Number theory : Digit Fundas  Vikas Saini

How to find unit digit :
Number
Unit digit
Repetition after(cycle)
1
1
Always 1
2
2,4,8,6,2,4,8,6
4
3
3,9,7,1,3,9,7,1
4
4
4,6,4,6
2
5
5,5,5
Always 5
6
6,6,6
Always 6
7
7,9,3,1,7,9,3,1
4
8
8,4,2,6
4
9
9,1,9,1
2
Main points :
1. unit digit of any power of 1,5,6 are 1,5,6 respectively.
1^n = 1.
5^n = 5.
6^n = 6.
2. unit digit of 2,3,7,8 repeat after power of 4.
3.unit digit of 4,9 repeat as per odd/even term.
4^odd = 4.
4^even =6.
9^odd = 9.
9^even = 1.
Q. Find unit digit of 97^473 .
Solution :
Unit digit is 7,forget other digits.
To get unit digit of 7, we use cycle of 4.
473 mod 4 = 1.
Remainder is 1.
7^1 = 7.
Unit digit is 7.
Q. What is unit digit of 16^48 + 17^48 + 18^48 + 19^48 ?
Solution :
We need to take just unit digit of every number.
48 is common as power of each number.
48 mod 4 = 0.
Although remainder is 0 here, we consider it power of 4.
6^(anything) = 6.
7^4 = 1.
8^4 = 6.
To get unit digit of 9, we need to check power is even or odd.
48 is even.
9^even = 1.
Sum of each unit digit number = 6+1+6+1 = 14.
Hence unit digit is 4.
Q. Find unit digit of 1^14 + 2^14 + 3^14 …100^14.
Solution :
14 mod 4 = 2.
We need to find unit digit number just from 1 to 9,then multiply by 10.
Unit digit of 1,5,6 is 1,5,6 respectively.
Unit digit of 2,3,7,8 is 4,9,9,4 respectively.
Unit digit of 4 and 9 is 6 and 1.
Add all = 1+5+6+4+9+9+4+6+1 = 45.
We will multiply it by 10 because unit digit will be as same as 1 to 9 of 11 to 19,21 to 29 and so on.
Now 45 x 10 = 450.
Hence unit digit 0.
Q. Find unit digit of ((1997)^41)^401.
Solution :
Unit digit is 7.We have to use cycle of 4.
41^401 mod 4 = 1.
7^1 = 7.
Unit digit is 7.
How to find last two digits :
To get unit digit number we see only unit digit whereas to get last two digits, we see only last two digits.
If any number’s unit digit is 1, then to get last two digits we apply this formula
(xyz1)^abc = (c x z)/1 (last two digit)
Then unit digit is nothing but 1.
Ten’s digit = z x c.
Find the last two digits of 21^3.
Unit digit is 1.
Ten’s digit = 2 x 3 = 6.
Hence last two digits of 21^3 is 61.
(21^3 = 9261).
Find the last two digit of 91^2.
Unit digit is 1.
Ten’s digit = 9 x 2 = 18 = 8.
Hence last two digits of 91^2 = 81.
(91^2 = 8281)
Shortcut approaches we will apply here :
1. Try to get unit digit 1, so that we could apply above written formula.
Though it’s possible in case of odd integers only.
2. We need to find remainder of power after dividing by 40 only.
3. In case of even number, we shall try to make power of 2.
2^10 = 24 (last two digits)
2^20k = 76(last two digits)
Q. Find the last two digits of 147^912.
Solution :
We need to focus on 47 from 147.
We know for 7, cycle of 4 works.
7^4 = 1(unit digit)
Hence 47^4 = 81(last two digits).
912 mod 40 = 32.
32 = 4 x 8.
Therefore 47^912 = 81^8.
We have unit digit 1,we can apply formula here.
Unit digit = 1.
Ten’s digit = 8 x 8 = 64 = 4.
Hence last two digits = 41.
Q. Find the last two digits of 296^144.
Solution :
We take only 96.
As we can’t get 1 in any power of 6.
So 144 mod 40 = 24.
Therefore 296^144
= 96^24
= 16^12
= ((2)^4)^12
= 2^48
= 2^(20k+8 )
= 2^20k x 2^8
= 76 x 56
= 56.
Hence last two digits are 56.
Q. Find last two digits of 488^488.
Solution :
We take 88 from 488.
488 mod 40 = 8.
