How to find the number of the largest pile

[IBstudent]

Hi,

This is a quesiton that is giving me a very hard time to solve... I know the answer, but could find a way of systematically approaching it...

Maria has 74 pebbles that she wawnts to divide into 25 piles. If the tenth pile is to have more pebbles than any other pile, what is the maximum and mininun numbers of pebbles Maria can put into the tenth pile?

I would really really really appreciate it if someone could show me a good method of approaching such quesitons

Thanks in advance

 

[pka]

Maria has 74 pebbles that she wawnts to divide into 25 piles. If the tenth pile is to have more pebbles than any other pile, what is the maximum and mininun numbers of pebbles Maria can put into the tenth pile?

The answer depends upon how one reads the question.
If we can have empty piles the \(\displaystyle \max=74\).
If we cannot have empty piles the \(\displaystyle \max=50\).

If we cannot have empty piles the \(\displaystyle \min=4\).

 

[IBstudent]

The answer depends upon how one reads the question.
If we can have empty piles the \(\displaystyle \max=74\).
If we cannot have empty piles the \(\displaystyle \max=50\).

If we cannot have empty piles the \(\displaystyle \min=4\).

That is correct... 4 is the mininum and 50 is the maximum... HOWEVER, I am not concerned with the answers but rather how you ended up getting it... how did you approach the question? what calculations did you use?

I would really appreciate it if you could tell me

 

[pka]

That is correct... 4 is the mininum and 50 is the maximum... HOWEVER, I am not concerned with the answers but rather how you ended up getting it... how did you approach the question? what calculations did you use? I would really appreciate it if you could tell me

Simple deductive logic. I thought about.

 

[IBstudent]

Simple deductive logic. I thought about.

That is a very scholarly answer, very detailed and informative...

(incase your mind is inept enough not to fathom the aforementioned insinuation, your answers is the antithesis of what I wrote; superficial and shallow, which might reflect either your incompetance of your sheer apathy, whatever it is, I'm disappointed)

 

[mmm4444bot]

At the intermediate-algebra level, what is an "empty pile" of pebbles? :???:

 

[IBstudent]

Relax, buddy! Your problem would be somewhat clearer if worded this way:

74 pebbles are to be divided into 25 piles, each pile with at least 1 pebble.
The first pile is to have more pebbles than any other pile.
What is the maximum number of pebbles in first pile?
What is the minimum number of pebbles in first pile?

Maximum is EVIDENT:
maximum occurs when piles 2 to 25 are at minimum,
or at 1 each: 74 - 24*1 - 74 - 24 = 50.

Minimum:
1? No (evident)

2? No (evident)

3? if 3, then piles 2 to 25 each have a maximum of 2:
3 + 24*2 = 3 + 48 = 51 ; so no.

4? if 4, then piles 2 to 25 each have a maximum of 3:
4 + 24*3 = 76 (so put 2 in 2 of the piles!) ; YES! *********

If you insist on impressing your drinking buddies with some
sort of "formula", feast your eyes on this:
let m = minimum (follow ********* above):
m + 24(m-1) => 74
m + 24m - 24 => 74
25m => 98
So lowest m = 4: 25*4 = 100 (greater than 98).

Capish?

Thanks

Your answer is explicit and succinct, I like it

Sorry, I seems I am an imbecile when it comes to common sense...