How have the right to you use 50 American coins to add up to a dollar? The coins available are a penny (1 cent), a nickel (5 cents), a dime (10 cents), a 4 minutes 1 (25 cents), and a half-dollar (50 cents).

You are watching: 50 coins to make a dollar

If you have not make the efforts to solve it, have a walk now. For anyone else, the price is after the break.You should use is 40 pennies, 8 nickels (40 cents) and 2 dimes (20 cents). Go you solve it? any other answers?

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Pete
says:

I did settle it, but only with trial and also error. Feels prefer there need to be a tidy formula to reach the result

tibo says:
Eddie says:
Smylers says:

I created a quick Perl regime to uncover all the ways of making \$1 out of those coins, then display screen just those i m sorry consist that 50 coins.

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There’s fixed anything to the algorithm: most of it it just separating it down right into simpler instances that have the right to be addressed trivially — view the comments because that details.

Running it finds both solutions:

\$ ./50_coins50 coins: 45 × 1¢ + 2 × 10¢ + 1 × 25¢ + 2 × 5¢50 coins: 40 × 1¢ + 2 × 10¢ + 8 × 5¢

Here’s the program, heavily commented:

#! /usr/bin/perl

# 50_coins## do \$1 out of 50 coins: https://rewildtv.com/blog-2/

use v5.10;use warnings;use strict;

use List::AllUtils qw;

# find all the methods 100 can be damaged down into the available coins:foreach my \$count (breakdown(100, (50, 25, 10, 5, 1)))

# count the complete coins used, and also display if there’s 50: mine \$coins = sum values %\$count; speak qq<\$coins coins: >, join q< + >, map qq<\$count->\$_ × \$_¢> sort tricks %\$count if \$coins == 50;

sub breakdown# returns all the methods of make \$target using
_;

# If the target is zero, then we can make the (using no coins): return if \$target == 0;

# Otherwise if there room no coins us can’t make anything: return if !
coin;

# If the target has actually gone below zero then it’s difficult to do anything: return if \$target # at the very least one). The the possible ways of make \$target, they either involve # the very first coin or castle don't — for example ways of making \$1 deserve to be split # into those that use at least one 25¢ and those that don't usage 25¢ at all. # work out every of those sets separately, and return the linked list. # # usage this duty again to solve each that the smaller troubles — which might # in turn involve invoking that again, and so on, until it reaches among the # simple cases above. Stop Perl worrying that it'll never end: no warnings qw;

(

# The means of make \$1 making use of at least one 25¢ coin is the very same as including # one 25¢ coin to each of the means of making 75¢ making use of those coins # (including potentionally further 25¢ coins). So find all the methods of # do that smaller sized target utilizing the exact same coins, and add one counting of the # very first coin right into each the the breakdowns: map \$_->\$coin<0>++; \$_ breakdown(\$target – \$coin<0>,
coin)),

# The means of making \$1 there is no using any 25¢ coins is similar to that coin # had actually never been contained in the perform of permitted coins in the an initial place. # So find all the ways of make the exact same target but without using the # very first coin: breakdown(\$target,