What are the last 2 digits of 7 to the power of 2018. This question was on a past exam paper.

Mathematics

1 answer:

Xelga [282]2 years ago

6 0

Answer:

Step-by-step explanation:

Try finding pattern of last two digits of 7 to the power of some numbers.

7^1=07

7^2=49

7^3=343

7^4= 2301

7^5= 16807

so , the pattern is, 07,49,43,01

so, if we start writing all numbers from 1 to 2018 , four number in each line, 2018 will fall in the second column.

so it will have 49 as the last 2 digits [by seeing this pattern : 07,49,43,01 ]

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What is the square root of -625

loris [4]

Answer:

25i

Step-by-step explanation:

Your calculator or your knowledge of powers of 5 will tell you that √625 is 25. The minus sign makes the root imaginary.

$\sqrt{-625}=\sqrt{(-1)(25^2)}=25\sqrt{-1}=25i$

6 0

3 years ago

Read 2 more answers

Please help please please

Grace [21]

$\Omega=\{1;\ 2;\ 3;\ 4;\ 5;\ 6;\ 7;\ 8;\ 9;\ 10\}\\\\|\Omega|=10\\\\A=\{1;\ 3;\ 6;\ 7\};\ B=\{2;\ 3\}\\\\A\ \cup\ B=\{1;\ 2;\ 3;\ 6;\ 7\}\\\\|A\ \cup\ B|=5\\\\P(A\ \cup\ B)=\dfrac{|A\ \cup\ B|}{|\Omega|}\to P(A\ \cup\ B)=\dfrac{5}{10}=\dfrac{1}{2}=0.5=0.50$