Use a variable

Which quantity could you name first? What could you then say about the other two?

Answer: Blue 35 cm, gold 50 cm and red 65 cm. No other set fits all three relationships.

Write it with a variable

Let \(t\) be the blue tail’s length in centimetres. We do not have to guess it yet. Gold has length \(t+15\); red has length \(2t-5\). Each expression describes a length, not a different kind of letter.

The three expressions must add to 150:

Check: gold is \(35+15=50\), red is \(2(35)-5=65\), and \(35+50+65=150\). All lengths are positive whole centimetres. Increasing the blue length by 1 increases the total by 4, so a different blue length cannot give the same total.

You could also draw four equal blue-length parts and account for the extra 10 cm. The letter keeps that same relationship available while the length is unknown.

Answer: No, the side lengths are not uniquely determined. Width 20 cm with length 30 cm, and width 10 cm with length 40 cm, both fit. Yes, equal sides are possible: 25 cm by 25 cm.

Let \(a\) be the width in centimetres and \(b\) the length in centimetres. The perimeter gives \(2a+2b=100\), or \(a+b=50\). That relation allows many pairs.

Check: each pictured rectangle has two copies of each side and perimeter 100 cm. All its sides are positive whole lengths. Two different fitting rectangles already disprove uniqueness. The square shows why different letters do not have to stand for different values.

A second independent measurement or relationship could select one rectangle. Introducing two letters cannot replace that missing clue.

Answer: A short rod is 50 cm; a long rod is 80 cm. This is the only pair that fits both records.

Let \(s\) be a short rod’s length and \(\ell\) a long rod’s length, both in centimetres. The first record gives \(2s+\ell=180\), so \(\ell=180-2s\). The second gives \(s+2\ell=210\).

Substitute

The long length and \(180-2s\) are equal. Replace each complete long-rod length in the second record with that expression. The parentheses keep the whole replacement together.

Then \(\ell=180-2(50)=80\).

Check both: \(2(50)+80=180\), and \(50+2(80)=210\). Both lengths are positive whole centimetres and \(50<80\). Any fitting pair must satisfy the substituted equation, which forces \(s=50\); the first record then forces \(\ell=80\).

Adding or comparing the original records offers another valid route. Substitution works because replacing a quantity by an equal expression preserves the relationship.

If you choose the wide-kit count, what else becomes determined? Remember that a partly filled last crate still counts as a crate.

Answer: There are nine allowed plans. Mini/wide counts are 21/3, 20/4, 19/5, 18/6, 17/7, 16/8, 15/9, 14/10 and 12/12. Their unused sheets are 27, 24, 21, 18, 15, 12, 9, 6 and 0; they need 9, 9, 10, 9, 10, 10, 10, 10 and 10 crates, respectively. The greatest allowed wide count is 12, with 12 mini kits.

Use a variable: let \(k\) count wide kits. Then \(24-k\) counts mini kits. The sheet total is \(2(24-k)+5k=48+3k\). The count and sheet rules reduce the possible whole values to \(3\le k\le12\).

Make an organized list

The crate rule does not increase smoothly: filling the last spare place in one crate may add no new crate. List each remaining whole value once, round up each type’s crate count separately, then add them. Checking just the largest sheet-feasible count would miss smaller working plans.

Check: every kept row totals twenty-four kits, meets both minimum counts and uses at most eighty-four sheets and ten crates. Rejected rows exceed the crate limit. All whole values from three through twelve were tested, so no allowed plan is missing. The largest surviving wide count is twelve; the list supplies its complete working witness.

How can one symbol keep the two dimensions connected? What could rule out whole groups of row counts before you multiply them?

Answer: 123 rows and 127 columns. This is the only allowed pair.

Use a variable: let \(r\) count rows. Then \(r+4\) counts columns and \(r(r+4)=15621\). This retains the four-column difference through every candidate check.

Bound the size

At one hundred rows or fewer, there are at most one hundred and four columns, giving at most \(100\times104=10400\) lights. At one hundred and fifty rows or more, the product is at least \(150\times154=23100\). Neither range can give 15,621. Only the forty-nine whole row counts from 101 through 149 remain.

Modular reduction

Remove complete tens from each factor. Any discarded ten multiplied by a whole number still gives complete tens, so the extra ones determine the product’s units digit. We are keeping remainders by ten, not claiming that the original factors equal their endings.

Check the units digit

The row ending fixes the column ending because the columns are four more. Check all ten endings:

Only a row count ending in three can produce a product ending in one. Combine that necessary condition with the bound: the candidates are 103, 113, 123, 133 and 143. The ending does not choose among those five.

Full check: \(127-123=4\) and \(123\times127=15621\). Both counts are positive and fill a complete rectangle. The bounds excluded every row count outside 101–149; the remainder test excluded all but five inside; full products test all five, of which only 123 works.

Listing factor pairs is a valid alternative. The coarse hundreds-based bounds and the ten-ending table give one accessible route: they leave five exact checks without needing a close square estimate. They are not the shortest possible route. If you notice that 15,621 is four below 125 squared, the relationship \(r(r+4)=(r+2)^2-4\) immediately gives \(r+2=125\), so \(r=123\). Tighter bounds from \(120\times124=14880\) and \(125\times129=16125\) also leave just four row counts. Use a useful relation you can justify; do not insist on a remainder test when a stronger observation has already settled the problem. A correct ending alone would not prove the answer.