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Q: How do you multiply by 2 or 3 digit numbers?

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There are 3 values (1, 2, 3) for each of the 4 digits. Therefore, there are 3*3*3*3 or 81 four digit numbers that can be formed.

If you can repeat a digit, there are 27. If you can't repeat a digit, there are only 6.

There are no two real numbers that will add to -9 and multiply to 81. Using complex numbers, the two numbers are: (-9/2 + i9/2√3) and (-9/2 - i9/2√3)

The 3 digit numbers under 500 are 100 through 499.

60. There are 5 different choices for the first digit, four for the second digit, and three choices for the third digit. According to the fundamental counting principle, you multiply them together to get the total number of possible ways, and 5 x 4 x 3 = 60 Another way to think about it is 5!/3!

Related questions

There are no such numbers.

you multiply 3 digit numbers by killing yourself and giving away body parts.

You multiply the one digit number on the bottom to every number on the top starting at the right and so on with every other number on the bottom.

A 3 or 4 digit number.

4 options for the first digit, 3 options for the second digit, 2 options for the third digit. Multiply the number of options together, and you find how many 3-digit numbers you can get.

Example: 222*12 222 3 digit * 13 2 digit ------- 666 Multiply 3 to all three twos +222 Multiply 1 to all three twos (skip spot) -------- 2886 Answer Add up

If you multiply 2*9, you get 18. Multiply that by 10 to get a three digit number, and you get 180.

Here are two examples

-1

3

Yes. By 1 digit, 2 digit and some even by other 3 digit numbers.

By using the 3 digits of a number we can form 3 different two digit numbers. 3C2 = 3!/[(3 - 2)!2!] = 3!/(1!2!) = (3 x 2!)/2! = 3

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