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Suppose you have a currency, named Miso, in three denominations: 1 Mis [#permalink]

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03 Aug 2017, 18:03

GTExl wrote:

Suppose you have a currency, named Miso, in three denominations: 1 Miso, 10 Misos and 50 Misos. In how many ways can you pay a bill of 107 Misos?

A) 16

B) 17

C) 18

D) 19

E) 20

I get 18.

There has to be a better way to do this problem; I just started with a denomination and used division to calculate possible ways, then considered two denominations (and checked with list of possibilities below).

For ones (1 Miso denomination)

107 can be divided by 1 in one way (107 ones "fit" into 107), that is:

107 ones = ONE way

For tens, first assume there are no fifties, and the remainders, composed of denomination one, will correspond with only one of the ways to use tens.

107 can be divided 10 ways (only 10 tens "fit" into 107), with whatever remainders:

Then fifties and ones: 107/50 = allows for two fifties, with whatever remainder:

50 + 57 100 + 7

= TWO ways

Ways total: 1 + 10 + 5 + 2 = 18

This is probably some horrible combinatorics problem, or a very easy arithmetic problem. Either way, or some other way, I can't see a method other than this one. I have no idea whether or not my method is harebrained. It yields the right answer. Please correct me if I've made a mistake.

Re: Suppose you have a currency, named Miso, in three denominations: 1 Mis [#permalink]

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03 Aug 2017, 20:15

genxer123 wrote:

I get 18.

There has to be a better way to do this problem; I just started with a denomination and used division to calculate possible ways, then considered two denominations (and checked with list of possibilities below).

Too time-consuming, I am looking forward to seeing a math based solution

Suppose you have a currency, named Miso, in three denominations: 1 Mis [#permalink]

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03 Aug 2017, 21:39

cbh wrote:

genxer123 wrote:

I get 18.

There has to be a better way to do this problem; I just started with a denomination and used division to calculate possible ways, then considered two denominations (and checked with list of possibilities below).

Too time-consuming, I am looking forward to seeing a math based solution

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