61 is prime. You cannot make 61 by multiplying four digits.
When multiplying by ten the digits shift to the left
If two decimal numbers have x and y digits after the decimal point respectively, then their product has (x + y) digits after the decimal point.
If two numbers have m and n significant digits, respectively, then then product can have at most m+n. However, the normally it is the minimum of m and n.
Because the number of digits after the decimal place in a product does not require that.
If the two multiplicands have X and Y digits after the decimal place then their product (before removing any trailing 0s) has (X+Y) digits after the decimal point.
You do the multiplication in exactly the same way. The only extra thing is that when multiplying decimals, you need to place the decimal point (or decimal comma - depending on your country) in the correct position. If one number has, for example, 3 digits after the decimal point, and the other 4, you need to place the decimal point in the result (BEFORE eliminating unnecessary zeros) in such a way that there are, in this example, 7 digits (3 + 4) to the right of the decimal point.
44. Also works for 36, 63.
Because when adding or subtracting, the operations must take account of the place values of individual digits in the numbers. When multiplying, only the overall order of magnitudes are relevant for placing the decimal point in the product.
It should be obvious (it is, isn't it? am I expecting too much of today's educational system?) that the largest possible result of multiplying two two digit numbers is 99x99 = 9801, which has four digits, and the smallest possible result is 10x10 = 100, which has three digits.
Mutilply the values of the digits.
The digits product of two digit has number four
Multiply vertically the extreme left digits is one pattern involved in multiplying algebraic expressions. Multiplying crosswise is another common pattern that is used.
When multiplying, the product is going to have the same number of significant digits as the least amount in the factors. So 8.08 has three significant digits and 5.232 has four significant digits. Remember that any zeros in-between significant digits is always significant. Because the least amount of sig digs in the factors is three (from 8.08), the product is going to have three sig digs.Multiplying out normally: 8.08 Ã— 5.232 = 42.27456Rounded to three places: 42.3
The least number of significant figures in any number of the problem determines the number of significant figures in the answer.
from 3 digits (10x10) to 4 digits (99X99)
The answer will depend on the order in which you do partial products. It is quite common in the UK for the first partial product to be the two digits in the tens' place and so that is often the largest. This ties in with the method for multiplying two binomials when they move on to algebra.
You had me until "product." The product of 4 digits can't be prime.
If you mean, "What is the largest number of digits possible in the product of two 2-digit numbers" then 99 * 99 = 9801, or 4 digits. Anything down to 59 * 17 = 1003 will have 4 digits.
52 is a number whose digits have a product of 10.