FINDING LAST DIGIT OF ANY POWER:
Last digits
of the powers of any number follow a cyclic pattern. If we find out after how
many steps the last digit of the powers of a number repeat then we can find out
the last digit of any power of any number. Let us check out the pattern of last
digits for different digits:
1
|
1
|
|||
2
|
2
|
4
|
8
|
6
|
3
|
3
|
9
|
7
|
1
|
4
|
4
|
6
|
||
5
|
5
|
|||
6
|
6
|
|||
7
|
7
|
9
|
3
|
1
|
8
|
8
|
4
|
2
|
6
|
9
|
9
|
1
|
No. ending with 1:
Any power on
number ending with 1 will have unit digit 1.
Ex: 25135
will have unit digit 1.
No. ending with 2:
Last digit
of 21 is2
Last digit
of 22 is 4
Last digit
of 23 is 8
Last digit
of 24 is 6
Last digit
of 25 is again 2
And the
whole cycle is repeated
So there are
only 4 possible unit digits in case of 2. These are 2,4,8,6.
Ex: Find the
unit digit of 298.
Since, in
case of 2, there are 4 possible unit digits as explained above. So the
cyclicity is 4.
Now we will
divide the power by 4 (cyclicity is 4) and check the remainder.
When 98 is
divided by 4 remainder is 2. The unit digit is same as that of 22
which is 4.
No. ending with 3:
Last digit
of 31 is3
Last digit
of 32 is 9
Last digit
of 33 is 7
Last digit
of 34 is 1
Last digit
of 35 is again 3
And the
whole cycle is repeated
So there are
only 4 possible unit digits in case of 3. These are 3,9,7,1.
Ex: Find the
unit digit of 373.
Since, in
case of 3, there are 4 possible unit digits as explained above. So the
cyclicity is 4.
Now we will
divide the power by 4 (cyclicity is 4) and check the remainder.
When 73 is
divided by 4 remainder is 1. The unit digit is same as that of 31
which is 3.
No. ending with 4:
Last digit
of 41 is4
Last digit
of 42 is 6
Last digit
of 43 is 4
Last digit
of 44 is 6
So in case
if 4, there are only 2 possible unit digits which are 4,6
We can also
say that
Unit digit
of 4odd no=4
Unit digit
of 4even no=6
No. ending with 5:
Last digit
of 51 is 5
Last digit
of 52 is 5
Last digit
of 53 is 5
Last digit
of 54 is 5
Unit digit
in case of 5 is 5 only.
No. ending with 6:
Last digit
of 61 is 6
Last digit
of 62 is 6
Last digit
of 63 is 6
Last digit
of 64 is 6
Unit digit
in case of 6 is 6 only.
No. ending with 7:
Last digit
of 71 is 7
Last digit
of 72 is 9
Last digit
of 73 is 3
Last digit
of 74 is 1
Last digit
of 75 is again 7
And the
whole cycle is repeated
So there are
only 4 possible unit digits in case of 7. These are 7,9,3,1.
Ex: Find the
unit digit of 7156.
Since, in
case of 7, there are 4 possible unit digits as explained above. So the
cyclicity is 4.
Now we will
divide the power by 4 (cyclicity is 4) and check the remainder.
When 156 is
divided by 4 remainder is 0. The unit digit is same as that of 74
which is 1.
No. ending with 8:
Last digit
of 81 is 8
Last digit
of 82 is 4
Last digit
of 83 is 2
Last digit
of 84 is 6
Last digit
of 85 is again 8
And the
whole cycle is repeated
So there are
only 4 possible unit digits in case of 8. These are 8,4,2,6.
Ex: Find the
unit digit of 855.
Since, in
case of 8, there are 4 possible unit digits as explained above. So the
cyclicity is 4.
Now we will
divide the power by 4 (cyclicity is 4) and check the remainder.
When 55 is
divided by 4 remainder is 3. The unit digit is same as that of 83
which is 2.
No. ending with 9:
Last digit
of 91 is9
Last digit
of 92 is 1
Last digit
of 93 is 9
Last digit
of 94 is 1
So in case
if 4, there are only 2 possible unit digits which are 9,1
We can also
say that
Unit digit
of 9odd no=9
Unit digit
of 9even no=1
Ques: Find
the unit digit of 152555x 9991000
Ans. Since
we are talking about unit digit we will take 2 in place of 152(unit digit)
And 9 in
place of 999
Unit digit
of 2555: divide the power by 4(cyclicity in case of 2) and check the
remainder
555 gives
remainder 3 when divided by 4. So unit digit is same as that of 23
which is 8
Unit digit
of 91000: We know that Unit
digit of 9even no=1
So unit digit
of 91000 is 1
Required
unit digit is 8 x 1 =8
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