There can be direct questions on finding last two digits of any power of a natural number
and this is very important for eliminating options in questions. Following are the four simple cases
with which we will be able to solve almost every problem.
Rules for Finding Last Two Digits of square of a number
In order to find the last two digits of square of any two digit number number, we can write it as difference or sum from 50 or 100, whichever
is closer. In general the last two digits of a number (50±k)^{2} or (100-k)^{2} will be determined by square of k.
e.g 47^{2} can be written as (50-3)^{2},
Now, the last two digits will be determined by square of 3 that is 09
Rules for Finding Last Two Digits of any power of a number
For a natural number N^{k}, where N is a natural number ending with 0 and K is a natural number greater than 2,
Last two digits will always be 00.
For a natural number N^{k}, where N is a natural number ending with 5 and K is a natural number greater than 2,
Last two digits will always be 25.
For a natural number (2×m)^{40k+1}, where m is a odd number not ending with 5 and K is a natural number,
Last two digits will always be (2×m +50).
For all the remaining cases, the last two digits of N^{40k+x},
Last two digits will always be equal to the last 2 digits of N^{x}
Question 1 Find the last two digits of 12345^{98745}.
05
00
25
50
C
Since the number ends with 5, thus rule 2 applies here
∴ the last two digits will be 25.
Question 2 Find the last two digits of 39^{2}.
21
81
79
67
A
Since 39^{2} can be written as (50-11)^{2}
∴ last two digits will be determined by 11^{2}=21
Question 3 Find the last two digits of 22^{482}.
84
48
92
64
A
Since no rule applies to this, thus we will apply rule 4 here
22^{482} can be written as 22^{40×12+2}
∴ the last two digits can be determined by 22^{2} and will be 84
Question 4 Find the last two digits of 42^{801}.
29
92
45
54
B
Since the number can be written as (2×21)^{40×20+1}, thus rule 3 applies
∴ the last two digits will be (2×21+50)=92
Question 5 Find the last two digits of 39^{900}.
01
81
27
34
A
Since no rule applies to this, thus we will apply rule 4 here
39^{900} can be written as 39^{40×22+20}
∴ the last two digits can be determined by 39^{20}
Now we will keep squaring 39 to reach closet to 39^{20}, we will consider only the last two digits.
39^{2}- Last two digits will be 21, (using rule to find last two digits of square of a number (50-11)^{2})
39^{4}- Last two digits will be 41, squaring 21
39^{8}- Last two digits will be 81, squaring 41 (using rule to find last two digits of square of a number (50-9)^{2})
39^{16}- Last two digits will be 61, squaring 81 (using rule to find last two digits of square of a number (100-19)^{2})
Now, last two digits of 39^{20}=Last two digits of (39^{4}×39^{16})
=Last two digits of (41×61)
=01
∴ last two digits of 39^{900}=01
Question 6 Find the last two digits of 53^{2}.
01
02
03
04
D
Since 81^{2} can be written as (50+2)^{2}
∴ last two digits will be determined by 2^{2}=04
Question 7 Find the last two digits of 81^{2}.
21
61
27
67
B
Since 81^{2} can be written as (100-19)^{2}
∴ last two digits will be determined by 19^{2}=61
Question 8 Find the last two digits of 80^{754}.
64
16
00
08
C
Since the number ends with 0, thus rule 1 applies here
No comments :
Post a Comment