in 6 exactly when its root ends in 4 or 6, because 4 × 4 = 16 and 6 × 6 = 36 are the only single-digit products ending in 6.
Check it yourself: 342 = 1156, 462 = 2116, 562 = 3136, 742 = 5476 — all end in 6, while 382 = 1444 and 822 = 6724 end in 4.
Q2.
Find more such patterns by observing the numbers and their squares from the table you filled earlier.
Answer

Here are four more patterns that the table of squares up to 302 makes visible.

  • Squares of numbers ending in 5. 152 = 225, 252 = 625 — the square always ends in 25, and the part before it is the tens digit times the next number: 1×2 = 2, 2×3 = 6.
  • Digit sums. The digit sum of a square, reduced repeatedly, is always 1, 4, 7 or 9. For 576: 5+7+6 = 18 → 9. For 441: 4+4+1 = 9. Never 2, 3, 5, 6 or 8.
  • Squares of numbers just below a round number. 192 = 361 = 400 – 39, 292 = 841 = 900 – 59 — subtract the two "neighbours" added together.
  • Nine-times pattern. Squares that are multiples of 3 are multiples of 9: 9, 36, 81, 144, 225, 324, 441, 576, 729, 900.
Why the "ending in 5" rule works: A number ending in 5 is 10a + 5, and (10a + 5)2 = 100a2 + 100a + 25 = 100 × a(a+1) + 25. So the last two digits are always 25, and the leading part is a × (a + 1). For 25: a = 2, a(a+1) = 6, giving 625.
Q3.
If a number contains 3 zeros at the end, how many zeros will its square have at the end?
Answer

Six zeros.

2000 = 2 × 1000
20002 = 4 × 1000 × 1000 = 4 × 1000000 = 4000000 → 6 zeros
Why it happens: A number ending in three zeros is m × 103, where m does not end in 0. Squaring gives m2 × 106, and m2 cannot end in 0, so exactly 6 zeros are left at the end. Squaring simply doubles the count of terminal zeros.
Q4.
What do you notice about the number of zeros at the end of a number and the number of zeros at the end of its square? Will this always happen? Can we say that squares can only have an even number of zeros at the end?
Answer

The square always has twice as many terminal zeros as the number.

102 = 100 → 1 zero becomes 2
202 = 400 → 1 zero becomes 2
1002 = 10000 → 2 zeros become 4
7002 = 490000 → 2 zeros become 4

Yes, this always happens, and yes, a square can only end in an even number of zeros.

Why it happens: Write the number as m × 10k, where m has no zero at the end. Then its square is m2 × 102k, and m2 also has no zero at the end. So the count of terminal zeros of the square is exactly 2k — always even. This gives a second quick test: a number ending in an odd number of zeros, such as 1000 or 490, cannot be a perfect square.
Q5.
What can you say about the parity of a number and its square?
Answer

A number and its square always have the same parity — even squares to even, odd squares to odd.

even : 42 = 16, 102 = 100, 262 = 676 → all even
odd  : 32 = 9, 72 = 49, 212 = 441 → all odd
Why it happens: An even number is 2k, so its square is 4k2 = 2(2k2) — a multiple of 2, hence even. An odd number is 2k + 1, so its square is 4k2 + 4k + 1 = 2(2k2 + 2k) + 1 — one more than an even number, hence odd.
Tip: Read this backwards too. If a perfect square is odd, its square root must be odd; if it is even, the root is even. That halves your search when you are hunting for a root.
Q6.
Let us explore the differences between consecutive squares. What do you notice? 4 – 1 = 3, 9 – 4 = 5, 16 – 9 = 7, 25 – 16 = 9. See if this pattern continues for the next few square numbers.
Answer

The differences are the consecutive odd numbers, and the pattern does continue.

36 – 25 = 11
49 – 36 = 13
64 – 49 = 15
81 – 64 = 17
100 – 81 = 19
Why it happens: The gap between n2 and (n+1)2 is
(n + 1)2 – n2 = n2 + 2n + 1 – n2 = 2n + 1,
which is exactly the (n+1)-th odd number. Adding these gaps one after another from 1 gives
1 + 3 + 5 + … + (2n – 1) = n2.
1 3 more 5 more → 32 → 22 12
Each new odd number is an L-shaped band that grows a square into the next square.
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