document.write( "Question 770986: Three monkeys ate a total of 25 nuts. Each monkey ate an odd number of nuts and the three odd numbers were all different and bigger than one. How many different solutions are ther? \n" ); document.write( "

Algebra.Com's Answer #469966 by KMST(5328) $\"\"$ $\"About$

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The monkey who ate the least may have eaten $\"3\"$ nuts.
\n" ); document.write( "That would leave $\"25-3=22\"$ nuts for the other two monkeys.
\n" ); document.write( "Along with the 3 for the least hungry monkey, we need two different odd numbers that add up to $\"22\"$ , and we cannot use $\"1\"$ or $\"3\"$ .
\n" ); document.write( " $\"5%2B17=22\"$
\n" ); document.write( " $\"7%2B15=22\"$ and
\n" ); document.write( " $\"9%2B13=22\"$ .
\n" ); document.write( "Since I cannot tell the monkeys apart, and I would not care which one ate the least and which one ate the most, the solutions I have found so far are the $\"3\"$ sets
\n" ); document.write( "{3,5,17}, {3,7,15} and {3,9,13}.
\n" ); document.write( "
\n" ); document.write( "If the monkey who ate the least eat more than $\"3\"$ nuts, he/she may have eaten $\"5\"$ nuts.
\n" ); document.write( "In that case the other two monkeys ate the other $\"25-5=20\"$ nuts. We need to odd numbers that add up to $\"20\"$ and we cannot use neither $\"1\"$ , nor $\"3\"$ , not $\"5\"$ .
\n" ); document.write( "It could be
\n" ); document.write( " $\"7%2B13=20\"$ or $\"9%2B11=20\"$ .
\n" ); document.write( "That accounts for $\"2\"$ more solutions, the sets

or more nuts, because the other two monkeys would have had to eat at least $\"9\"$ and $\"11\"$ nuts, and then the total would be at least $\"7%2B9%2B11=27\"$ nuts.
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\n" ); document.write( "So there are $\"3%2B2=highlight%285%29\"$ solutions, unless you can tell the monkeys apart, and care which one ate more or less than the others.
\n" ); document.write( "If 3, 5, and 17 for monkeys A, B, and C respectively is considered different from 3, 5, and 17 for monkeys B, A, and C respectively, then there are 6 different ways that we can order the set {A,B,C}, and that multiplies times 6 the number of solutions. \n" ); document.write( "

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