the product of two consecutive integers is always divisible by?
103253
Thus the product of two consecutive integers has three '2's as factors and hence 8 as factors. Thus the product of any two consecutive integers is divisible by 8.
426167
If A and B are independent, P(exactly one of them occurs) = P(A) + P(B) — 2P(A)*P(B)
298888
Assume True: 5n – 1 is divisible by 4. This means that 5n – 1 = 4a for an integer a. Then Prove: 5n+1 – 1 is divisible by 4. This means to prove that 5n+1 – 1 = 4b, for some integer b.
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.
812457
yes , the quotient of two integer is always a rational number. therefore it can always be represented in the form of p/q where p and q are relatively co- prime.
573172
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.
923511
Understand that integers can be divided, provided that the divisor is not zero, and every quotient of integers (with non-zero divisor) is a rational number. If p and q are integers, then -(p/q) = (-p)/q = p/(-q).
1055175
The product of two consecutive even integers is 14 more than their sum. Find the integers. Hi Terry, Since the smaller of the two integers is even it is divisible by 2 and hence it can be written 2n for some integer n.
656497
There are proofs on this site, but they are either wrong or overcomplicated. Since one of those first numbers had to have been divisible by 3, this new product (n)(n2−1) must also be divisible by 3. ... If one of those has to be divisible by 3, then n2+1 cannot be.
584018
Given two consecutive numbers, one must be even and one must be odd. Since the product of an even number and an odd number is always even, the product of two consecutive numbers (and, in fact, of any number of consecutive numbers) is always even.
1065212
Prove by induction that 8n−1 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. ... If n=1 and k=2 for example, then we end up with 7=14 which is obviously invalid.
1415904
We have calculated all the numbers that 29 is evenly divisible by. The numbers that 29 is divisible by are 1 and 29 and 29.
705014
Two consecutive natural numbers are those positive integers, which are next to each other. Therefore, there is no gap or difference between any two of these consecutive natural numbers, simply these numbers are next to each other.
1233425
The degree of the remainder is always less than the degree of the divisor in case of polynomial division. ... But degree of (y+5) is 1. So it can't be the remainder.
1176550
Is the integer number 0 divisible by 2? 0 is divisible by 2 (2 | 0). Zero is divisible by any number other than zero.
1394903
6n+2+72n+1 is divisible by 43. Since both forms in this sum are divisible by 43(the first because it is 43 times an integer and the second by the assumption of the induction step), it follows that 6k+2−72k+1 is also divisible of 43. Thus, by the principle of mathematical induction, 6n+2—72n+1 is a divisible of 43.
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0 is divisible by 11 (11 | 0). Zero is divisible by any number other than zero.