NCERT Solutions for Class 8 Maths Chapter 1 A Square and a Cube (PDF) – 2027

He counted factors. Person k toggles locker n only when k is a factor of n, so each locker is toggled once for every factor of its number.

A locker ends open only after an odd number of toggles. Factors come in partner pairs (1 × 6, 2 × 3), so the count is even — unless one pair has equal partners, as in 3 × 3. That happens only for square numbers. So the lockers left open are

No. Most numbers do, but the squares do not.

So: every number except a perfect square has an even number of factors.

Yes — take any number times itself.

Their factor counts are 1, 3, 3, 5, 3, … — all odd.

Every one of these is a square, and no number that is not a square appears on such a list.

Yes, 36 has 9 factors — an odd number. Here is the check, pair by pair.

Four pairs of distinct factors give 4 × 2 = 8 factors, and the lone 6 adds one more:

The lockers with square numbers between 1 and 100:

That is 10 lockers, namely 1², 2², 3², …, 10². The next square, 11² = 121, is beyond 100.

The passcode lockers are the ones touched exactly twice, so their numbers have exactly two factors — the prime numbers.

The first five such lockers are 2, 3, 5, 7 and 11, so the code is 2-3-5-7-11.

Yes. A side length need not be a whole number; squaring works for fractions and decimals in exactly the same way.

The completed table of the first 30 perfect squares:

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.
• Same last digit: numbers ending in 1 or 9 have squares ending in 1; 2 or 8 → 4; 3 or 7 → 9; 4 or 6 → 6; 5 → 5; 0 → 0.
• Differences: consecutive squares differ by consecutive odd numbers: 3, 5, 7, 9, …
• Parity: even numbers have even squares; odd numbers have odd squares.
• Zeros: 10² = 100, 20² = 400 — one zero at the end of the number gives two zeros at the end of the square.

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

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

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

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

Both end in 1, as expected.

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

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

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

• Numbers ending in 5: 15² = 225, 25² = 625 — the square ends in 25, and in front of it is (tens digit) × (tens digit + 1): 1 × 2 = 2, 2 × 3 = 6.
• Digit sums: add the digits of a square again and again until one digit is left; you always get 1, 4, 7 or 9 (576 → 18 → 9; 289 → 19 → 10 → 1).
• Just below a round number: 19² = 400 – 39 = 361 and 29² = 900 – 59 = 841 (subtract 19 + 20, or 29 + 30).
• Multiples of 3: a square that is a multiple of 3 is also a multiple of 9: 9, 36, 81, 144, 225, …

Six zeros.

Squaring doubles the number of zeros at the end: 3 zeros → 6 zeros.

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

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

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

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

1225 is the sum of the first 35 odd numbers, so 36² is that sum plus the 36th odd number.

List the odd numbers with their positions and the rule appears at once.

Each odd number is one less than twice its position: 2 × 6 – 1 = 11. So

The n-th odd number is 2n – 1.

Between n² and (n+1)² there are exactly 2n numbers.

Blockwise count of the perfect squares up to 1000:

Total = 10 + 4 + 3 + 3 + 2 + 2 + 2 + 2 + 2 + 1 = 31 squares below 1000, and the largest is

Yes — two consecutive triangular numbers always add to a square.

So the next picture is a 5 × 5 square of dots split by a staircase into a triangle of 10 dots and a triangle of 15 dots.

The side is the number that gives 49 when multiplied by itself.

We call 7 the square root of 49, and write √49 = 7.

8 × 8 = 64, and (–8) × (–8) = 64 as well.

Use the tests in order — cheapest first.

1. Units digit: 327 ends in 7, so it is not a perfect square. 576 ends in 6, so it may be one — we must check further.
2. List the squares: 20² = 400, 21² = 441, 22² = 484, 23² = 529, 24² = 576 (slow for large numbers).
3. Subtract odd numbers 1, 3, 5, … in turn; reaching exactly 0 at step k means the number is k².
4. Prime factorisation: split the prime factors into two identical groups.

Yes, but it is time-consuming — it takes 27 subtractions.

Yes. If the prime factors can be divided into two equal groups, the product of one group is the square root.

Yes, 324 = 18².

No.

Therefore 156 is not a perfect square.

1156 is a perfect square; 2800 is not.

So 2800 is not a perfect square.

(i) 2032, (ii) 2048 and (iii) 1027 are not perfect squares. 1089 is.

Only the units digit of the base matters.

Two of them qualify: 108² and 292².

(iv) 15625 + 251

The side is √441.

First find the LCM, then top it up to a square.

Factorise 9408 and pair up the prime factors.

Every factor pairs up except the single 3. So multiply by 3.

Between n² and (n+1)² there are 2n numbers.

Fourth row first:

Now read the rule from the completed rows. In each one the third base is the product of the first two, and the answer is one more than that product.

Count in two stages — blocks, then tiny squares inside a block.

Prime factorisation:

8 cubes.

27 cubes.

Because each of them counts the unit cubes that fill a cube.

No.

So 9 is not a perfect cube — and neither is any number from 10 to 26.

64 unit cubes.

The completed table of cubes:

Patterns worth naming:

• Parity: odd numbers have odd cubes; even numbers have even cubes.
• Last digits: cubes can end in any digit from 0 to 9, and the last digit of a cube tells the last digit of the number: 1→1, 8→2, 7→3, 4→4, 5→5, 6→6, 3→7, 2→8, 9→9, 0→0.
• Zeros: 10³ = 1000, 20³ = 8000 — one zero at the end of the number gives three zeros at the end of the cube.
• Cubes grow fast: there are only 20 cubes up to 8000, but 89 squares.

All ten digits, 0 to 9, are possible.

Count them directly from 1³, 2³, 3³, … and compare with the squares.

Observation: cubes are much rarer than squares. There are only 9 cubes below 1000, against 31 squares.

No. A cube can end with 0 zeros, 3 zeros, 6 zeros … but never exactly 2.

Search the cubes below each number and look for two that add up.

1000, which is 10³.

Yes, 3375 = 15³.

No.

Therefore 500 is not a perfect cube.

Factorise each into triplets.

Start from the cubes and keep differencing.

The differences become constant at level 3, and the constant is 6.

Because the Sanskrit word mula — the root of a plant, and also basis, cause, origin — was the term ancient Indian mathematicians used for this operation.

• varga = square (the figure, its area, and the second power); varga-mula = square root.
• ghana = cube (the solid and the third power); ghana-mula = cube root.
• pada (foot, basis, origin) was another word for root. Brahmagupta (628 CE) writes, “The pada (root) of a krti (square) is that of which it is a square.”

Factorise into triplets.

Factorise and see which prime falls short of a multiple of 3.

The 3s already form a triplet; the 7s are one short. So multiply by 7.

All five are false. Here is why in each case.

Yes — use the last digit to fix the units digit, and the size to fix the tens digit.

(i) 67³ – 66³ is the greatest.

Reason: the gap between consecutive cubes (7, 19, 37, 61, …) grows much faster than the gap between consecutive squares (3, 5, 7, 9, …), and it is larger for larger numbers. So the cube difference with the larger numbers, (i), is the greatest.

Here is an arrangement that works:

Every adjacent pair sums to 9, 16 or 25:

No — apart from reading the same row backwards, the arrangement is the only one.

Yes. Here is one circular arrangement of 1 to 32 in which every neighbouring pair adds to a square:

The sums going round are all 9, 16, 25, 36 or 49: