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

Two natural numbers a and b are given in base 10. The number a can be written

as 212 in base b and 128 in base b + 2.

13.

The value of a + b in base 10 is

1. 219

2. 125

3. 114

4. 107

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.

These questions are very difficult to solve. Not very student is capable enough to do such question. Thats the reason students opting for Online Assignment Writing Services these days to complete their assignments.

Great thread ,i tried to do these but not able to do..thanks for sharing this answer.

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?

**In a test match in cricket, the scores of Rohit and Virat in the first innings are in the ratio of 13 : 16. In the second innings as compared to the first innings, their scores increase by the same number of runs and their scores are in the ratio of 53 : 62 in the second innings. What is the ratio of Virat’s second innings score and his first innings score?**

**A. 5: 4B. 31: 24C. 30: 23D. 31 : 25**