umg
What is the last non-zero digit of expression
\(\frac{(25!)}{(2 * 14!)}\)
Where a! = a * (a-1) * (a-2) * (a-3) * ... * 1
A. 2
B. 3
C. 4
D. 5
E. 6
bond001 and for others looking for a quicker way...
The answer depends on the multiplication of units digit, so we have to find a way that eases the calculations..
If pure guess, there are more number of 2s than the number of 5s, so in xyz000, z will be even. Eliminate odd choices, and you are left with 3 options. Now, to be sure of your answer, you will have to multiply all units digit after removing all 5s and an equal number of 2s.
\(\frac{(25!)}{(2 * 14!)}=\frac{25*24*...*20*....*16*15}{2}=25*24*23*22*21*10*19*18*17*16*15\)
Remove the 5s by multiplying with the same number of 2s => 25*4*6*23*22*21*10*19*18*17*8*2*15=100*3*22*21*10*19*18*17*8*3*10
Take only the units digits now...
3*2*1*9*8*7*8*3
Remove 3*3*9 as we will get 1 as the units digit => 2*8*7*8
It's easy to see now that we are getting 8 from 2*8*8, and finally 6 from 8*7
there are various means to cancel out terms to get to singe units digit. you can multiply 3*7 or 9*9 to cancel out those terms.