gremather wrote:
Hi, I don't quite understand your last sentence. What do you mean by the limiting number?
amorphous wrote:
realize that for a 0 to occur there has to be a multiplication of 5 and 2
simplify the 1st term 33445566= 33∗(22)4∗55∗(2∗3)6 = 3^9* 2^1^4* 5^5
Similarly simplify the 2nd term that should come out to be 3^9* 2^1^3* 5^4
Subtracting 2nd term from 1st term:take the common term which is whole of the 2nd term
3^9*2^1^3*5^4(10−1)
now we have to find out the number of zeros in the common term because non common term is 9
2 and 5 multiply to 10. Here the limiting number is 5 which is equal to 4 hence 4 zeros
since for a '
0' to occur
5 has to be multiplied by
2. The number of zeros will depend on the minimum power raised of either of the two numbers
For eg.
100=52∗22=2 zeros at the end (because both the terms have power raised to 2)
125=53∗20=0 zeros at the end (because 2 is raised to a power of 0 hence 2 becomes the limiting number)