Question

prove the product of 4 consecutive integers is always divisible by 24 using the principles of math induction

prove the product of 4 consecutive integers is always divisible by 24 using the principles of math induction. Could anyone help me on this one? Thanks in advance!


Sure
For induction we want to prove some statement P for all the integers. We need:
P(1) to be true (or some base case)
If P(k) => P(k+1) If the statement's truth for some integer k implies the truth for the next integer, then P is true for all the integers.
Look at the first four integers 1,2,3,4. The product 1*2*3*4 is true for the base case n=1,n+1=2,n+2=3 and n+1=4
Suppose now the statement is true for all integers less than or equal to k, so k*(k+1)(k+2)(k+3) is divisible by 24. We want to show that this implies the statement is true for (k+1)(k+2)(k+3)(k+4)
We should observe that with four consecutive numbers 4 will divide one of them say n, and two will divide n-2 or n+2. Three will divide at least one of four consecutive numbers.
You should be able to see that k and k+4 have the same remainder when divided by 4 or 2. This means 8 will divide 4 consecutive numbers. If three does not divide k then it divides one of the next three numbers. If three divides k then it divides k+3. In any case, 8=4*2 and 3 will divide (k+1)(k+2)(k+3)(k+4)
You might be able to simplify my reasoning a little. I didn't proof it closely, so make sure I covered all cases.


thank u very much!
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Answer #1

it just is


source: me
answered by: real.jiffles
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Answer #2

prove the product of 4 consecutive integers is always divisible by 4

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Answer #3

Since the product of n consecutive integers viz. n(n-1)(n-2) ... 3.2.1 = n! is divisible by n!, so the product of any four consecutive integers must be divisible by 4! = 24.

answered by: Dr. Khalid Naeem
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