Question

Divisibility of 7and 8

please answer...​

Answer 1

Step-by-step explanation:

Divisibility by 8

If the last three digits of a number are divisible by 8, then the number is completely divisible by 8. Example: Take number 24344. Consider the last two digits i.e. 344.

To check if a number is evenly divisible by 7: Take the last digit of the number, double it Then subtract the result from the rest of the number If the resulting number is evenly divisible by 7, so is the original number. ... The last digit is 3, double that to make 6, subtract from 6 from the remaining digits.

Answer 2

Answer:

Test #1. Take the digits of the number in reverse order, from right to left, multiplying them successively by the digits 1, 3, 2, 6, 4, 5, repeating with this sequence of multipliers as long as necessary. Add the products. This sum has the same remainder mod 7 as the original number! Example: Is 1603 divisible by seven? Well, 3(1)+0(3)+6(2)+1(6)=21 is divisible by 7, so 1603 is.

Test #2. Remove the last digit, double it, subtract it from the truncated original number and continue doing this until only one digit remains. If this is 0 or 7, then the original number is divisible by 7. Example: 1603 -> 160-2(3)=154 -> 15-2(4)=7, so 1603 is divisible by 7.

THESE WERE DIVISIBILITY RULE FOR 7

Test#1. Numbers are divisible by 8 if the number formed by the last three individual digits is evenly divisible by 8. For example, the last three digits of the number 3624 is 624, which is evenly divisible by 8 so 3624 is evenly divisible by 8.

THIS IS DIVISIBILITY RULE FOR 8