Make an Organized List

Try plushie counts from zero and check the leftover. Zero leaves \(53\), one leaves \(41\), two leave \(29\), three leave \(17\), and none of those are multiples of \(5\). Four plushies cost \(48\), leaving \(53 - 48 = 5\), which is exactly one keychain, so \(\boxed{4}\) plushies.

Five plushies would already cost \(60\), more than the total, so the list ends there and the answer is unique.

Walk the smoothie count upward and test each leftover. Zero leaves \(60\), which is \(10\) juices. One leaves \(51\), not a multiple of \(6\). Two leave \(42\), which is \(7\) juices. Three leave \(33\), no. Four leave \(24\), which is \(4\) juices. Five leave \(15\), no. Six leave \(6\), one juice. Seven would cost more than \(60\). That is \(\boxed{4}\) orders.

Marching the smoothie count from \(0\) to \(6\) without skipping is what guarantees nothing was missed.

Digit pairs multiplying to \(12\) must be factor pairs with both factors at most \(9\), so \(2 \times 6\) and \(3 \times 4\). The pair \(1 \times 12\) fails, since \(12\) is not a digit.

Each surviving pair builds two numbers, \(26\) and \(62\), \(34\) and \(43\).

$$\boxed{4}$$

The two-digit multiples of \(9\) are \(18, 27, 36, 45, 54, 63, 72, 81, 90, 99\). Both digits must be prime, so both come from \(2, 3, 5, 7\), and only \(27\) and \(72\) survive. The ones digit has to be larger than the tens digit, which leaves \(\boxed{27}\).

Starting from the multiples of \(9\) meant checking ten numbers instead of all ninety.

Go in alphabetical order. Pairs starting with A are AB, AC, AD, AE, AF, five of them. Pairs starting with B add BC, BD, BE, BF, four new ones, since AB is already counted. Each letter after that adds one fewer.

$$5 + 4 + 3 + 2 + 1 = \boxed{15}$$

This is the pairs count from the counting chapter, rebuilt by hand. If the formula slips your mind mid-problem, an organized list rebuilds it for you.

Write every split largest pile first. The largest pile can be at most \(5\), since the other two trays need a ring each.

Largest \(5\) forces \(5, 1, 1\). Largest \(4\) gives \(4, 2, 1\). Largest \(3\) gives \(3, 3, 1\) and \(3, 2, 2\). A largest pile of \(2\) could hold at most \(2 + 2 + 2 = 6\) rings, too few.

$$\boxed{4}$$

Sorting each split before writing it is what stops \(4, 2, 1\) and \(1, 2, 4\) from sneaking in as two different entries.

Numbers that leave remainder \(2\) when divided by \(7\) come every seven steps, so between \(30\) and \(60\) they are \(30, 37, 44, 51, 58\). Their remainders under \(4\) run \(2, 1, 0, 3, 2\), and only \(37\) leaves the required \(1\).

$$\boxed{37}$$

With two remainder conditions, list out the one that gives fewer candidates and let the other one filter.

Case on the hundreds digit. If it is \(1\), the remaining digits sum to \(3\), giving \(103, 112, 121, 130\), four numbers. If it is \(2\), they sum to \(2\), giving \(202, 211, 220\), three numbers. If it is \(3\), two numbers, \(301, 310\). If it is \(4\), just \(400\).

$$4 + 3 + 2 + 1 = \boxed{10}$$

The case counts step down \(4, 3, 2, 1\), which is the kind of pattern an organized list often hands you for free.

One box gives a single packing, all \(8\) together.

Two boxes, written larger pile first, give \(7{+}1\), \(6{+}2\), \(5{+}3\), \(4{+}4\), and stopping there matters, since \(3{+}5\) would repeat \(5{+}3\). That is \(4\) packings.

Three boxes, marching the largest pile downward, give \(6{+}1{+}1\), \(5{+}2{+}1\), \(4{+}3{+}1\), \(4{+}2{+}2\), \(3{+}3{+}2\). That is \(5\) more.

$$1 + 4 + 5 = \boxed{10}$$

Case one, the code ends in \(0\). The first two digits add to \(12\), strictly decreasing into the \(0\), giving \(93, 84, 75\), so the codes \(930, 840, 750\).

Case two, the code ends in \(5\). The first two digits would add to \(7\) while each stays above \(5\), but two digits above \(5\) already add to at least \(13\). The case is empty.

$$3 + 0 = \boxed{3}$$

An empty case is not wasted work. Proving nothing lives there is exactly what makes the final count trustworthy.

Marching muffins from zero, the leftovers run \(67, 59, 51, 43, 35, 27, 19, 11, 3\). Only \(35\) ends in \(0\) or \(5\), at four muffins, leaving \(7\) cookies.

$$\boxed{4}$$

The multiples run \(14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98\). Checking digit sums, only \(28\) and \(91\) reach \(10\).

$$\boxed{2}$$

One bag, just \(6\). Two bags, largest first, \(5{+}1\), \(4{+}2\), \(3{+}3\). Three bags, \(4{+}1{+}1\), \(3{+}2{+}1\), \(2{+}2{+}2\).

$$1 + 3 + 3 = \boxed{7}$$

Remainder \(3\) under \(8\) gives the list \(3, 11, 19, 27, 35, 43\). The only multiple of \(5\) on it is \(\boxed{35}\).

For each tens digit \(1\) through \(9\), the ones digit must be \(9\) minus it, giving \(18, 27, 36, 45, 54, 63, 72, 81, 90\), one number per tens digit.

$$\boxed{9}$$

Choosing which digit to leave out gives every possibility, \(123, 124, 134, 234\).

$$\boxed{4}$$

With the flavors ordered, the first starts \(6\) pairs, the second \(5\) new ones, then \(4, 3, 2, 1\).

$$6+5+4+3+2+1 = \boxed{21}$$

Palindromes \(1x1\) end in \(1\), which is odd, so no multiples of \(4\) live there. The \(2x2\) family is \(202, 212, 222, 232, 242, 252, 262, 272, 282, 292\).

A number is a multiple of \(4\) exactly when its last two digits are, so test \(02, 12, 22, \ldots\) The survivors are \(212, 232, 252, 272, 292\).

$$\boxed{5}$$

Marching rides, the leftovers are \(100, 88, 76, 64, 52, 40, 28, 16, 4\), and only \(28\), at six rides, is a multiple of \(7\), giving four games.

$$6 + 4 = \boxed{10}$$

Case on tens. With two tens, \(5\) gems remain, allowing zero or one five, \(2\) ways. With one ten, \(15\) remain, allowing zero through three fives, \(4\) ways. With no tens, \(25\) remain, allowing zero through five fives, \(6\) ways.

$$2 + 4 + 6 = \boxed{12}$$

The singles handle the cleanup, filling any remainder in exactly one way, so the whole count comes down to the tens and fives.