ATM with strange policy

Both 306 and 221 are multiple of 17, you can withdraw only multiple of 17. The largest multiple of 17 not exceeding $1000 is $986. Or You will withdraw $986 leaving $14 in your account $1000 \equiv 14\mod 17.$ Modular Arithmetic mod

Then the maximum amount of money that you can withdraw is $= 1000 - 14 = 986$ . This can be achieved as follow:-

$\begin{aligned} (\text{Withdraw/Deposit actions} &= \text{current blance in the account })\\ 1000 - 306 - 306 - 306) &= 82 \\ 82 + 221 + 221 -306 &= 218 \\ 218 +221 -306 &= 133 \\ 133 + 221 - 306 &= 48 \\ 48 + 221 +221 &= 490 \\ 490 -306 &= 184 \\ 184 + 221 &= 405 \\ 405 - 306 &= 99 \\ 99 + 221 &= 320 \\ 320 -306 &= 14 \end{aligned}$

Maximum amount of money that you will get after 9 withdrawals and 8 deposits $= 1000 -14 = \$986$

HCF of 306 and 221 is 17. The closest multiple of 17 to 1000 is 986.

$306=18\times17$ , $221=13\times17$ . So only multiples of $17$ can be handled.

$1000=58+\frac{14}{17}$ , so the absolute best would be to leave only $\$14.$

To accomplish this, we need to find $a$ and $b$ such that $18a-13b=58$ .

I did this by trial and error, with a little help of Excel. I made a table with multiples of $18$ over the top in line $1$ , multiples of $13$ down column $A$ and then into $B2$ placed the formula $=B\$1-\$A2-58$ . Copying that into the table made that zero jump right at me. So I found that $a=9$ and $b=8$ .

So if we take $9\times 306$ and deposit $8\times 221$ , we should be fine.

We can only add money when we have it, so that requires collecting enough cash for the purpose. If everything is handled in $20 and $1 bills, there should be no problem with the currency. We can simply alternate between taking out and putting back 8 times, then take out one more withdrawal.

306 - 221 = 85. So, you have two functions you can perform: add 306 or add 85. 85, being the smaller number, and subsequently approaching 1000 with less difference, will become 85x < 694, because 85x + 306 cannot be higher than 1000. 85* 8 = 680. 680 + 306 = 986 .

I did a little cheating in there. I defined it as an optimization problem, setting the cost function as

$\max{306m-221n}$

and with the restrictions

$306m-221n <= 1000$

$m-n > 0$

and $m$ and $n$ must be integers.

The optimization returns $m=9$ and $n=8$ , which gives us a maximum amount of money of $986$

Just follow this simple Python code

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13

    bank = 1000
    me = 0
    x = []
    while bank > 0 and me < 1000:
        me += 306
        bank -= 306
        x += [me]
        if not (bank > 0 and me < 1000):
            break
        me -= 221
        bank += 221
        x += [me]
    print sorted(x)[:-1] 

It can be viewed like this You can remove either 85 ( difference between two numbers ) or remove 306 ...now maximising 85x + 306y < 1000 for values of integral x and ys we get the max as 986 Where x and y are no. Of 85s and number of 306s taken out respectively... I'm new please point out if I'm wrong