show that 2^3n-1 is divisible by 7?
15626
Now, P( 1): 23 - 1 = 7, which is divisible by 7. ... = 7(8m + 1), which is divisible by 7. Thus, P(k + 1) is true whenever P(k) is true. So, by the principle of mathematical induction P(n) is true for all natural numbers n.
1025563
Originally Answered: How can you prove that product of any 2 consecutive even integers is divisible by 8? P =4×n ×(2m ) =8×m×n. Thus proved.
1522
Proving 6n−1 is always divisible by 5 by induction Prove 6n−1 is always divisible by 5 for n≥1. ... Base Case: n=1: 61−1=5, which is divisible by 5 so TRUE. Assume true for n=k, where k≥1: 6k−1=5P.
750877
8 + 7 = 15, a multiple of 3, so 87 is divisible by 3. 87 is a composite number. 87 = 1 x 87 or 3 x 29. Factors of 87: 1, 3, 29, 87.
741123
The numbers from 1 to 70 which are divisible by 2 and 3 are: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, and 66.
948859
While the number should be divisible by 7, since we do not want these numbers to be multiple of 3, we have to get rid of numbers 21, 42, 63 and 84 i. e. 4 numbers out of 14 numbers. 14 - 4 = 10. Thus there are 10 numbers (between 1 and 100) that are divisible by 7 but not 3.
634135
All numbers divisible by 4 are divisible by 2 . But all numbers divisible by 2 are not divisible by 4.
531909
Thank you. n(n^2-1 ) is always divisible by 24 for all odd numbers..
193511
['The prime factorization of 7 is: 7.', 'The prime factorization of 8 is: 2 x 2 x 2.', 'The prime factors and multiplicities 7 and 8 have in common are: 1.', '1 is the gcf of 7 and 8.', 'gcf(7,8) = 1.']
1134217
For the number 3 there are 33 numbers divisible by 3 from 1 to 100. If you want the number divisible by both 3 and 4 it is 58 .
76501
For 7, the answer is: yes, 7 is a prime number because it has only two distinct divisors: 1 and itself (7). As a consequence, 7 is only a multiple of 1 and 7.
1359508
The orange divisor(s) above are the prime factors of the number 30,030. If we put all of it together we have the factors 2 x 3 x 5 x 7 x 11 x 13 = 30,030. It can also be written in exponential form as 21 x 31 x 51 x 71 x 111 x 131.
1033008
Prove by induction that 8n−1 is divisible by 7 Prove by induction that 8n−1 for any positive integer n is divisible by 7. Hint: It is easy to represent divisibility by 7 in the following way: 8n−1=7⋅k where k is a positive integer.
1290692
['Step III: Divide the first divisor by the first remainder. Step IV: ... ', 'Step II: The first divisor is 18 and the remainder is 12, so we need to divide 18 by 12. [Divide the first divisor by the first remainder]. ... ', 'Step I: Here we need to divide 180 by 75. [Divide the larger number by the smaller one].']
1469293
3, 6, 9, 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, 93,96,99 are divisible by 3. 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96 are divisible by 4.
1321220
We know that if the sum of the numbers that make up 25 is divisible by 3, then 25 is divisible by 3. Note: If you divide 25 by 3, you get 8.33 which is not a whole number.
1088485
Explanation: 12−1=0 which is divisible by 8 (base case for a=1 ). Since a is odd, write a=2n−1,n∈N . Assume (2n−1)2−1 is divisible by 8.