math kangaroo contest question

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??????????? "natural numbers" ?????????

em thangs be INTEGER

produkt av 1*2*3*... is kalled the FAKTORIAL

50!=30,414,093,201,713,378... 65 digits

it hav 12 zeroes at rite end
by

To calculate the number of zeroes all we need to do is to consider the different numbers that generate at least one zero at the end of their product. 10=2*5 is the main generator, so (2*5)^n generates n zeroes at the end of the product. So we have between 1 and 50: 2, 4, 8, 16 and 32; and 5 and 25; but we also have 10, 20, 30, 40 and 50, as well as 15, 35 and 45 which generate a zero with even numbers: 6, 12, 14, 18, 22, 24, 26, 28, 34, 36, 38, 42, 44, 46, 48. We can't use any natural number more than once, so we need to pair up just three even numbers with the three ending in 5. No other numbers between 1 and 50 will generate any more zeroes.

So let's collect the generators.

2*5, 4*25, 8*15, 16*35, 32*45 generate 6 zeroes between them.

10*20*30*40*50=12000000 generate a further 6 because 20*50=1000, giving an extra zero.

Therefore, there must be 12 zeroes at the end of 50! (shorthand for 1*2*3*4*...*50).

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