Consecutive Integer

[grayfalcon89]

Prove that any five consecutive integer is divisible by 5 such that the least term is > 0.

 

Prove that any two integer's sum is NOT always divisible by 3. State why this is and tell in what case is this possible.

 

Prove that:

 

[math]\binom {n}{k}[/math] equals [math]\binom {n}{n-k}[/math]

[matt grime]

the first question makes no sense. i think the word produc is missing.

 

the third one is obvious from the definition.

[bloodhound]

for the second one i dont even know what there is to prove actually. 0 + 0 anyone?

[JaKiri]

I presume the 2nd one refers to 'consecutive integer'.

 

Let us set our numbers as x (divisible by 3), x+1 and x+2. For example, 3, 4 and 5.

 

x+1 + x+2 = 2x+3, divisible by 3. However, any sum involving the multiple of 3 cannot be divisible by 3, because you end up with either 2x+1 or 2x+2.

 

Re: the first one,

 

Basically it's the same answer as the 2nd one. Taking 5 consecutive numbers, one MUST be divisible by 5, and thus so must the product. It's fairly clear where this assertion comes from.

[matt grime]

0+0=0 which is divisible by three, bloodhound. I dont' see why the "general" disproof was offered, since it only requires one counter example, eg 0+1.

 

i'm also mystified as to why the original 1st question limits us by some positvity condition - if the question is to make sense there is no reason for this.

[bloodhound]

0+0=0 which is divisible by three' date=' bloodhound. I dont' see why the "general" disproof was offered, since it only requires one counter example, eg 0+1.

 

[/quote']

yes sorry, i meant exactly that. i should have made that a bit clear hehe.

[Primarygun]

Prove that any five consecutive integer is divisible by 5 such that the least term is > 0.

For 5 con. integer, the reminder is 1,2,3,4,0 = 1,2,-2,-1,0 ( By definition of mod ) Or

Sum of reminder = 1+2+3+4+0= 10 which is divisible by 5 [Two approach]

[Primarygun]

Prove that any two integer's sum is NOT always divisible by 3. State why this is and tell in what case is this possible.

Let x and y be two integers

So x+y=3n ( n : integer)

(x+1)+y=3a (a: integer)

[Math]---> 3(n-a)=1 [/ Math]

[Math]--->n-a=1/3 [/ Math]

Do it with (x+2)----> n-a=2/3

Since n and a are integers , the different of them is integers.

----------->Not the sum of any two integers is divisible by 3.

The sum of two integers is divisible by 3 occurs when the sum of the two reminders of the integers is divisible by 3.