AMC 10 specimen paper 2

Factor the numerator to obtain 32^n.

The ratio is therefore 3(2/3)^n, which strictly decreases for positive integer n.

At n = 2 it is 4/3 > 1, while at n = 3 it is 8/9 < 1.

Thus only n = 1 and n = 2 satisfy the inequality; every later ratio is smaller than the already failing n = 3 ratio.

Both lie within the given upper bound.

The answer is 2.

The fraction before reduction is 3(2 n + 3) / (n (n + 3)).

If n = 3 m, it reduces to (2 m + 1) / (m (m + 1)); its numerator is coprime to each denominator factor.

The coprime consecutive positive factors cannot both be squares, so the denominator is not a square.

If 3 does not divide n, the original numerator is coprime to n and n + 3, and those two factors are coprime.

A square denominator therefore requires n = u squared and n + 3 = v squared.

Then (v - u)(v + u) = 3, whose positive factors force u = 1 and v = 2.

Thus only n = 1 occurs; its reduced fraction is 15 / 4.

Answer 1.

Reduction must be completed before testing the denominator.

For a nondegenerate polygon, the longest side must be shorter than the sum of the others; for positive side lengths this criterion also permits a closed polygon.

Here every side is an integer at most 17.

Remove the mandatory lengths 3 and 7, and write the other three as x < y < z, with x + y + z = 25 and none equal to 3 or 7.

The possible x are 1, 2, 4, 5, 6: any x at least 8 would make the sum at least 8 + 9 + 10 > 25.

Enforce z = 25 - x - y, z <= 17, y < z and the exclusions.

For x = 1 the possible y are 8, 9, 10, 11; for x = 2 they are 6, 8, 9, 10, 11; for x = 4 they are 5, 6, 8, 9, 10; for x = 5 they are 6, 8, 9; and for x = 6 they are 8, 9.

These disjoint cases contribute 4 + 5 + 5 + 3 + 2 = 19 sets.

Sorting the three extra sides counts every set once.

The successive net gains in a complete three-minute cycle are 3, 3 and 1, so each completed cycle contributes 7 tokens.

At the start of minute 3 k + 1 the state is 7 k; the three peaks in that cycle are 7 k + 5, 7 k + 8 and 7 k + 11.

For k = 4, the final peak at minute 15 is 39, and all earlier cycles have smaller peaks.

For k = 5, the first peak at minute 16 is 40.

Therefore the first threshold event occurs at minute 16, before its removal.

Let the hundreds, tens and units digits be a, b, c.

Subtraction of the reversed number gives 99(a - c) = 198, so a = c + 2.

The product condition becomes c(c + 2) b = 24, with 1 ≤ c ≤ 7 and b an integer from 1 through 9.

For c = 1, 2, 3, 4 the possible b values are 8, 3, 8/5, 1; for c ≥ 5 the factor c(c + 2) already exceeds 24.

Hence the valid triples are (3, 8, 1), (4, 3, 2), (6, 1, 4).

Their digits are nonzero and distinct.

The integers are 381,432,614, so there are 3.

The square's centre is distance rsqrt(2) from every disk centre.

The enclosing circle centred there has radius r(1 + sqrt(2)).

It is minimal: two opposite disks have outward points on a diagonal separated by 2 r(1 + sqrt(2)), so every containing circle has at least that diameter.

Thus r = 6/(1 + sqrt(2)) = 6sqrt(2) - 6, with r^2 = 108 - 72sqrt(2).

Neighbouring disks are tangent and diagonal disks are disjoint, so their areas add.

The requested area divided by pi is 36 - 4 r^2 = -396 + 288sqrt(2).

At minute 55 printer A has completed nine scheduled packets but missed one, giving 32 items.

At minute 60 it has completed ten scheduled packets but missed one, giving 36.

Printer B has floor((t - s)/5) completed packets at either deadline.

The minute-60 condition is 36 + 7 floor((60 - s)/5) >= 100, equivalent to at least ten B packets and hence s <= 10.

The minute-55 condition is 32 + 7 floor((55 - s)/5) < 100, equivalent to at most nine B packets and hence s > 5.

Thus s = 6, 7, 8, 9, 10: five starts.

For each, the totals are 95 at minute 55 and 106 at minute 60.

Whole packets and inclusive completion deadlines are essential to these floor bounds.

Each equal side has length t √(1 + t²), so perimeter is 2 t + 2 t √(1 + t²).

For a feasible t, the residual 30-2 t must be positive before squaring.

At t = 3 the comparison is 24² ≥ 4 × 9 × 10, which holds; at t = 4 it is 22² ≥ 4 × 16 × 17, which fails.

Both terms of the perimeter increase for positive t, so all t ≥ 4 fail.

The largest integer is 3, and t = 1, 2 are feasible as well.

The selected divisors are 3, 6, 12, 15, 24, 30, 60, 120.

More structurally, each is 3 e where e divides 40; pairing e with 40/e pairs the selected divisors to product 360.

There are four such pairs and no fixed point because 40 is not a square.

Thus P = 360^4.

Modulo 101, 360 is 57, 57^2 is 17 and 17^2 is 87.

