A number theory problem by Tammu Triveni

${7}^{1}$ the unit digit is $7$

${7}^{2}$ the unit digit is $9$

${7}^{3}$ the unit digit is $3$

${7}^{4}$ the unit digit is $1$

${7}^{5}$ the unit digit is $7$

${7}^{6}$ the unit digit is $9$ ... This is a pattern. We must to divide 96 by 4 and use the remainder left . $96 \equiv 0 \left( mod\quad 4 \right)$ . Thus the remainder been 0, the answer is 1 because if 7 is raised by a power $x$ and $x$ is a multiple by 4 the unit digit be 1. ${2387}^{96}$ the unit digit is 1.

Gotta gets some of dat $MoDuLaR$ $aRiThMeTiC$

Firstly, notice that no matter how big a number, the units digit is always dependent on another units digit in multiplication. For example, the units digit of 123456789 times 987654321 will be nine.

So then, since we are raising a number to a power, we see that the ones digit will be multiplied to that same ones digit. In general, for every ones digit "x" in a real integer "y" and for every real integer exponent "z", we see that y^z will result in a ones digit equal to the ones digit of x^z

The pattern of ones digits for multiplying 7 by itself multiple times is as follows

(only the ones digit of each operation will be shown)

7^1 -> 7

7^2 -> 9

7^3 -> 3

7^4 -> 1

7^5 -> 7

7^6 -> 9

There is a pattern. We can define the ones digit for x^z for x = 7 based on z mod(4).

if z mod(4) is congruent to 1 mod(4), then the ones digit of x^z will be 7

if z mod(4) is congruent to 2 mod(4), then the ones digit of x^z will be 9

if z mod(4) is congruent to 3 mod(4), then the ones digit of x^z will be 3

if z mod(4) is congruent to 4 mod(4), then the ones digit of x^z will be 1

z = 96, and 96 is congruent to 4 mod(4) (because 96 is a multiple of 4)

Therefore the ones digit of 2387^96 is 1.

It holds that 7^(1+4x), for any positive integer of x, the number will have a 7 in its units digit. So 1+4x=96 and x=23 remainder 3. Then just multiply 7 times 7 cubed and the units digit is 1.

Our concern area is last digit, So 2387^96 = 2387^45 *2387^45 *2387^6. First two no. will gv last digit 5 and last no. Will give 9 So 9×7×7 =81 here 1 is last digit (Concept ofCyclicity of number is used to find solution)