How this instrument works
A three-number ratio is in lowest terms when no whole number bigger than 1 divides all three terms at once. Finding that shared divisor and dividing it out is exactly what g = gcd(gcd(a, b), c) does: it folds the third number into an ordinary two-number GCD, then divides a, b, and c by whatever survives. The result — first term, second term, third term — describes the identical proportion in the smallest whole numbers that still carry it.
GCD is strictly a two-argument operation — there is no separate three-number GCD formula, only the ordinary one applied twice. Because GCD is associative, gcd(gcd(a, b), c) and gcd(a, gcd(b, c)) always agree, so it makes no difference which pair gets combined first; this sheet resolves a and b first purely as a fixed convention. The two-argument version itself traces back to antenaresis, the reciprocal-subtraction procedure Euclid sets out in Elements, Book VII, Proposition 2 — nesting it twice is the natural, not arbitrary, way to reach three numbers.
The genuinely surprising case is that the three-way GCD is not the smallest of the three pairwise GCDs. Take 6, 10, and 15: pairwise, gcd(6,10) = 2, gcd(10,15) = 5, and gcd(6,15) = 3 — none of them zero, and the smallest is 2. Yet no number bigger than 1 divides all three at once, since 2 fails on 15, 3 fails on 10, and 5 fails on 6, so the true combined GCD is 1 and 6:10:15 is already in lowest terms. Chaining catches this correctly — gcd(gcd(6,10),15) = gcd(2,15) = 1 — where eyeballing the smallest pairwise GCD would give a wrong answer of 2.
- Enter your three quantities into Number 1, Number 2, and Number 3, keeping them in the order your ratio should read.
- Read GCD used to simplify — the one factor common to all three numbers, found by chaining gcd(a, b) with c.
- Read Simplified ratio: first term, Simplified ratio: second term, and Simplified ratio: third term for the reduced a:b:c.
- If GCD used to simplify comes back as 1, the ratio you entered was already in lowest terms — nothing bigger divides all three.
Worked example — reducing a 12:18:24 pigment mix
A workshop measures three pigments for a paint batch: 12 ml, 18 ml, and 24 ml, entered as Number 1, Number 2, and Number 3. The sheet first finds gcd(12, 18) = 6, then chains the third number in: gcd(6, 24) = 6, so GCD used to simplify reads 6. Dividing each quantity by that shared factor gives Simplified ratio: first term = 12 ⁄ 6 = 2, second term = 18 ⁄ 6 = 3, and third term = 24 ⁄ 6 = 4 — the batch is really a 2:3:4 mix at heart.
The reduction can be checked by reversing it: multiply 2, 3, and 4 back up by the same factor of 6 and the original 12, 18, and 24 return exactly. No smaller whole-number triple than 2:3:4 says the same thing, since gcd(2,3,4) — computed the identical chained way as gcd(gcd(2,3),4) = gcd(1,4) = 1 — confirms there is nothing left in common to cancel.
Questions
How do you compute the GCD of three numbers instead of two?
Chain two ordinary two-number GCD calls: first find gcd(a, b), then feed that result back in against the third number, gcd(gcd(a, b), c). GCD only ever takes two arguments at a time, so a three-number version has no separate formula — it is the same two-argument operation applied twice, in sequence. For 12, 18, and 24 that is gcd(12,18)=6, then gcd(6,24)=6.
Does the order I chain the three numbers in change the answer?
No. GCD is associative and commutative, so gcd(gcd(a,b),c), gcd(gcd(a,c),b), and gcd(a,gcd(b,c)) always agree — nested any way, or with the three numbers reordered, the chain lands on the identical shared divisor. This sheet always resolves a and b first purely for consistency, not because another order would answer differently.
Is the three-number GCD just the smallest of the three pairwise GCDs?
No, and this is the mistake that trips people up. Take 6, 10, and 15: the pairwise GCDs are gcd(6,10)=2, gcd(10,15)=5, gcd(6,15)=3, and the smallest of those, 2, is not the answer. The true three-way GCD is 1, because no single number divides all three at once — chaining gcd(gcd(6,10),15) = gcd(2,15) = 1 finds that correctly, where the pairwise minimum does not.
What if the three numbers share no common factor greater than 1?
Then g comes out as 1 and the ratio you entered is already in lowest terms — the sheet returns the same three numbers unchanged, just confirming there is nothing left to cancel. Three numbers like 5, 7, and 11 behave this way: no pair, let alone all three together, shares a divisor above 1.
How does simplifying three numbers extend the two-number ratio case?
A two-number ratio a:b simplifies with one gcd(a,b) call. Extending to a third number is not a cosmetic change — GCD is strictly a two-argument operation, so covering three quantities means running it twice and passing the first result into the second, gcd(gcd(a,b),c). Skipping that second call and using gcd(a,b) alone would ignore c and could leave the third term unreduced.
Where do real three-number ratios show up?
Anywhere a proportion is split three ways: a concrete mix of cement, sand, and gravel; an NPK fertilizer blend; three pigments measured for a paint batch; or a recipe scaled across three ingredients at once. Reducing 12:18:24 to 2:3:4 states the same underlying proportion in the smallest whole numbers that still describe it.