The remainder is 87.

The pairing is internal to the selected divisor slice; pairing every divisor of 360 would use a different set.

The total item size is 2*6 + 3*4 + 4*3 = 36, so at least four capacity-10 bins are necessary.

A four-bin witness is (6 + 4), (6 + 4), (4 + 3 + 3), (3 + 3).

Each 6-item occupies its own bin and can be alone (E), with a 4-item (F), or with a 3-item (G).

The unordered types of the two 6-bins are EE, EF, EG, FF, FG, GG.

EE and EG leave more than 20 units for the other two bins and are impossible.

EF leaves exactly (4 + 3 + 3), (4 + 3 + 3): one arrangement.

FF leaves one 4 and four 3 s, giving (4 + 3), (3 + 3 + 3) or (4 + 3 + 3), (3 + 3): two.

FG leaves two 4 s and three 3 s, giving (4 + 4), (3 + 3 + 3) or (4 + 3), (4 + 3 + 3): two.

GG leaves three 4 s and two 3 s, forcing (4 + 4), (4 + 3 + 3): one.

These disjoint classes give 1 + 2 + 2 + 1 = 6 optimal arrangements.

Let the long and short sides have a, b spacing intervals.

The closed perimeter has 24 gaps, giving 2 a + 2 b = 24 and a + b = 12.

Side post counts including corners are a + 1 and b + 1; hence a + 1 ≥ 2(b + 1), or a ≥ 2 b + 1.

Substitution gives b ≤ 3.

The areas for b = 1, 2, 3 are 4 b (12-b) = 44, 80,108 square metres.

The maximum is 108, attained with a = 9, b = 3, and every positive case has been considered.

Column 1 has four cells and must contain all four letters, with its first A already fixed.

Each later row must contain B, because three B tokens must occupy three distinct rows and row 1 is A.

Now consider the six permutations of B, C, D in the remaining column-1 cells, and fill each remaining row with distinct letters while observing the column bans and remaining inventory.

Writing the three later rows as words, the full possibilities are BA/CBA/DCB, BA/DCB/CBA, BC/CAB/DBA, BC/CBA/DAB, BC/DAB/CBA, BC/DBA/CAB, CB/BCA/DAB, CB/DAB/BCA, DB/BCA/CAB, DB/CAB/BCA.

Each has the required inventory and no repeated row or column letter.

This exhaustive six-column-order classification yields 10 fillings.

Letters are indistinguishable, so no extra token factorial appears.

Let A(n) be the amber amount just after the addition on day n.

Then A(1) = 2 and A(n) = 2 + (2/3) A(n − 1).

Summing the finite geometric progression gives A(n) = 6(1 − (2/3)^n).

Blue always has amount 3 at that moment.

Amber exceeds 60% precisely when A(n)/(A(n) + 3) > 3/5, or A(n) > 9/2.

Thus (2/3)^n < 1/4.

For n = 3 this power is 8/27 > 1/4, while for n = 4 it is 16/81 < 1/4.

The powers strictly decrease, so day 4 is the first.

Every altitude equals twice the same area divided by its corresponding side length.

The longest altitude is therefore 168/13 cm and the shortest is 168/15 cm.

Their ratio is (168/13)/(168/15) = 15/13.

The intermediate side gives an altitude between these two, so the extremal choices are justified.

For each a, b, c, the sum condition fixes d modulo 3 and permits exactly three values among 0 through 8.

Thus there are 9³ times 3 = 2187 equally likely conditioned quadruples.

Partition them by a + b modulo 3.

Each of the three rows has 729 outcomes.

Exact residue enumeration of ad − bc modulo 9 gives favourable row counts 297, 81, 81, for total 459.

The required probability is 459/2187 = 17/81.

Divisibility modulo 9 cannot be replaced by a field argument modulo 3, because the higher-power condition carries additional information.

A path moves between the two colours of a checkerboard at every step.

The board has five cells of one colour, namely its corners and centre, and four of the other.

Since there are five odd labels and four even labels, the corners and centre must hold the odd labels.

Those labels total 11+13+15+17+19=75.

Removing the corner total leaves centre label 75−60=15.

The conditions are attainable: the rows (11,12,13), (16,15,14), (17,18,19) give a consecutive path and the required corner sum.

Every operation reduces the number of entries by 2.

Starting with four evens, a ternary sum or product can remove at most two even entries.

One operation therefore cannot yield all odds, while two operations can; the maximum remaining size is 4.

Both operations must remove exactly two evens.

If just one original odd changes, its replacement is o + q, where o is one of 1, 3, 5, 7 and q is in {20, 56, 70,100,194 }.

Here 20 is the sum of all four evens; the other q values arise by multiplying three evens and adding the fourth.

These give 20 distinct multisets.

Otherwise two different original odds change, and the evens split into pairs with sum pairs (6, 14), (8, 12), (10, 10).

Assign these sums to two distinct original odds, retaining the other two.

Removing duplicate numerical outcomes leaves 21 multisets.

