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In my previous blogs, I wrote on how to
find the last digit or the units digit of a
number raised to a power, if you hadn't
gone through it, then you may read it from
here.
In this blog, I am writing on how to find the
tens digit of a number when it is raised to
some power.
For example,
If a question is to determine the units place of say, (32)89 then you can
determine that with the help of 'cyclicity of numbers', but what if you
have to determine the second last or the ten's digit of (32)89.
So, for that you can follow the following approach1. The ten's digit of any number ending with 1, i.e.,
having units digit of 1 can be determined as- Tens
digit of number * Units digit of the exponent
For example,
Take an example of say (45151)98
Tens digit of number(i.e., 45151) is 5
Units digit of exponent(i.e., 98) is 8
So, tens digit of (45151)98 = Tens digit of number * Units digit of the
exponent
=5 * 8
=40
So, the tens digit in this case will be 0.
But, what if the number is not ending with 1 or
the units place of the number is not 1.
In that case, try to break it into a number whose units place is
1.
For example, say we have to determine the tens place of (13)89
Here, the number(i.e., 13) has 3 at its units place.
So , we have to obtain a power of 13, which has units place of 1.For
thatUnits place of (13)1 = 3
Units place of (13)2 = 9
Units place of (13)3 = 7
Units place of (13)4 = 1
We have (13)4 = 28561, and it has 1 at its units place.
Now , we can change the original question like this(13)89 = ((13)4)22 * 13
= (28561)22 * 13
Tens digit of (28561)22 = Tens digit of number * Units digit of the
exponent
=6 * 2
=12
Now, Tens digit of (13)89 = Tens digit of (28561)22 * 13
=12 * 13
=156
Thus, tens digit of (13)89 is 6.
2. Any number ending with 24, i.e., whose last 2 digits
are 24 when raised to an odd power, will yield a number
whose last 2 digits are also 24.
3. Any number ending with 24, i.e., whose last 2 digits
are 24 when raised to an even power, will yield a
number whose last 2 digits are 76.
For example, say we have to find the tens digit of (14)33
The number(i.e., 14) is not ending with 24, so we have to break it into
a number which will end in 24.
Now, (14)33 = (2)33 * (7)33
(2)33 = ((2)10)3 * (2)3
(2)33 = (1024)3 * (2)3
1024 is ending with 24 and we have seen, that any number ending
with 24 when raised to an odd power will yield a number having last
2 digits as 24,
So, (1024)3 = (.....24), no need to calculate the complete number
=>(2)33 = (......24) * (2)3
=>(2)33 = (......24) * 8
=>(2)33 = (.........92)
And, last 2 digits of (7)33 can be calculated as shown above, i.e.,
=> (7)33 = ((7)4)8 * (7)
=> (7)33 = (2401)8 * (7)
=> (7)33 = (......01) * (7)
=>(7)33 = (.....07)
Thus, (14)33 = (2)33 * (7)33
=>(14)33 = (.........92) * (.....07)
=>(14)33 = (.........44)
So, the tens digit of (14)33 is 4.
4. Any number ending with 2 or 4 or 6 or 8, i.e., having
units place as 2 or 4 or 6 or 8, when raised to a power of
20k, where k is an integer, then the last 2 digits are always
76.
5. Any number ending with 1 or 3 or 7 or 9, i.e., having
units place as 1 or 3 or 7 or 9, when raised to a power of
20k, where k is an integer, then the last 2 digits are always
01.
For example, say we have to find last 2 digits of (82)41
We can break it like this=>(82)41 = (82)40 * (82)1
=>(82)41 = (.....76) * (82)
=>(82)41 = (.....32)
Thus, last 2 digits are 32.
If you want to practice questions on
finding the tens digit, when a number is
raised to a power, then you can practice
them from here.