Đề : Tìm chữ số tân cùng của : \(2^{2003}.4^{2004}.6^{2017}\)
\(2^{2003}=\left(2^4\right)^{500}.2^3\)
\(2^4\) có chữ số tận cùng bằng 6
2^3 có chữ số tận cùng =8
=> 2^2003 có chữ số tận cùng = 8
\(4^{2004}=\left(4^2\right)^{1002}\) có chữ số tận cùng = 6
\(6^{2017}=\left(6^2\right)^{1008}.6\) có chữ số tận cùng =6
=> Tích có chữ số tận cùng = 8
cho mình hỏi 2 câu này:
Question 1:
The difference of the greatest and the least 4 – digit number can you make using all digit 9;3;0;7 is ......
Question 2:
When X divided by 234, the remainder is the greatest possible remainder and the quotient is 5.
The value of X is ..........
how many 3 digit numbers that are multiples of 6 in which the sum of hundreds digit and units and units digit is twice tens digit ?
What is the units digit of 82017 ?
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\(3-3^2+3^3+.....................+3^{2003}-3^{2004}\)
Giúp mình nhé ai nhanh nhất mình tick
Chứng minh rằng: (n+20032004)×(n+20042005) chia hết cho 2.Giải ra nha! Ai nhanh nhất mà đúng nhất mình sẽ tick cho.
