If a number is divisible by 5 and 7, it's divisible by 35.
Double the last digit and subtract the last digit from the remaining digits.
The divisibility rule for 42 isdivisibility by 2, ANDdivisibility by 3, ANDdivisibility by 7.Divisibility by 2 requires the number to be even. That means it must end in 0, 2, 4, 6 or 8.Divisibility by 3 requires that the digital root of the number is divisible by 3. That is, the sum of the digits is divisible by 3. If the first sum is large, you can calculate the digital root of the digital root (and again, if necessary) and check that for divisibility by 3.The divisibility rule for 7 is more difficult.Take the last digit (units).From the number formed by the remaining digit, subtract twice the last digit.If the answer is divisible by 7 (including zero or negative numbers), then the original number is divisible by 7.For large numbers, repeat the process to bring the number down to a manageable size.
no it does not and the smallest missing factor that would allow for divisibility is 13
The special divisibility rule of 17 is "Take the difference of the rest of the digits and the last digit times 5." For instance: 17 1 - 7 * 5 = 1 - 35 = -34 which is divisible by 17.
It is somebody talking about divisibility.
2,3,9,6
7 is only divisible by 1 and itself.
Yes. 210 is evenly divisible by seven.
The number must be divisible by 2 and by 7.
2 x 7 x 7 = 98
If the last digit doubled subtracted from the rest is a multiple of 7, the whole number is divisible by 7.
It is: 7
All Factors of 91:1, 7, 13, 91
If a number is divisible by 5 and 7, it's divisible by 35.
Double the last digit and subtract the last digit from the remaining digits.
The divisibility rule for 42 isdivisibility by 2, ANDdivisibility by 3, ANDdivisibility by 7.Divisibility by 2 requires the number to be even. That means it must end in 0, 2, 4, 6 or 8.Divisibility by 3 requires that the digital root of the number is divisible by 3. That is, the sum of the digits is divisible by 3. If the first sum is large, you can calculate the digital root of the digital root (and again, if necessary) and check that for divisibility by 3.The divisibility rule for 7 is more difficult.Take the last digit (units).From the number formed by the remaining digit, subtract twice the last digit.If the answer is divisible by 7 (including zero or negative numbers), then the original number is divisible by 7.For large numbers, repeat the process to bring the number down to a manageable size.