The formula this page uses
GCF(a, b) = the product of every prime both numbers share, taking the smaller power each time GCF(a, b) × LCM(a, b) = a × b
Use the free greatest common factor calculator to work through the number relationship, preserve exact values, and check the result with estimation. Choose a focused operation below to calculate, inspect the method, and connect the result to a visual model.
The visual updates with your calculation.
GCF(a, b) = the product of every prime both numbers share, taking the smaller power each time GCF(a, b) × LCM(a, b) = a × b
Find GCF(48, 60)
Use the remainder chain. Divide the big number by the small one, keep the remainder, then divide the old small number by that remainder, and keep going until the remainder is 0. The last remainder you saw before the 0 is the answer, and it usually takes three or four lines.
Dividing the top and the bottom by the same number does not change the value, and the GCF is the biggest number that divides both without a remainder. So 36/60 with GCF 12 goes straight to 3/5, with no second round of tidying.
Yes, whenever the smaller number divides the bigger one. GCF(6, 24) = 6, because 6 divides itself and also divides 24. That is a useful sign that one number is a multiple of the other.
If you are cutting something into equal groups or equal-sized pieces, you want the GCF, because the answer must divide both amounts. If you are waiting for two repeating events to line up again, you want the LCM, because the answer has to be a multiple of both.
This workspace keeps the formula and the meaning together. Decimal results are rounded for display; retain full precision when you continue a calculation.
Match each input to the quantities in the problem and keep units consistent.
The result panel identifies the formula or algorithm and shows the main substitutions.
Use the diagram, plot, or data display to check scale, direction, and plausibility.
Move to targeted questions once you can explain why the method applies.
A calculator confirms an answer. Working the method yourself is what makes the next problem faster.