Question

Fool's problem of the day,

A father wishes to distribute \( \$1400\) among this three sons such that the eldest one gets the maximum amount, the middle one gets more money than the youngest one. In how many ways can the father distribute the money if each of his sons gets an amount in a multiple of \(100\)?

Answer 1

we are partying now @lalaly right ? :D so later :D

Answer 2

hahaha i can multitask

Answer 3

""today i dont feel like doing anything"" :D

Answer 4

Multiple of 100 -> neglect the 2 '0's
c<b<a
When c get 1, b at least need to get 2 then a would get 11
c get 2, b at least need to get 3 then a would get 9 ...
The max amount b can get when c gets 1 = 6 (a gets 7 in that case)
2, 3, 4, 5, 6 => 5 cases

When c get 2, b at least need to get 3 then a would get 9
The max amount b can get when c gets 2 = 5 (a gets 7 in that case)
3, 4, 5 => 3cases

When c get 3, b at least need to get 4 then a would get 7
The max amount b can get when c gets 3 = 5 (a gets 6 in that case)
4, 5 => 2 cases

When c get 4, b at least need to get 5 then a would get 5 (rejected)

Total number of ways = 5+3+2 = 10ways

Answer 5

the youngest must get either 100,200, or 300

yes solution is 10
@Callisto beat me to it

Answer 6

@dumbcow can you do it for n 100$ packs?

Answer 7

Well done calli!

Answer 8

One more point... To get max amount b can get, use (14-amount c gets)/2 . not integer -> round down ; integer -> the number before it (since a and b get equal amount if it is an integer). ie, in second case, it is (14-2)/2 = 6, number before it =5, so count from 3 to 5, you'll get 3 cases.

Answer 9

@Ishaan94 , if total was 100n
then number of ways would look like this
\[\lfloor \frac{n-4}{2} \rfloor+\lfloor \frac{n-7}{2} \rfloor+\lfloor \frac{n-10}{2} \rfloor ...\]
for (n-a) >=2