How to Solve Any Board

Most people solve their first few boards by trial and error, then plateau. This guide replaces the guessing with a method: find an anchor near 37, work out the gap, and spend what is left closing it.

Start with the one fact that shapes everything

37 is a prime number. It has no factors except 1 and itself, which has a direct and practical consequence at the board: you can never reach 37 as a single product. There is no a*b using dice faces that lands on it. The largest product two dice can make is 6*6 = 36, and it misses by one.

So every solution has the same shape — a big term, plus or minus a smaller correction. Once you see that, you stop hunting for a magic combination and start doing two much easier jobs in sequence: pick the big term, then close the gap.

The method in one line

Anchor near 37 with a product or a power, work out the gap, then spend your remaining dice closing it exactly.

Step 1 — find your anchor

An anchor is the one term that does the heavy lifting. You are looking for something that gets you within a handful of 37 using as few dice as possible, because every die the anchor consumes is one you no longer have for the correction.

The anchors worth memorising

6*6 = 36 — the single most useful fact in the game. Gap of 1.
6*5 = 30 — gap of 7, which is easy to build from two or three dice.
2^5 = 32 — gap of 5, and it uses only two dice.
6^2 = 36 — the same 36, but built with the red power die.
3^3 = 27 — gap of 10.
5*5 = 25 — gap of 12.
4^2 = 16 or 5^2 = 25 — useful when your board is short on sixes.
6*7 is impossible (no 7 face), but 6*(4+3) = 42 overshoots to a gap of −5.

Notice that the last one runs the other way. Overshooting is a legitimate plan and often the better one, because subtracting is just as cheap as adding and a slightly-too-big anchor is usually easier to build.

Step 2 — count the gap, then count your dice

Once the anchor is chosen, the puzzle stops being about 37 and becomes about a much smaller number. If you anchored on 6*6 = 36, you need exactly 1 more from the five dice you have left. If you anchored on 6*5 = 30, you need exactly 7.

This is the moment most solves are won or lost, because you must use all seven dice. Not "may" — must. A correction that reaches the gap but leaves two dice unplaced is not a solution, and this is the single most common reason a board that "should" work will not submit.

6×6+1+1−1−1+1 Anchor 36, gap 1 — and the gap is closed with five dice rather than one, because all seven have to appear.

That example looks wasteful and it is exactly right. +1+1-1-1+1 nets to 1 while consuming the five dice the rules demand.

Step 3 — spend the dice you do not need

You will constantly finish the arithmetic with dice still in hand. There are four standard ways to make a die disappear without changing your total, and knowing them turns "I am stuck" into "I am finished".

Divide a pair into 1. 3/3, 5/5, 6/6 — any face over itself is 1. Two dice, one unit of value.
Subtract a pair into 0. 5-5, 4-4 — two dice for nothing at all. Attach it to a term you want to keep unchanged.
Multiply by 1. If a 1 is on the board, *1 absorbs it with no effect.
Raise to the power of 1. x^1 = x. This is the quiet workhorse — it spends a die *and*, if the die is your red power die, uses it correctly.

2^5+2×5−6+3÷3 Anchor 32, then +10 −6 gets to 36, and the leftover pair 3/3 supplies the final 1.

The x^1 trick deserves special attention in the power-dice modes — see the power-dice guide for why it so often turns a Valid solve into a Perfect one.

The four solution families

Almost every board you will ever solve falls into one of four shapes. When you sit down to a new board, try them in this order.

1. Thirty-six plus one

Build 36 — as 6*6, as 6^2, as 4*9, as (6-1)*7-style combinations — then find a single unit anywhere. This is the most common family by a wide margin because 36 is so easy to reach.

2. A big multiple, plus a run

Group several dice into a sum, multiply the whole group, then add or subtract the remainder. The parenthesis does the work.

(2+3−1)×6+4+4+5 (2+3−1) = 4, times 6 is 24, then +4+4+5 = 13 closes it. Four dice inside the bracket, three outside.

3. Overshoot and come down

Deliberately build past 37, then subtract. Useful when your board is heavy with large faces and light on small ones.

(6+5+4)×3−2×1−6 15 × 3 = 45, then −2 and −6 bring it back to 37.

4. A power, plus a correction

Lead with an exponent — 2^5, 3^3, 6^2, 5^2 — and correct from there. This family is mandatory reading in the power-dice modes because the red die wants to be an exponent anyway.

(1+5+5)×4+2−3−6 Not every board needs a power: 11 × 4 = 44, then −7. Try the simple families first.

When you are genuinely stuck

Re-count your dice. More stuck boards are unplaced sevenths than impossible arithmetic.
Change the anchor. If 36 + 1 will not close, try overshooting to 42 and coming down.
Look for a pair you can neutralise. Two identical faces are a free / into 1 or - into 0, which frees the rest of the board.
Try a bracket you have not tried. Grouping three small dice and multiplying is the move people forget.
Check the parity. If your anchor is even and every remaining die is even, you cannot make an odd correction — you will need a bracket or a division to break the pattern.

And if a board defeats you, read 16 worked boards — every one shows both a Valid and a Perfect solution, so you can see the same method applied sixteen different ways.