= 88^8
= (2^3 x 11 )^8
= 2^24 x 11^8
= 2^(20K+4) x 11^8
= 2^20k x 2^4 x 11^8
= 76 x 16 x 81
= 16 x 81
= 96
Hence last two digits 96.
Q . Find the last two digits of 297^455.
Solution :
We take 97 from 297.
455 mod 40 = 15.
Therefore 297^455
= 97^15
= (97)^4k + 3
= (97^4)^3 x 97^3
=81^3 x 97^2 x 97
= 41 x 09 x 97
= 69 x 97
= 93.
Hence 93.
Miscellaneous
Q. In first 252 natural numbers, how many times we need to write number 4?
Solution :
Unit place = 10+10+5=25.
Ten’s place = 10+10+10 = 30.
Total times = 25 + 30 = 55.
Q. In first 252 natural numbers, how many numbers contain digit 4?
Solution :
Unit digit = 10+10+5 = 25.
Ten’s digit = 10+10+10 = 30.
Unit & Ten’s digit both = 3(44,144,244)
Total numbers = 25+303 = 52.
Q. In book has 100 pages, how many digits have been used to number the pages ?
Solution :
1 9 = 9 digits
10 – 99 = 2 x 90 = 180 digits.
100 = 3 digits.
Total = 9 + 180 + 3 = 192.
Q. There are 3 consecutive natural numbers such that square of the second is 19 more than 9 times of sum of first & third number. Find second number.
Solution :
Let 3 consecutive numbers are n1, n and n+1.
= > n^2 – 19 = 9(n 1+n+1)
= > n^2 – 19 = 9 x 2n
= > n^2 – 18n + 19 = 0.
= > (n19) (n+1) = 0.
= > n = 19,1.
Second number is 19.
Q. Let N = 999999………………….36 times.
How many 9’s are there in N^2 ?
Solution :
99^2 = 9801 (one 9)
999^2 = 998001 (2 times 9)
9999^2 = 99980001 (3 times 9)
Therefore (99999……….36 time)^2
= 35 times 9.
Q. How many positive numbers are there which is equal to 12 times sum of their digits ?
Solution :
For 2 digit number
Let the number is 10a+b.
10a + b = 12(a+b)
= > 10a + b = 12a + 12b
= > 2a = 11b
None two digit number exist here.
For 3 digit numbers
Let the number is 100a+10b+c.
100a + 10b + c = 12(a+b+c)
= > 100a + 10b + c = 12a + 12b + 12c
= > 88a – 2b = 11c.
b = 0.
= > 88 a = 11 c.
= > 8a = c.
a = 1, c = 8.
Number 108.
For 4 digit number
1000a+100b+10c+d
1000a+100b+10c+d = 12(a+b+c+d)
= > 1000 a + 100 b + 10 c + d = 12 a + 12 b + 12 c + 12 d
= > 988 a + 88b = 2c+11d
Not possible.
Hence only positive number is 108.
Q. If we arrange 4 different color balls like 1 red,2 black,3 green,4 white,5 red,6 black, 7 green, 8 white,9 red balls in a row.
What is number of 100^{th} black ball ?
Solution :
Number\Balls
Red
Black
Green
White
1
1
2
3
4
2
5
6
7
8
3
9
10
11
12
4
13
14
15
16
5
17
18
19
20
6
21
22
23
24
7
25
26
27
28
8
29
2


Total
120
100
105
112
Total = 120+100+105+112 = 437.
437^{th} ball is the 100^{th} black ball.
Q. Volume of a cuboid is 231 cubic unit. Find the minimum sum of all the edges.
Solution :
A x B x C = 231.
231 = 3 x 7 x 11.
Minimum sum = 3 + 7 + 11 = 21.
Q. If A,B,C,D,E represent distinct digit from 1 to 5.
ABE = E x BE.
CDA = CA x AA.
BCA + CCD = EEE.
Find the value of AB x CE.
Solution :
Here given E x E = E and A x A =A.
So E = 1 or 5, A = 5 or 1.
Let’s put E = 1 and A = 5.
5B1 = 1 x B1
Not possible.
Let’s put E=5 and A=1.
1B5 = 5 x B5
So B = 2.
BBA + CCD = EEE.
221 + CCD = 555.
1+D = 5.
D = 4.
2+C=5.
C = 3.
AB x CE = 12 x 35 = 420.