A Square and A Cube Class 8 Solutions

No. Factors usually occur in pairs, but for a perfect square the square root pairs with itself. That unpaired factor makes the total number of factors odd.

Exactly the perfect squares have an odd number of positive factors. For example, 36 has the pairs 1 × 36, 2 × 18, 3 × 12, 4 × 9, and the unpaired factor 6 × 6, giving 9 factors.

Locker n is toggled once for every factor of n. It remains open only when it has an odd number of factors, so the open lockers are the perfect squares: 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100.

A square ends in 6 when its base ends in 4 or 6. Therefore, the required numbers are 34^2, 46^2, 56^2, and 74^2.

Write the number as m \times 10^3. Its square is m^2 \times 10^6, so it has 6 zeros at the end.

If a number has k trailing zeros, it contains a factor 10^k. Squaring gives a factor 10^{2k}, so the number of trailing zeros doubles. A non-zero perfect square therefore has an even number of trailing zeros.

An even number is 2k, whose square 4k^2 is even. An odd number is 2k+1, whose square 4k^2+4k+1 is odd. A number and its square always have the same parity.

Their difference is (n+1)^2-n^2=2n+1. Removing the two square endpoints leaves 2n whole numbers between them.

1089=33^2. The other three lie between consecutive squares and are not squares. Therefore, 2032, 2048, and 1027 are not perfect squares.

A square ends in 4 when its base ends in 2 or 8. Hence, the required numbers are 108^2 and 292^2.

Use (125+1)^2=125^2+2\times125+1. Thus 126^2=15625+250+1=15625+251=15876. The correct choice is (iv).

If the side is s, then s^2=441. Since 21\times21=441, s=21 m.

The least common multiple is 180=2^2\times3^2\times5. The factor 5 needs a pair, so multiply by 5: 180\times5=900=30^2. The smallest such square is 900.

9408=2^6\times3\times7^2. Only the factor 3 is unpaired, so multiply by 3. Then 9408\times3=2^6\times3^2\times7^2=(8\times3\times7)^2=168^2. The answers are 3 and 168.

For consecutive squares n^2 and (n+1)^2, there are 2n numbers between them. With n=16, the answer is 2\times16=32.

Using the same rule with n=99, the number of integers between the squares is 2\times99=198.

The completed lines are 1^2+2^2+2^2=3^2, 2^2+3^2+6^2=7^2, 3^2+4^2+12^2=13^2, 4^2+5^2+20^2=21^2, and 9^2+10^2+90^2=91^2.

No. In a cube, every prime exponent is a multiple of 3. If a cube is divisible by 100=2^2\times5^2, it must actually contain at least 2^3\times5^3=1000, so it ends with at least three zeros.

The pattern says that the sum of the next n consecutive odd numbers is n^3. There are 10 terms here, so the sum is 10^3=1000.

Since 4^3=64, \sqrt[3]{64}=4.

Since 8^3=512, \sqrt[3]{512}=8.

Since 9^3=729, \sqrt[3]{729}=9.

27000=27\times1000=3^3\times10^3=30^3. Therefore, \sqrt[3]{27000}=30.

22^3=22\times22\times22=484\times22=10648. Therefore, \sqrt[3]{10648}=22.

1323=3^3\times7^2. Multiplying by 7 completes the third factor of 7: 1323\times7=3^3\times7^3=21^3. The answer is 7.

False. An odd number is 2k+1, and its cube remains odd. For example, 3^3=27.

False. For example, 2^3=8, and cubes of numbers ending in 2 also end in 8.

False. The smallest two-digit number is 10, and 10^3=1000, already a four-digit number.

False. The largest two-digit number is 99, and 99^3=970299, which has six digits.

False. Only perfect squares always have an odd number of factors. For example, 8=2^3 is a cube and has 4 factors: 1, 2, 4, and 8.

11^3=11\times11\times11=1331, so \sqrt[3]{1331}=11.

The cube root must end in 7 because the cube ends in 3. Since 17^3=4913, \sqrt[3]{4913}=17.

The cube root must end in 3 because the cube ends in 7. Checking 23 gives 23^3=12167, so the cube root is 23.

32^3=32\times32\times32=1024\times32=32768, so \sqrt[3]{32768}=32.

For consecutive numbers, n^3-(n-1)^3=3n^2-3n+1, which grows as n grows. Cube differences also exceed the corresponding square differences here. Thus 67^3-66^3=13267 is the greatest.