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Solution. Verified by Toppr. We know, first two digit number divisible by 3 is 12 and last two digit number divisible by 3 is 99. Thus, we get.
Three digit numbers divisible by 3 are 102, 105,....., 999 It is nothing but a series of AP whose common difference is 3 and first term is 102 ⇒ A.P with a = 102 , d = 3 , a n = 999
Are all numbers divisible by 3 also by 9 give example? All numbers divisible by 3 are NOT divisible by 9. As an example, 6, which is divisible by 3, is not divisible by 9.
Divisibility Test by 7. 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. If the sum is divisible by 7 then it passes the divisibility test of 7.
Open in App. Solution. Verified by Toppr. A number is divisible by 3 if the sum of its digits is divisible by 3. 18=1+8=9. 60=6+0=6. 24=4+2=6. Therefore, all are divisible by 3. So, option D is correct.
Because a − c = (a − b) + (b − c) = 3 k + 3 p = 3 (k + p) which is divisible by 3. Therefore, the given relation R is a transitive relation. Since, the given relation R satisfies the reflexive, symmetric, and transitive relation properties, Therefore, it is an equivalence relation.
∀ n ∈ N, 4 n − 3 n − 1 is divisible by. View Solution. Q5. Prove that 2 4 n ...
We know the divisibility rule for 3: If the sum of the digits of a number is divisible by 3, then the number is also divisible by 3. Here, 9 + 9 + 6 = 24 is multiple of 3. So, 996 is the largest number divisible by 3.
Click here:point_up_2:to get an answer to your question :writing_hand:find the sum of all odd integers between 2 and3
Find the sum of. (i) the first 15 multiples of 8. (ii) the first 40 positive integers divisible by (a) 3 (b) 5 (c) 6. (iii) all 3 − digit natural numbers which are divisible by 13. (iv) all 3 − digit natural numbers, which are multiples of 11. (v) all 2 − digit natural numbers divisible by 4. (vi) first 8 multiples of 3.