Unit Digits and Last Two Digits

(a5 )^even always ends with 25 

( a5)^Odd => 

If a is odd then last two digits 75

and if a is even then , it ends with 25. 

1st method 

Divide the number by 100 . Remainder will be last two digits of that number 

1785^95 mod 100 

Using Chinese Remainder Theorem 1785^95 mod 4 = 1 and 

1785^95 mod 25 = 0 

So , 25a = 4b + 1 gives the last two digits : 25. 

Alternate Approach 

Last two digits of [1785^95 ] 

= Last two digits of 85^95 

Last two digits of= 17^95 × 5^95 

Last two digits of17 × [(17)²]^47 × 25

Last two digits of [ 17 × 25 × 89^47 ] 

Last two digits of 17 × 25 × ( -11)^47 

Last two digits of 17 × 25 × (-71 ) 

Last two digits of 17 × 25 × 29 = 25 

Hello abhi,

Find the solution below

Let two digit number is ab

(10a + b) – (10b +a) =36

9a -9b = 36

a – b = 4

For last three digits in binary divide the number by 2^3 i.e 8 . 

So 43211724 = 724 mod 8 

4 mod 8 . 

4 in base 2 => 100. 

(1/5)¹ = 2/10 =  0.2 => same as the last digit of 2¹
(1/5)² = 1/25= 4/100 = 0.04 => same as the last digit of 2²
(1/5)³ = 1/125 = 8/1000 = 0.008 => same as the last digit of 2³
(1/5)⁴ = 1/ 625 = 16/10000 0.0016 => same as the last digit of 2⁴

…………………

thus, the last digit of ( 1/5)²⁰⁰⁰ is same as the last digit of 2²⁰⁰⁰

2^(4k) ends with 6 so last digit of ( 1/5)²⁰⁰⁰ => 6.

46^23 = 23^23*2^23
= 23^3*2^3
= 67*08 = 36

similarity last 2 of 23^46 = 23^6 = 67^2 = 89

hence answer is x36-89 = 47 are last 2 digits

Hi,
It is given that a= (212)b .
Forming an equation using this
a= 2b2+b+2
Similarly, a= (128)b+2. Forming an equation for it also.
a= (b+2)2+ 2(b+2)+8
Equate both equations we get,
b= 2 or 7, b cannot be equal to 2 for obvious reasons. Therefore, b=7.
Now, if b=7, using this value and putting it in the first equation.
a= 98+9=107
a + b= 114.

@neha-singh 

@mayanksri 

7¹.7².7³.7⁴.……………..7²⁰²⁰

= 7^{(1 + 2+ 3+ ………….2020)}

=7^{2020(2020+1)/2} = 7^{1010 x 2021}

When 1010 x 2021 is divided by 4 gives remainder = 2 x 1 =2

Unit digit = 7² =>9