Problem
GCD and LCM in tiling problems
In (1) a floor 20 cm by 12 cm is covered by identical squares with no gaps. The largest such square's side is wanted. In (2) tiles 2 cm by 3 cm are used to build a square. Find the side length in each part.
Reasoning · Grade 5-1 Greatest Common Divisor and Least Common Multiple
Your answer
Watch out The parts look alike, but one hunts a divisor and the other a multiple.
How to solve
Strategy Make a Systematic List — The two parts look alike but pull in opposite directions, so I treat them as two subproblems and settle each one separately. In each part I first draw the picture and ask what a side length has to do to work: in part (1) a row of squares must land exactly on the far edge, so the side has to divide 20 and divide 12; in part (2) a wall of tiles must reach exactly as high as it is wide, so the side has to be a multiple of 2 and a multiple of 3. Once that is clear, the search is short enough to just write out an honest list — every divisor of 20 and of 12 for part (1), the first few multiples of 2 and of 3 for part (2) — and read the answer off the list. Listing beats guessing here because the list also proves nothing bigger (or nothing smaller) was missed.
1STEP 1
Split the question into two separate jobs
Treat the parts as two different problems.
Deciding first whether the answer will be smaller or bigger than the numbers given stops the two parts from being mixed up, and that decision needs nothing more than reading the picture.
4.OA.A.3Identify Subproblems2STEP 2
Part (1): see why the square's side must divide both 20 and 12
In (1) the side must be a common divisor of 20 and 12.
20 = (number of columns) × s, 12 = (number of rows) × s
Tiling a rectangle with equal squares makes a rectangular array, and an array's row lengths are multiplication facts — so the side length has to be a factor of each side of the floor.
3.MD.C.7Draw A DiagramA square that tiles the floor exactly must have a side that divides both 20 and 12 with nothing left over.
Why?
Laying squares along a wall covers whole squares plus a leftover, and any leftover would mean a strip the tiles cannot reach.
🧱Division with remainderAny number splits into a whole count of equal groups plus a leftover, and the leftover is always smaller than the group size.
Why?
The sides that divide both numbers can be listed in pairs, so the largest common one can be picked with none overlooked.
🧱Divisors come in pairsEach divisor hands over its partner for free, so divisors pair off two by two unless one is its own partner.
3STEP 3
Part (1): list the divisors and pick the largest common one
The greatest common divisor is 4.
20: 1, 2, 4, 5, 10, 20 12: 1, 2, 3, 4, 6, 12 common: 1, 2, 4
Both lists are short enough to write out in full, so the largest shared number can be read straight off — no formula needed, and nothing can have been skipped.
4.OA.B.4Make A Systematic List4STEP 4
Part (1): check the 4 cm square really covers the floor
Fifteen 4 cm squares cover it exactly.
20 ÷ 4 = 5, 12 ÷ 4 = 3, 5 × 3 × 16 = 240 = 20 × 12
Multiplying rows by columns and comparing with length times width is the standard area check for a rectangle, and it catches any leftover strip at once.
4.MD.A.3Draw A Diagram5STEP 5
Part (2): see why the square's side must be a multiple of 2 and of 3
In (2) the side must be a common multiple of 2 and 3.
width = 2 × (tiles across), height = 3 × (tiles up)
Repeatedly adding the same length is multiplication, so the side of the built square is automatically a multiple of the tile's width and of its height.
3.MD.C.7Draw A Diagram6STEP 6
Part (2): list the multiples and take the first one they share
The least common multiple is 6.
2: 2, 4, 6, 8, 10, 12 3: 3, 6, 9, 12 first common: 6
Counting by 2s and by 3s until the same number turns up is exactly how a skip-counting list finds a least common multiple.
4.OA.B.4Make A Systematic List7STEP 7
Part (2): build the 6 cm square and rule out anything smaller
Six tiles build the 6 cm square exactly.
3 × 2 = 6 cm, 2 × 3 = 6 cm, 6 × 6 = 36 = 6 × 6 tiles × area
Showing one arrangement that works and checking the areas of all the smaller squares turns a plausible answer into a certain one.
6.NS.B.4Make A Systematic ListAnswer
4, 6 cm
20 ÷ 4 = 5, 12 ÷ 4 = 3
Both answers are lengths in centimetres, as asked. In part (1) the covering square, 4 cm, is smaller than both sides of the floor and it divides them exactly (20 divided by 4 = 5 and 12 divided by 4 = 3), which is what breaking a big rectangle into equal squares should give. In part (2) the built square, 6 cm, is bigger than both sides of the tile and both 2 and 3 go into it exactly, which is what stacking small tiles into a bigger square should give. The two answers also fit the familiar check on 2 and 3, and on 20 and 12: greatest common divisor times least common multiple equals the product of the pair — 4 times 60 = 240 = 20 times 12, and 1 times 6 = 6 = 2 times 3.
Takeaway
Cutting a big shape into equal squares asks for the biggest shared factor; building a big square out of small tiles asks for the smallest shared multiple!
- Split the question into two separate jobs
- Part (1): see why the square's side must divide both 20 and 12
- Part (1): list the divisors and pick the largest common one
- Part (1): check the 4 cm square really covers the floor
- Part (2): see why the square's side must be a multiple of 2 and of 3
- Part (2): list the multiples and take the first one they share
- Part (2): build the 6 cm square and rule out anything smaller