Bunuel wrote:
Official Solution:
What is the last digit of the following number \(2^{22} * 3^{15} * 5^{16} * 7^1\)?
A. 6
B. 5
C. 2
D. 1
E. 0
Since we have \(2*5=10\) in the product, the last digit will definitely be 0. The rest of the prime factors don't make any difference to the last digit under these circumstances.
Answer: E
this is a special case with 2 and 5 but to solve similar questions, its better to use cyclic values and then taking unit values.
2 has a cyclic value of 4 .. ie - the values of exponents repeats after 4 increments
ie-
\(2^1\)= 2 (last digit 2)
\(2^2\)= 4 (last digit 4)
\(2^3\)= 8 (last digit 8)
\(2^4\) = 16 (last digit 6)
\(2^5\)= 32 (last digit 2)
\(2^6\) = 64 (last digit 4)
\(2^7\)= 128 (last digit 8)
\(2^8\) = 256 (last digit 6)
\(2^9\)= 512 (last digit 2)
so 2 has a cyclic value of ( 2,4,8,6 )
so unit digit of \(2^{22}\) = \(2^{(4*5)+ 2}\) = unit digit of \(2^2\)= 4
similarly 3 also has cyclic value of 4 ( it repeats every 4 exponents) ie ( 3,9,7,1)
5 has a cyclic value of 1 ( it repeats every 1 exponent) ie (5)
7 has a cyclic value of 4 ( it repeats every 4 exponent) ie (7,9,3,1)
you can calculate the same for every other integers.
so last digit of \(2^{22} * 3^{15} * 5^{16} * 7^1\) = \(2^{20+2} * 3^{12+3} * 5^{15+1} * 7^1\) = \(2^{2} * 3^{3} * 5^{1} * 7^1\) = 4*7*5*7= 49*20 =980 so the
ans is E