Cyclicity of digits
2-2,4,8,6
3-3,9,7,1
4-4,6
5,6-only nos,
7-7,9,3,1
8-8,4,2,6
9-9,1,
Prime nos
1-100=25 prims nos
1-200,=46 prime nos
201-400=32 prime nos
401-600=31 pirm nos
601-800=30 prime nis
PRIMES NOS =6K+/-1 which are not are multiples of 5 and 7
801-1000=29
FIND TOTAL NOS FROM A-B
BETWEEN A-B
B-A+1
(B-A-1)-2
REMAINDER THERORM EG
7327/11 =75/11=2
1,22,333…… FIND 150 TERM
use n(n+1)/2=150 hit and trail 17 answer
a^n+b^n and if n is odd
then a+b is always factor of the expression
least 3 didgit no divided by 3,5,6,7 remainder=1,3,4,5 largest no below 4000
lcm(3,5,6,7)=210k-2 and n is 19
how to find no of zeros
in trailing factoer like 5! find 5 and 2
in genral nos=200=remove lcm
No of factors odd
Product of factors
Odd factors then the number is a perfect square and middle factor is root of the number
P=n^a. Divide power by 2
Find 1889 digit no
Till 2 -189
1889-189=1,700
1700/3=567. The. Number 566+100=666.
1700/3 =566.667
Then. Point is 2/3. The ans 2nd digit
Is 6
Question n=2^5 3^45^2
Total factors =90. All powers +1
Even factors = don’t take 2^0 so answer =75
Odd factors=take 2^0 only the =15
Divisible by 10 = take factors 2*5
Take minimum power so remove power of zero for both number
Then and =50
Perfect square factors =take all even powers and 0=332=10
Perfect cube factors =take 3 power and 0so =4
Nit divvisble by 30=take divisible then minus =50
When 1000 consecutive no starting form 1000!+2 are prime ?
Since 1001!+2 is always an even number
And for all odd number like divide by 17,19 they all there inside so divisible
Prefect numbers
6,28,496 ,8128 …….
420 factorial when divide by 13^n then maximum value of n
For another nos 24 divid 360factorial
Divide 420/13. Quotient =32
Remainder 32/13. Quotient =2
Then n =34 this only works in prime nos
24=2^3 * 3
No of 2=356. So we find 3 2s so 118
And 3s = 178 And =118
Fin no of zeros in 2200! Base 38
38=2*19. The find pairs in 2200!
Then ans=121
Product of factors no real no
Perfect square no
N= N^no of factors /2.
36=36^9/2. So 6^2*9/2. So 6^9
Sum of reciprocal of factors
Sum of factors /N
No of co primes
100 = prime factor 2 and 5
100*(1-1/2) *(1-1/5) =40
Sum of Cl primes less than 100
S=N*Q/2
No of factors of 7200 perfect square
2^53^25^2.
No of factors = (5/2+1) (2/2+1) (2/2+1). Take quotient 322=12
Cube = divide be 3 and =2
For both divide by 6 =find ans
Ulenus theorem
a^n quotient /N. r =1. A,n=co prime N natural nos
17^20|25=1
If divide 4^n/6.
Then Learn r=4 always
If 24 power odd
24 power even
76 any power
1- last 2 digit =24
2.last 2 digit =76
2 power 10=1024
6 power 5 =7776
Divisible rule of 11-7-13.
No 2387643. Make three groups
002-387-643. The. 645-387=258 divide by 11-7-13 not divisible
Abab
abcabc/
1-101
2.1001