I want to Know any Short tricks to solve HCF Question easily in exam
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The method I use is to take each number and break it down to its prime factors. For example: 60=2×2×3×5. This gives you an individual list of factors for each number. Note that some factors may be repeated as in this example where 2 is repeated.

When you've done this, write down the factors common to all the numbers. To make sure you don't duplicate anything, cross off the factors you note from the individual lists of factors. If a factor is repeated in a list, only cross off one of its appearances.

Keep going through the lists until there are no more common factors. All you have to do now is to multiply all the factors you made a note of. This product is the HCF.

Some calculators reduce fractions to their lowest form, so by dividing one number by another you can discover if there is a common factor between two numbers. That's a trick you can use to help you, but in an exam, you're probably not allowed to use a calculator. For example, if you have 52 and 91, the fraction would be 52/91 which reduces to 4/7. 52/4=91/7=13, so 13 is the HCF. This trick becomes tedious if you a have to find the HCF of a group of numbers.

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