How can the following definition be written correctly as a biconditional statement? An odd integer is an integer that is not divisible by two. (A+ answer) An integer is odd if and only if it is not divisible by two
Every number is divisible by any non-zero number. 527 is not evenly divisible by 3.
Every number is divisible by any non-zero number. However, 1023 is not evenly divisible by 2.
It's very easy to test a number to see if it is divisible by 4 or by 9. If it passes both tests, then it is divisible by 4x9=36.To test for divisibility by 9, add the digits of the number. If the sum is divisible by 9, then the number is divisible by 9.To test for divisibility by 4, look at the last two digits. If they are a multiple of 4, then the number is divisible by 4.
There is no number, no matter the number of digits, that is only divisible by 2.
1,234
Any number that ends in a 0 or a 5.
Four of them.
See the following: http://www.artofproblemsolving.com/Wiki/index.php/Divisibility_rules/Rule_for_3_and_9_proof
11
No. 189 is only evenly divisible by 3 and 9 (from the set provided). Using the following rules of divisibility on the number 189: Divisible by 2? No - the number is not even Divisible by 3? Yes - the sum of the digits (1 + 8 + 9 = 18) is divisible by 3 Divisible by 4? No - the last two digits are not evenly divisible by 4 Divisible by 5? No - the last digit is not a 0 or a 5 Divisible by 6? No - the number is not even Divisible by 9? Yes - the sum of the digits is divisible by 9 Divisible by 10? No - the number is not divisible by 2 or 5
Yes because 5624/2 = 2812
How can the following definition be written correctly as a biconditional statement? An odd integer is an integer that is not divisible by two. (A+ answer) An integer is odd if and only if it is not divisible by two
For a number to be divisible by 105 it must be divisible by 3, by 5 and by 7. So, divisibility by 3 requires all three of the following to be satisfied:Sum the digits together. Repeat if necessary. If the answer is 0, 3, 6 or 9 the original number is divisible by 3.If the final digit of the number is 0 or 5, the original number is divisible by 5.Take the number formed by all but the last digit. From it subtract double the last digit. Keep going until there is only one digit left. If it is 0 or 7 then the original number is divisible by 7.
Converse:If a number is divisible by 3, then every number of a digit is divisible by three. Inverse: If every digit of a number is not divisible by 3 then the number is not divisible by 3? Contrapositive:If a number is not divisible by 3, then every number of a digit is not divisible by three.
1,845 is divisible by the following numbers: 1, 3, 5, 9, 15, 41, 45, 123, 205, 369, 615 and 1845.
999 is divisible by 9, but not by six; the next lower number divisible by 9 is 990, which is also divisible by 6, so that's the answer. Some shortcuts for divisibility: 0 is divisible by any number. If the last digit of a number is divisible by 2, the number itself is divisible by 2. If the sum of the digits of a number is divisible by 3, the number itself is divisible by 3. If the last TWO digits of a number are divisible by 4, the number itself is divisible by 4. If the last digit of a number is divisible by 5, the number itself is divisible by 5. If a number is divisible by both 2 and 3, it is divisible by 6. If the last THREE digits of a number are divisible by 8, the number itself is divisible by 8. If the sum of the digits of a number is divisible by 9, the number itself is divisible by 9. 990: 9+9+0=18, which is divisible by 9, so 990 is divisible by 9. 18 is also divisible by 3, so 990 is divisible by 3, and since 990 ends in 0 it's also divisible by 2, meaning that it's divisible by 6 as well.