To discover whether a number is divisible by 7, double the last digit of the number, and subtract it from the rest of the number. If the result is either 0 or divisible by 7, then the number is divisible by 7. For example:
672 - last digit is 2, double 2 to make 4, subtract from 67 equals 63, 63/7 = 9, therefore, 672 is divisible by 7.
Dividing by 7 (2 Tests)
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If a number is divisible by 5 and 7, it's divisible by 35.
No, 87 is not divisible by 7. To determine if a number is divisible by 7, you can use the rule that states a number is divisible by 7 if the difference between twice the digit in the ones place and the number formed by the other digits is either 0 or a multiple of 7. In this case, the number formed by the other digits is 8, and twice the digit in the ones place is 14. The difference between 14 and 8 is 6, which is not a multiple of 7, so 87 is not divisible by 7.
To find out, if a number is divisible by 7, take the last digit, double it, and subtract it from the rest of the number. If you get an answer divisible by 7 (including zero), then the original number is divisible by 7.If you don't know the new number's divisibility, you can apply the rule again.Eg: Take the number 343The units digit is 3.Doubling this 3,we get 6.Subtracting 3 from 34(that is the remaining 2 digits in the number), we get 34-6=28.As 28 is divisible by 7, The whole number, i.e., 343 is divisible by 7.Proof of Divisibilty Rule for 7Let 'D' ( > 10 ) be the dividend.Let D1 be the units' digitand D2 be the rest of the number of D.i.e. D = D1 + 10D2We have to prove(i) if D2 - 2D1 is divisible by 7,then D is also divisible by 7and (ii) if D is divisible by 7,then D2 - 2D1 is also divisible by 7.Proof of (i) :D2 - 2D1 is divisible by 7 ⇒ D2 - 2D1 = 7k where k is any natural number.Multiplying both sides by 10, we get10D2 - 20D1 = 70kAdding D1 to both sides, we get(10D2 + D1) - 20D1 = 70k + D1⇒ (10D2 + D1) = 70k + D1 + 20D1⇒ D = 70k + 21D1 = 7(10k + 3D1) = a multiple of 7.⇒ D is divisible by 7. (proved.)Proof of (ii) :D is divisible by 7 ⇒ D1 + 10D2 is divisible by 7 ⇒ D1 + 10D2 = 7k where k is any natural number.Subtracting 21D1 from both sides, we get10D2 - 20D1 = 7k - 21D1⇒ 10(D2 - 2D1) = 7(k - 3D1)⇒ 10(D2 - 2D1) is divisible by 7Since 10 is not divisible by 7,(D2 - 2D1) is divisible by 7. (proved.)
If it is divisible by 2 and 3, it is divisible by 6.
A number is divisible by 3 if the sum of its digits is a multiple of 3. For example, 27 = 2 + 7 = 9/3 = 3. Therefore, 27 is divisible by 3.If the sum of the digits of the number in question is divisible by three, the whole number is divisible by three.
If a number is divisible by 5 and 7, it's divisible by 35.
a number is divisible by 12 if the number is also divisible by both 3 and 7. You mean a number is divisible by 21 if it is divisible by 3 and 7.
The number must be divisible by 2 and by 7.
The sum of the factors are divisible by 9. Ex: 738 7+3+8=18 18/9=2
No, 87 is not divisible by 7. To determine if a number is divisible by 7, you can use the rule that states a number is divisible by 7 if the difference between twice the digit in the ones place and the number formed by the other digits is either 0 or a multiple of 7. In this case, the number formed by the other digits is 8, and twice the digit in the ones place is 14. The difference between 14 and 8 is 6, which is not a multiple of 7, so 87 is not divisible by 7.
To find out, if a number is divisible by 7, take the last digit, double it, and subtract it from the rest of the number. If you get an answer divisible by 7 (including zero), then the original number is divisible by 7.If you don't know the new number's divisibility, you can apply the rule again.Eg: Take the number 343The units digit is 3.Doubling this 3,we get 6.Subtracting 3 from 34(that is the remaining 2 digits in the number), we get 34-6=28.As 28 is divisible by 7, The whole number, i.e., 343 is divisible by 7.Proof of Divisibilty Rule for 7Let 'D' ( > 10 ) be the dividend.Let D1 be the units' digitand D2 be the rest of the number of D.i.e. D = D1 + 10D2We have to prove(i) if D2 - 2D1 is divisible by 7,then D is also divisible by 7and (ii) if D is divisible by 7,then D2 - 2D1 is also divisible by 7.Proof of (i) :D2 - 2D1 is divisible by 7 ⇒ D2 - 2D1 = 7k where k is any natural number.Multiplying both sides by 10, we get10D2 - 20D1 = 70kAdding D1 to both sides, we get(10D2 + D1) - 20D1 = 70k + D1⇒ (10D2 + D1) = 70k + D1 + 20D1⇒ D = 70k + 21D1 = 7(10k + 3D1) = a multiple of 7.⇒ D is divisible by 7. (proved.)Proof of (ii) :D is divisible by 7 ⇒ D1 + 10D2 is divisible by 7 ⇒ D1 + 10D2 = 7k where k is any natural number.Subtracting 21D1 from both sides, we get10D2 - 20D1 = 7k - 21D1⇒ 10(D2 - 2D1) = 7(k - 3D1)⇒ 10(D2 - 2D1) is divisible by 7Since 10 is not divisible by 7,(D2 - 2D1) is divisible by 7. (proved.)
If it is divisible by 2 and 3, it is divisible by 6.
7 is not divisible by 104 and 104 is not divisible by 7.
To determine if a number is divisible by 28, it must be divisible by both 4 and 7, since 28 is the product of these two numbers. To check for divisibility by 4, the last two digits of the number must form a number that is divisible by 4. For divisibility by 7, you can either perform direct division or apply the rule of subtracting twice the last digit from the rest of the number until you reach a manageable number. If the resulting number is divisible by 7, then the original number is also divisible by 28.
Not only is 119 divisible by 7 but it is evenly divisible by 7.
No it is not divisible by 7
The divisible rule for 2 is if the last digit of the number is 0,2,4,6, or 8. The divisible rule for 3 is if the last 2 digits of the number is divisible by 3,03,06,09,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,57,60,63,66,69,72,75,78,81,84,87,90,93,96, or 99.