Hey can someone help me with this math riddle? It's a hard problem for me. puzzle?

Question

Ok here it goes...

"Where are those valuable Indian-head pennies I left on the table this morning, James? I put them in a square array and now there are only 2 left. You didn't take them, did you?" "Well sir," replied the bulter, "shortly after you left, three burglars came in. They divided the pennies equally among themselves, but left these 2 because they couldn't divide them equally."

Is James telling the truth?

Any help would be GREATLY appreciated.

Answer 1

If you divide every perfect square by 3 you get an interesting pattern:

4/3=1 remainder 1

9/3=3

16/3=5 remainder 1

25/3=6 remainder 1

36/3=12

49/3= 16 remainder 1

64/3=21 remainder 1

81/3=27

100/3= 33 remainder 1

So it looks like every perfect square divided by 3 either gives us a whole number (ie no remainder) or a remainder of 1. So the butler must be lying since we cannot get a remainder of 2.

Answer 2

Absolutely not. If the pennies were in a square array, then that means there were n² pennies for some n. Now, either n≡0 mod 3, n≡1 mod 3, or n≡2 mod 3. If n≡0, then n²≡0 and the three burglars would be able to divide them equally, and if n≡1 or n≡2, then in both cases n²≡1 so there would be only one penny left. The fact that there were two pennies left means that events could not have transpired in the manner James said they did.

That's the mathematical solution, anyway. But it makes an implicit assumption, however, that the pennies filled the entire array. If the pennies did not fill the array, then James might very well be telling the truth. Personally, I would be sure to ask that question before assuming that James is lying.

Answer 3

Basically, you would need a number that can have a square route, thus the reason its in a square array. However, this number can't be divisable by 2. So, those numbers with square routs, 4, 9, 16, 25, 36, 49, 64, etc. You need to take this number, subtract 2 (thats what's still left) and see if that number is divisable by another number (number of robbers). This isn't possible, so I'd say he's lying.

Answer 4

lying.

if the pennies were laid out in a square, let's say its side equaled X.

Then X = 3k, 3k+1, or 3k+2 for some integer k.

in other words, x is divisible by 3, or it has a remainder of 1, or it has a remainder of 2.

if x=3k, then x^2 = 9k^2 - divisible by 3, so robbers would divide all the pennies and nothing would be left

if x=3k+1, x^2=9k^2+6k+1 = 3(3k^2+2k) + 1 - meaning, if robbers divided pennies into 3, only 1 would be left

if x=3k+2, x^2=9k^2+12k+4 = 3(3k^2+4k + 1) + 1 - meaning, if robbers divided pennies into 3, only 1 would be left

so no way 2 coulda been left

Answer 5

Assuming the square shaped array is in a 3x3 array, the bulter is lieing seeing how there would be 9 coins so the 3 burglars should have been able to equally divide them, not leave 2.