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Penny Piles (Posted on 2008-04-09) Difficulty: 3 of 5
Susan gave her nephew a number of pennies, as well as a mathematical challenge: to figure out how many ways there were of dividing the pennies into three piles. The pennies are indistinguishable, so the identity of the pennies doesn't matter, nor does the order of the piles. For example, if there had been nine pennies, the piles could have been arranged in any of seven ways: 1+1+7, 1+2+6, 1+3+5, 1+4+4, 2+2+5, 2+3+4, 3+3+3.

There were actually more pennies than this, and in fact, the number of ways was a four-digit number.

However, the nephew misunderstood the instructions. He thought that no two of the piles could be equal, and so came up with a smaller number. For example, if the number of pennies were nine, as above, only three of the arrangements into piles consisted of unique sizes: 1+2+6, 1+3+5, 2+3+4, and the nephew would have reported that, incorrectly.

As mentioned the actual number of ways was a four-digit number. The number reported by the nephew was also a four-digit number, and as a result of his misunderstanding, the only difference between his reported number and Susan's expected answer was that the middle two digits were reversed.

How many pennies did Susan give to her nephew?

See The Solution Submitted by Charlie    
Rating: 3.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re:partitions | Comment 2 of 6 |

IntegerPartitions
IntegerPartition         [n, {k}]
gives partitions into exactly
k integers.

For three piles k=3 so

 

[n, {3}] A=
partitions into exactly
3 piles.      Is the number of

This number equals to number  OF partitions of  N+3  pennies into 3 piles with distinct number of pennies in each pile.

 

 Proof :    Arrange in ascending order each of  all A possible

partitions and then  for each partition   add 1 penny to the second pile and two pennies to the 3rd pile to assure  they are distinct.

To clarify in the problem posted  

 [6, {3}]  =  3   >>>   (114,123,222),

By adding 3 pennies  we get the same number of

distinct  partitions of   9   >>>  (126,135,234)  

          

SOLUTION :

GET(OR GENERATE) A LIST 0F [n, {3}] 

WITHIN A 4 DIGIT RANGE COMPARE[X,{3}] with [X+3,{3}]

FIND THE INVERSED DIGITS AS REQUIRED

DONE   

 

 I am too lazy to do it   ,     but I looked up the posted answer   and it appears to be OK

 

Edited on April 10, 2008, 6:39 pm
  Posted by Ady TZIDON on 2008-04-10 18:34:20

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