Question

please tell me answer of question number 28 with solution

Answer 1

7^2015 = 7(7^2014) = 7(49^1007)= 7(50-1)^(1007)

using binomial expansion it can be observed that all but the last term will contain 50. So all of them will be divisible by 25. The remaining last term will be -

7[1007C1007(50)^0(-1)^1007] = -7

Now go back to the basics of long division method from primary classes. When a dividend by some divisor, in order to find out the remainder, we write the immediate number smaller than the dividend which is a multiple of divisor and then subtract the two.

so when -7 is divided by 25, we need to write the immediate multiple of 25 (ie -25) smaller than -7 and then subtract the two numbers. So,

-7-(-25) =18

So the answer when 7^(2015) is divided by 25 is 18

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Look at this pattern:

When divided by 25,

71 leaves remainder 7

72=49 leaves 24

73=343 leaves 18

74=2401 leaves 1

75=16807 leaves 7

The remainders repeat (after every fourth power of 7). Now, 2015 can be written as 4∗503+3.

7^2015=7^(4∗503+3) will leave a remainder 18.

Thus, the remainder will be 18.

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my friend,there is an easy and beautiful function which will help you to crack any problem on modulo arithmetic…-the euler totient function-you can refer it in the net to know more …,it is noted by phi..

so here, phi of 25 is 20 ,and 7^2015=7^2000 .7^15=(7^20)^100 .7^15,by using the function we can say all we need to find is 7^15 % 25. so (7^2)^7 . 7 % 25

=49^7 . 7 % 25

=-1^7 . 7 % 25 (as 49 %25 =-1,also by using ab % c =a%c * b%c)

=-7 % 25=25 -7 =18…

maths is the singularity of the black hole beauty…..enjoy maths!!

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18

7^2, when divided by 25 it gives a remainder -1

(7^2)^1007*7 = 7^2015

Therefore, (-1)^1007*7 is the remainder.

-1*7 = -7

Hence, 25–7=18 should be the answer.