Dear students,
Solve this question.
What is the unit digit of the product
71.72.73.74.……………..72020
71.72.73.74.……………..72020
= 7(1 + 2+ 3+ ………….2020)
=72020(2020+1)/2 = 71010 x 2021
When 1010 x 2021 is divided by 4 gives remainder = 2 x 1 =2
Unit digit = 72 =>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 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.
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.
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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: 4
B. 31: 24
C. 30: 23
D. 31 : 25
