A Square and A Cube

A locker is toggled once for every factor of its number.

Factors pair as a×b=n, except a square has one unpaired factor √n.

So exactly the square-numbered lockers have an odd number of factors.

A locker n is toggled as many times as n has factors.

A prime number has exactly the two factors 1 and itself.

Most factors pair as a and n/a.

Only a perfect square has a middle factor paired with itself.

A square with side n units contains n×n unit squares.

1²=1×1.

2²=2×2.

3²=3×3.

4²=4×4.

5²=5×5.

10²=10×10.

11²=11×11.

12²=12×12.

13²=13×13.

14²=14×14.

15²=15×15.

20²=20×20.

21²=21×21.

29²=29×29.

30²=30×30.

(3/5)²=3/5×3/5=9/25.

2.5²=2.5×2.5=6.25.

The possible square endings are 0, 1, 4, 5, 6 or 9.

A square such as 16 ends in 6, but 26 also ends in 6 and is not square.

Squares never end in 2, 3, 7 or 8.

A square ends in 6 when its root ends in 4 or 6.

The roots 34, 46, 56 and 74 end in 4 or 6.

If n ends in k zeros, n=10ᵏm where m does not end in zero.

n²=10²ᵏm².

(2k)²=4k² is even.

(2k+1)²=4k²+4k+1 is odd.

n²−(n−1)²=n²−(n²−2n+1)=2n−1.

Pairing the first n odd numbers gives n².

There are six terms, so the sum is 6².

Each next odd number adds one new outer L-shaped layer.

The sixth partial sum is 6².

(n+1)²−n²=2n+1.

Exclude both square endpoints: 2n+1−1=2n.

Multiply each number by itself. Grouping the answers by tens makes the table easier to check.

Numbers ending in 1 or 9 have squares ending in 1; continue the listed base numbers.

Add the next counting number to get each triangular number, then add neighboring terms.

Check the last digits, differences between consecutive squares, and sums of consecutive odd numbers.

Possible patterns: square units digits are 0, 1, 4, 5, 6 or 9; consecutive squares differ by odd numbers; 1+3+⋯+(2n−1)=n².

Subtract 1, 3, 5, … until the remainder is zero.

The first 27 odd numbers sum to 27²=729.

40²=1600 and 50²=2500, so the root lies between 40 and 50.

45²=2025, so the root is below 45.

44²=(40+4)²=1600+320+16=1936.

15²=225 and 16²=256, so 15<√250<16.

250 is 6 below 256 and 25 above 225.

15²=225>125, so a 15 cm square is too large.

11²=121≤125 and 12²=144>125.

A square cannot end in 2, 3, 7 or 8.

2032 ends in 2; 2048 ends in 8; 1027 ends in 7.

1089=33².

A square ends in 4 when its root ends in 2 or 8.

108 ends in 8; 292 ends in 2.

(n+1)²=n²+2n+1.

126²=125²+2×125+1=15625+251.

21×21=441.

LCM(4,9,10)=2²×3²×5.

A square needs even prime exponents, so one more factor of 5 is needed.

2²×3²×5²=900.

9408=2⁶×3×7².

Only the exponent of 3 is odd, so multiply by 3.

9408×3=28,224=2⁶×3²×7²=(2³×3×7)².

17²−16²=289−256=33.

Exclude the two endpoints: 33−1=32.

100²−99²=199.

Exclude one endpoint step: 199−1=198.

For consecutive n,n+1: n²+(n+1)²+[n(n+1)]²=[n(n+1)+1]².

For n=4, the last two values are 20 and 21. For n=9, they are 90 and 91.

A number is a perfect square when every prime factor has an even exponent.

Count the 9×9 tiles, then count the 5×5 tiny squares inside each tile.

If x²=y, then x is the square root of y; if x³=y, then x is the cube root of y.

The next square is found by adding the next odd number, 2×36−1=71.

A perfect square reaches exactly 0 after subtracting 1, 3, 5, … .

Use the squares table to list the roots whose squares fall in each hundred-wide interval.

A positive times itself and a negative times itself can both equal 64, but the radical symbol names the positive one.

List squares around 576. For 327, use the rule that a square cannot end in 7.

Find the positive number whose square is 49.

7²=49, so the side is 7 cm.

For a cube of side n, there are n layers of n² unit cubes.

2³=8; 3³=27.

Each layer has 4×4=16 cubes.

There are 4 layers: 16×4=64.

If a number has k terminal zeros, its cube has 3k terminal zeros.

The number of terminal zeros in a cube must be a multiple of 3.

10³=10×10×10.

4³=64; 8³=512; 9³=729.

30³=27×1000=27000.

22³=22×22×22=10648.

1323=3³×7².

A cube needs prime exponents divisible by 3, so multiply by 7.

1323×7=3³×7³=21³.

Test the claim using the definition or a counterexample.

Test the claim using the definition or a counterexample.

Test the claim using the definition or a counterexample.

Test the claim using the definition or a counterexample.

Test the claim using the definition or a counterexample.

11³=1331; 17³=4913; 23³=12167; 32³=32768.

For cubes, n³−(n−1)³=3n²−3n+1.

67³−66³=13,267; 43³−42³=5,419.

The square differences are 133 and 85.

A perfect cube is a number n×n×n for a whole number n.

Multiply each whole number by itself three times; the table includes the key values used later in the chapter.

The units digit of n³ repeats according to the units digit of n. Compare the listed cubes with powers of 10.

Cube each factor three times. For a fraction, cube the numerator and denominator; a negative number cubed stays negative.

Check each equality by cubing the two numbers and adding.

There are 10 consecutive odd terms, and the pattern shows that each such group sums to a cube.

Keep subtracting adjacent values. For cubes, the third differences become constant.

For a perfect cube, every prime exponent must be a multiple of 3.

Check cubes of digits 0 through 9, and compare cubes of even and odd numbers.

Cube units digits for 0–9 are 0,1,8,7,4,5,6,3,2,9; even bases give even cubes and odd bases give odd cubes.

Each adjacent sum in the chain is a square. The endpoints are forced because 16 and 17 each have only one possible partner in this set.

Check each neighboring sum, including the last number paired with the first.