Submitted by Bhuvan Sharma

i solve this question just by using pattern

first, I divide 7 with 25 and i find the remainder is 7

second, I divide 49 with 25 and i find the remainder is 24

third, I divide 343 with 25 and i find the remainder is 18

fourth, I divide 2401 with 25 and i find the remainder is 1

fifth, i divide 16807 with 35 and i find the remainder is 7

then, if you look at the first and the fifth, it is the same remainder

if you continue it you will find the second is the same with the sixth, the third is the same with the seventh and the fourth is the same with the eight. this is call by pattern

so the pattern is: 7 , 24 , 18 , 1

because there are 4 pattern you divide 103 with 4 and the remainder is 3, so the remainder is the third number =18

$7^{103} = 7^{102} (7) = (7^2)^{51} ( 7) = 49^{51} ( 7)$ 49 ≡ -1 (mod 25), so by the Modular Arithmetic Lemma, $49^{51} ≡ (-1)^{51} = -1 (mod 25)$ Thus, $7^{103} ≡ -1(7) ≡ 18 (mod 25)$

Note that $7^2=49$ which is congruent to $-1$ mod 25. Factor $7^{103}$ as $7^{102}*7 \equiv -26*7 \equiv -1 \equiv 18 mod 25$ .

$7^{2}$ = -1 mod 25

$7^{100}$ = 1 mod 25

$7^{103}$ = $7^{3}$ mod 25

$7^{103}$ = 343 mod 25

$7^{103}$ = 18 mod 25

The problem above can be solved as:

$7^{103} \pmod{25}$

Yes, this looks challenging, but if you try to raise 7 to certain powers:

$7^1 = 7$

$7^2 = 49$

$7^3 = 343$

$7^4 = 2401$

There. Since $7^4 = 2401, 7^4 \pmod{25} \equiv 1$ . We can proceed now by:

$(7^{100})(7^3) \pmod 25$

$\equiv (7^4)^{25}(7^3) \pmod{25}$

$\equiv (1)^{25}(7^3) \pmod{25}$

$\equiv 7^3 \pmod 25$

Since $7^3 = 343$ ,

$= 343 \pmod{25}$

$\equiv \boxed{18}$ (343 when divided by 25 gives a quotient of 13 and a remainder of 18.)

We are looking for $7^{103}$ mod $25$ . Since $103$ is congruent to $3$ mod 25 the problem simplifies to $7^{3}$ mod $25$ . $7^3 = 343$ and the remainder when $343$ is divided by $25$ is $\boxed{18}$

a cycle can be formed by 7^1 up to 7^5 same as 7^96 up to 7^100 7^3 has the same remainder of 7^103 343 divided by 25 has a remainder of 18

$7^{103} = {(7^4)}^{25} 7^3 = {2401}^{25} \times 343 \\ Now, {2401}^{25} \times 343 \equiv 1^{25} \times 18 \equiv 18 \pmod {25}$

103 / 25 = 4,12

103 - ( 25 * 4 ) = 3

7 7 7 = 343

343/25 = 13,72

25*13 = 325

343 - 325 = 18

since 7^103 = 7 x 7^102

then 7^102 = 7^51 x 7^2

logically, 7^103 is logically = to 7^ 3

therefore dividing 7^3 to 25

the remainder is 18.

The remainder can be find by repeating the following loop for j = 1, 2, 3, 4, ..., 103

Initially, we have 7^0 = 1 , we divided by 25, gives a remainder of 1. So set r = 1

R = 1

For j = 1, 103

R = 7*R MOD 25

Next j

After the loop you will have R = 18

I just solved it by taking power 3 apart and done 7^3 it is 343 and I divided it with 25 I got 13 point something then I multiplied 25 with 13 then I got value I subtracted with 343 and I got 18

• 7^ 1= 7
• 7^2 =49
• 7^ 3= 343
• 7^ 4= 2401
• 7^ 5 = 16807
• _ Therefore the cyclicity of 7 is 4 _
• _Therefore 103 / 4 leaves remainder 3. _
• _ Therefore 7 ^103 is the same as 7 ^3 = 343 _
• _ Therefore 7 ^103 is the same as 7 ^3 = 343 _
  
  • _ Therefore 343/25 leaves remainder _
  
  
• 18

7^103 = 7^102 × 7

= 7^2^51 × 7

49 ÷ 7 remainser is 49 = (– 1)

Total remainder when = 7^2^51 × 7 divided by 25 = (– 1)^51 × 7 = – 7 = 25 – 7 = 18

For 7 remainder is 7 when divided by 25 for 7 7- 16 for 7 7 7- 18 for 7 7 7 7- 1 for 7 7 7 7 7- 7 and it is repeated for further divisions here the exponent of 7 is 103= 4(25) + 3 i.e for 100- remainder is 1 remains 7 7 7 whose remainder is 18

<-> (7^103) mod (25) = (7^3 . 7^100) mod (25) <-> (343 . (7^4 . 25)) mod (25) = (343 . (2401)^25) mod (25) <-> (343 . (96.25 + 01)^25) mod (25) = 343 . (1)^25 mod (25) <-> 343 mod 25 = 18. Answer : 18

By Euler's Theorum :both are co-prime ,hence 7^5 is divisible by 25 hence left part is 7^3 ,which in tern leaves 24 7 as left remainder , which in other terms is -1 7 or 18 :),kindly suggest some other way exept euler ,its cliched