Question

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1. How many integer numbers n such that n^2+3 is divisible by n-1?

2.Caculator A=4+2^2+2^3+...+2^20

3. The last digit of the value of A= 2x4x6x...x48 - 1x3x5x...x49

Answer 1

\(a,n^2+3⋮n-1\)

\(\Rightarrow n\left(n-1\right)+3+n⋮n-1\)

\(\Rightarrow n+3⋮n-1\)

\(\Rightarrow n-1+4⋮n-1\)

\(\Rightarrow4⋮n-1\)

...........You make the table, If i make the table, it will spent my times

That is answer:6 intergers

\(b,A=4+2^2+2^3+...+2^{20}\)

Put \(B=2^2+2^3+...+2^{20}\)

\(\Rightarrow2B=2^3+2^4+...+2^{21}\)

\(\Rightarrow2B-B=\left(2^3+2^4+...+2^{21}\right)-\left(2^2+2^3+...+2^{20}\right)\)\(\Rightarrow B=2^{21}-4\)

`\(\Rightarrow A=2^{21}-4+4=2^{21}\)

c,Consider the range of values

Put B=2.4.6.8...48

Because of B=2.4.6...10....48

=> The last digit of B is 0

Put C=1.3.5...49

Because of C= 1.3.5...49

=> The last digit of C is 5

=> B-C will have the last digit is 5

=> A have the last digit ís 5