They are (1, 3, 11, 21), (1, 3, 13, 19), (1, 3, 15, 17), (1, 5, 9, 21), (1, 5, 11, 19), (1, 5, 13, 17), (1, 5, 15, 15), (1, 7, 9, 19), (1, 7, 11, 17), (1, 7, 13, 15), (3, 5, 7, 21), (3, 5, 9, 19), (3, 5, 11, 17), (3, 5, 13, 15), (3, 7, 7, 19), (3, 7, 9, 17), (3, 7, 11, 15), (3, 7, 13, 13), (5, 7, 7, 17), (5, 7, 9, 15), (5, 7, 11, 13).

The classes share exactly (3, 5, 7, 21).

Thus the answer is 20 + 21 - 1 = 40.

The classification covers both disjoint and nested operations and counts final values, not histories.

Let u be the leftover part of the leg parallel to the rectangle side of length 6, and v the leftover part of the other leg.

The corner triangles have legs (u,5) and (6,v).

Their hypotenuses lie on the same line, so similarity gives 5/u = v/6, or uv = 30.

Their area product is (5u/2)(6v/2) = 30uv/4 = 225, which also equals one quarter of the square of the rectangle’s area.

Thus 9T = 225 and T = 25, regardless of which corner has area nine.

Fix A by rotational symmetry.

A chord meets the disk precisely when its endpoints have smaller angular separation at least 90°, since its distance from the centre is 10 cos(θ/2), and the threshold distance is 5√2.

Consequently B and C must both lie in the half-circle opposite A, whose angular length is 180°.

This has probability (1/2)² = 1/4.

Conditional on that event, their angular coordinates are independent and uniform in an interval of length 180°.

Their own chord meets the disk when their coordinate difference in absolute value is at least 90°.

In the square of ordered coordinate pairs, the two admissible corner triangles have combined area (180° − 90°)², one quarter of the square's area.

Thus the conditional probability is 1/4, and the required probability is (1/4)(1/4) = 1/16.

Coincident endpoints and tangencies have probability zero and do not affect the result.

Equal squared lengths give 100 + e^2 = (10 - f)^2 + 100, so e = 10 - f because both are positive.

The triangle area is (100 - 10 e + ef)/2 = (100 - e^2)/2.

Requiring it to exceed 30 gives e^2 < 40.

For integer e from 1 through 9 this allows exactly 1 through 6, and f = 10 - e stays within its bounds.

Hence there are 6 pairs.

Choose a library badge that is bronze; its existence is given.

It is metal because every bronze badge is metal.

This one badge proves II.

Statement I need not hold: there could also be a metal badge that is not a library badge.

Statement III need not hold: the bronze badges could consist only of the given library badge.

Both counterexamples preserve the premises.

Therefore only II must hold.

Divide vertical coordinates by √3, leaving hexagon area 6 and band half-height k.

Its six vertices are (-2 + k, -k), (2-k, -k), (2, 0), (2-k, k), (-2 + k, k), (-2, 0).

Decomposition or shoelace gives band area 8 k-2 k².

If this is twice the remaining caps, it must equal 4.

Thus k²-4 k + 2 = 0 and k = 2±√2.

Only 2-√2 lies between 0 and 1.

Squaring gives k² = 6-4√2, so u + v = 6 + 4 = 10.

Substituting k=2−√2 into 8k−2k² gives band area 4 and combined caps area 2 in the scaled coordinates.

A sequence of distinct digits with no local valley must increase and then decrease: a decrease followed later by an increase would create a valley at the first change back to increasing.

The maximum is the unique peak, possibly an endpoint.

If the last digit is 0, choose the other five digits from 1 through 9.

Each of the four selected nonmaximum positive digits chooses left or right of the peak, giving C(9, 5)2^4 = 2016.

If the last digit is 5, zero cannot occur because the minimum must be an endpoint and cannot lead.

Let g selected digits exceed 5.

Choose these from four options and choose 5-g lower digits from {1, 2, 3, 4}; after fixing the maximum at the peak, each of the g-1 other greater digits chooses a side, while every lower digit is forced left.

For g = 1, 2, 3, 4 the counts are 4, 48, 96, 32, totalling 180.

The final total is 2196.

The equal distances BE = CE = 5 place E above the midpoint of BC.

Its perpendicular height is sqrt(25 -4^2) = 3.

The inner and outer triangles have the same altitude, so the area ratio 3 makes AD = 3*BC = 24.

As AB = CD, each extension has length (24 -8)/2 = 8.

E is also above the midpoint of AD; each outer equal side therefore has squared length 12^2 + 3^2 = 153 = 9*17.

The outer perimeter is 24 + 2*3*sqrt(17) = 24 + 6*sqrt(17).

Positivity and the stated point order are satisfied.

The two index rules determine every term after a_1 because its parent index is smaller.

Evaluate by successive binary levels: indices 1; 2 through 3; 4 through 7; 8 through 15; 16 through 31.

Their exact sums are 2, 9, 74, 1185, 38315.

For each parent n, the even child contributes 3 n a_n and the odd child contributes a_n + 1, so their combined contribution is (3 n + 1) a_n + 1.

Applying this across each full level corroborates the five level sums.

Adding them gives 39585.