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Q: How can you test to determine if a factor is a factor of another number?

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if it has no other number that can divide into itself but itself ================================== Another contributor clarified: You have to determine whether there is any number ... besides '1' and the number under test itself ... that can divide evenly into the number under test. If you find even one, then it's not a prime. If there are none, then it's a prime.

To keep it simple: Write a main loop that goes through all the numbers, starting with 2, and incrementing one at a time. Determine whether each number is a prime number. If it is, increment a counter. To determine whether each number is a prime number, either use an inner loop, or a separate function. Test divisibility of the number "n" by every number from 2 to n-1. If you find a factor, then it is not a prime number. Note that you can test divisibility by using the "%" operator. For example: if (number % factor == 0) // number is divisible by factor else // it isn't

a streak test is a test wheree you rub a mineral across a streak plate to see the color of its streak, which is a better indentifying factor of the mineral than the external color. A scratch test is when you scratch a mineral to find out its hardness on the Mohs Scale of Hardness. This is also another useful identifying factor

A prime number is a positive integer with two factors: one and the number itself. If you test the numbers up to the square root and your number is not divisible by any of them, it's prime.

A prime number is a positive integer with two factors: one and the number itself. If you test the numbers up to the square root and your number is not divisible by any of them, it's prime.

Related questions

Divide the smaller number into the bigger number. If the answer comes out even with no remainder, it's a factor.

Divide the larger number by the smaller. If the result has no remainder (no decimal) then the smaller number is a factor of the larger.

If N is the number, and f is the number that you want to test as a possible factor, then first of all:test N > f (this must be true, the factors are always smaller in magnitude)next perform N ÷ f (N divided by f). If the quotient (answer to a division problem) is a whole number with no remainder or fractional part, then f is a factor of N.If the quotient is not a whole number (meaning there is a remainder), then f is not a factor.

it's simple, you divide the potential factor by the original number.For example, if I wanted to find out if 12 was a factor of 36;36/12= 3Because this number is an integer and not a decimal we know that in fact, 12 is a factor of 36.I hope that helped :)

if it has no other number that can divide into itself but itself ================================== Another contributor clarified: You have to determine whether there is any number ... besides '1' and the number under test itself ... that can divide evenly into the number under test. If you find even one, then it's not a prime. If there are none, then it's a prime.

paper bag test

Yes.

To keep it simple: Write a main loop that goes through all the numbers, starting with 2, and incrementing one at a time. Determine whether each number is a prime number. If it is, increment a counter. To determine whether each number is a prime number, either use an inner loop, or a separate function. Test divisibility of the number "n" by every number from 2 to n-1. If you find a factor, then it is not a prime number. Note that you can test divisibility by using the "%" operator. For example: if (number % factor == 0) // number is divisible by factor else // it isn't

Paper Bag test :D

number of rings

When finding the factors of 841, the largest number you would test is 29. No prime number higher than 29 could be a factor because the square of that number would exceed 841.

By using tensile test.

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