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If the hundreds digit is odd, the number obtained by the last two digits must be 4 times an odd number. 352: 52 = 4 x 13. Add the last digit to twice the rest. The result must be divisible by 8. 56: (5 × 2) + 6 = 16. The last three digits are divisible by 8. [ 2][ 3] 34,152: Examine divisibility of just 152: 19 × 8.
Every terminating decimal representation can be written as a decimal fraction, a fraction whose denominator is a power of 10 (e.g. 1.585 = 1585 / 1000 ); it may also be written as a ratio of the form k / 2 n ·5 m (e.g. 1.585 = 317 / 2 3 ·5 2 ).
Long division is the standard algorithm used for pen-and-paper division of multi-digit numbers expressed in decimal notation. It shifts gradually from the left to the right end of the dividend, subtracting the largest possible multiple of the divisor (at the digit level) at each stage; the multiples then become the digits of the quotient, and the final difference is then the remainder.
This is denoted as 20 / 5 = 4, or 20 / 5 = 4. [2] In the example, 20 is the dividend, 5 is the divisor, and 4 is the quotient. Unlike the other basic operations, when dividing natural numbers there is sometimes a remainder that will not go evenly into the dividend; for example, 10 / 3 leaves a remainder of 1, as 10 is not a multiple of 3.
Long division. In arithmetic, long division is a standard division algorithm suitable for dividing multi-digit Hindu-Arabic numerals ( positional notation) that is simple enough to perform by hand. It breaks down a division problem into a series of easier steps. As in all division problems, one number, called the dividend, is divided by another ...
v. t. e. In mathematics and computing, the hexadecimal (also base-16 or simply hex) numeral system is a positional numeral system that represents numbers using a radix (base) of sixteen. Unlike the decimal system representing numbers using ten symbols, hexadecimal uses sixteen distinct symbols, most often the symbols "0"–"9" to represent ...
Casting out nines is any of three arithmetical procedures: [ 1] Adding the decimal digits of a positive whole number, while optionally ignoring any 9s or digits which sum to 9 or a multiple of 9. The result of this procedure is a number which is smaller than the original whenever the original has more than one digit, leaves the same remainder ...
The most significant digit (10) is "dropped": 10 1 0 11 <- Digits of 0xA10B ----- 10 Then we multiply the bottom number from the source base (16), the product is placed under the next digit of the source value, and then add: 10 1 0 11 160 ----- 10 161 Repeat until the final addition is performed: 10 1 0 11 160 2576 41216 ----- 10 161 2576 41227 ...