Problem 18 worked answer

Shedletsky’s Self-Checking Formula Cabinet

Designed for Grade 12 · Uses HS CALC math · About 25–55 minutes

Earn one jewelBrilliant-Cut DiamondLevel 4

At a glance

Damaged drawer: J
Printed value of J: $56$
Corrected label: $⌊ \sqrt 3026 ⌋$
Corrected value of J: $55$
Combination: $261$
Built-in four-drawer checks: 16

1. Evaluate the printed labels

Drawer	Printed value	Exact evaluation
A	96	$∣ 102210 ∣ = 10 (10) - 2 (2) = 96$ , and $⟨ 10, 2 ⟩ \cdot ⟨ 10, - 2 ⟩ = 100 - 4 = 96$ .
B	43	The integers from $- 21$ through $21$ give $21 + 1 + 21 = 43$ .
C	39	If $(5 1 / 3) 𝑦 = 513$ , then $𝑦 / 3 = 13$ , so $𝑦 = 39$ .
D	83	Since $- 9 < 0$ , $𝑓 (- 9) = (- 9) 2 + 2 = 83$ .
E	51	The complex modulus is $\sqrt 242 + 452 = \sqrt 2601 = 51$ .
F	71	$702 < 5000 < 712$ , so $70 < 50 \sqrt 2 < 71$ , and its ceiling is 71.
G	76	$𝑑 𝑑 𝑥 (𝑥 3 + 𝑥) = 3 𝑥 2 + 1$ , which is 76 at $𝑥 = 5$ .
H	63	$𝐴 𝐵 ⟂ 𝐵 𝐶$ makes the angle $90 \circ$ , and $70 %$ of 90 is 63.
I	67	$(110) + (111) + (112) = 1 + 11 + 55 = 67$ .
J	56	$552 = 3025 < 3026 < 3136 = 562$ , so the printed ceiling is 56.
K	59	$58 . ―― 9 = 58 + 0 . ―― 9 = 58 + 1 = 59$ .
L	80	$\int 40 (𝑥 3 + 4) 𝑑 𝑥 = [𝑥 4 4 + 4 𝑥] 40 = 64 + 16 = 80$ .
M	47	$2.35 \times 104 m m = 23, 500 m m$ , while $12 m = 500 m m$ ; $23, 500 / 500 = 47$ .
N	92	A uniform distribution on the integers 1 through 183 has mean $(1 + 183) / 2 = 92$ .
O	87	The two standard small-angle limits contribute $44 + 43 = 87$ .
P	35	$1612 \geq 0$ , so $𝑓 (1612) = 2 (1612) + 2 = 33 + 2 = 35$ .

The printed value grid is

⎡ ⎢ ⎢ ⎢ ⎢ ⎣ 96433983517176636756598047928735 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦

2. Use the cabinet’s checks

Three rows agree:

96 + 43 + 39 + 83 = 261, 51 + 71 + 76 + 63 = 261, 47 + 92 + 87 + 35 = 261 .

The third row does not:

67 + 56 + 59 + 80 = 262 .

Three columns agree, while the second column fails:

96 + 51 + 67 + 47 = 261, 39 + 76 + 59 + 87 = 261, 83 + 63 + 80 + 35 = 261,

43 + 71 + 56 + 92 = 262 .

The descending diagonal totals 261, while the rising diagonal totals 262:

96 + 71 + 59 + 35 = 261, 83 + 76 + 56 + 47 = 262 .

Two more natural groups fail:

67 + 56 + 47 + 92 = 262

for the lower-left $2 \times 2$ chamber, and

71 + 76 + 56 + 59 = 262

for the central $2 \times 2$ chamber.

Every failed check contains J: the third row, second column, rising diagonal, lower-left chamber, and central chamber. No other drawer belongs to all five failures. The overlapping numerical checks therefore locate J without relying on its appearance.

3. Repair J

The printed label is

⌈ \sqrt 3026 ⌉ = 56 .

Replace the matched ceiling brackets with floor brackets:

⌊ \sqrt 3026 ⌋ = 55 .

Only the orientation of the enclosing marks changes. The repaired grid is

⎡ ⎢ ⎢ ⎢ ⎢ ⎣ 96433983517176636755598047928735 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦

4. Open the cabinet

The combination is

261 .

The repaired cabinet produces this number in sixteen visually natural ways.

Four rows

$96 + 43 + 39 + 83 = 261$
$51 + 71 + 76 + 63 = 261$
$67 + 55 + 59 + 80 = 261$
$47 + 92 + 87 + 35 = 261$

Four columns

$96 + 51 + 67 + 47 = 261$
$43 + 71 + 55 + 92 = 261$
$39 + 76 + 59 + 87 = 261$
$83 + 63 + 80 + 35 = 261$

Two long diagonals

$96 + 71 + 59 + 35 = 261$
$83 + 76 + 55 + 47 = 261$

Four corner chambers

Upper left: $96 + 43 + 51 + 71 = 261$
Upper right: $39 + 83 + 76 + 63 = 261$
Lower left: $67 + 55 + 47 + 92 = 261$
Lower right: $59 + 80 + 87 + 35 = 261$

Two more structural checks

Central four drawers: $71 + 76 + 55 + 59 = 261$
Four outer corners: $96 + 83 + 47 + 35 = 261$

These are the four rows, four columns, two diagonals, four corner chambers, center chamber, and outer corners: $4 + 4 + 2 + 4 + 1 + 1 = 16$ checks.

Post-lock extensions

Extension A: all sixteen

The classification above lists all sixteen deliberately built-in appearances of the key.

Extension B: five readings of a vertical stroke

A: determinant bars
B, outer bars: set cardinality
B, inner bars: absolute value
E: complex modulus
G: evaluation at $𝑥 = 5$

The same visual stroke takes its meaning from context.

Extension C: the two half-charges

Opposite drawers through the center pair as follows:

𝐴 + 𝑃 = 96 + 35 = 131, 𝐵 + 𝑂 = 43 + 87 = 130, 𝐶 + 𝑁 = 39 + 92 = 131, 𝐷 + 𝑀 = 83 + 47 = 130, 𝐸 + 𝐿 = 51 + 80 = 131, 𝐹 + 𝐾 = 71 + 59 = 130, 𝐺 + 𝐽 = 76 + 55 = 131, 𝐻 + 𝐼 = 63 + 67 = 130 .

The totals alternate between 130 and 131. Every built-in four-drawer check is one 130 pair plus one 131 pair, so $130 + 131 = 261$ .

Optional hint ladder

Keep these teacher-controlled or reveal them one at a time.

Finish the complete $4 \times 4$ value grid before searching for damaged notation.
Look for simple groups suggested by the physical arrangement.
Compare complete horizontal and vertical groups. Most agree; a small number do not.
Check the two long diagonals as well.
The four $2 \times 2$ corner chambers, the central $2 \times 2$ chamber, and the four outer corners provide additional checks.
List every failed check. Which single drawer belongs to all of them?
The damaged drawer’s printed value is exactly one greater than its intended value. Change only the orientation of its enclosing marks.

Check

After the repair, all sixteen listed four-drawer groups total 261. Reversing the repair makes J equal 56 and raises exactly the five groups containing J from 261 to 262, reproducing the printed cabinet’s failure pattern.

Technical fit and rating

MJ HS:CALC.7 · C4 · W4 Insight challenge · Extended workload

Calculus and analysis · Patterns and algebra

How the rating works →

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