Dear students,

Solve this question.

What is the unit digit of the product

7^{1}.7^{2}.7^{3}.7^{4}.……………..7^{2020}

7^{1}.7^{2}.7^{3}.7^{4}.……………..7^{2020}

= 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 = 7^{2} =>9

Sir, How to find last two digits of (1785)^95 and how do we know whether it will end in 25 or 75

(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 sir, please explain

The difference between a two digit number and the number obtained by

interchanging the positions of its digits is 36. What is the difference between the

two digits of that number?

(a) 4 (b) 9

(c) 3 (d) Cannot be determined

Hello abhi,

Find the solution below

Let two digit number is ab

(10a + b) – (10b +a) =36

9a -9b = 36

a – b = 4

Find the last three digits in the binary expansion of the number 43211724?

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.

What is the unit digit of (1/5)^2000

(1/5)^{1} = 2/10 = 0.2 => same as the last digit of 2^{1}

(1/5)^{2} = 1/25= 4/100 = 0.04 => same as the last digit of 2^{2}

(1/5)^{3} = 1/125 = 8/1000 = 0.008 => same as the last digit of 2^{3}

(1/5)^{4 }= 1/ 625 = 16/10000 0.0016 => same as the last digit of 2^{4}

…………………

thus, the last digit of ( 1/5)^{2000} is same as the last digit of 2^{2000}

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

how many perfect squares of natural numbers between 30000-50000 are there whose last two digits end with 81?

The 111th digit of the series 149162536496481........ is?

what are the last two digits of 46^23 - 23^46?

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