Twos and Fives

Number Theory Level 3

$\large \color{#302B94}{2222}^{\color{#20A900}{5555}} \bmod (\color{#302B94} 2+ \color{#20A900} 5 ) = \ \color{grey}?$

1 5 3 2 6 4

4 solutions

Otto Bretscher
Apr 9, 2015

By Fermat we have $3^6\equiv1(\mod7)$ so $2222^{5555}=(317\times7+3)^{5555}\equiv3^{5555}$ $=3^{6\times925+5}=(3^6)^{925}3^5\equiv3^5\equiv5(\mod7)$

Samuel Ayinde
Apr 8, 2015

Using $Fermat little theorem$ ,

$a^{p-1}=1 mod$ $p$ , where $p$ is a prime.

therefore, $2222^{6(925)}=1 mod$ $7$ .

so, $2222^{5555} mod$ $7$ = $2222^{5} mod$ $7$ = $5$

Where did you get the 925?

Edwin Espinosa - 6 years, 2 months ago

he just multiplied it to get closer to 5555, then there is a remaining 5 so he multiplied it again by 2222^5

Ojna Eugitaraba - 6 years, 2 months ago

There is a very simple algorithm you can follow to produce the remainder of a number following a division: (In this case)

step 1) divide by 7

step 2) subtract the integer part of the answer

step 3) Multiply by 7

Step 4) Celebrate :P

Curtis Clement - 6 years, 2 months ago

Gamal Sultan
Apr 9, 2015

(2222)^(5555) = (7 X 317 + 3)^(6 X 925 + 5) =

(3)^(5) (mod 7) = 243 (mod 7) = (7 X 34 + 5) = 5 (mod 7)

Somesh Singh
Apr 9, 2015

powers of 2222 mod 7, cycles after an interval of 6 in the sequence 3,2,6,4,5,1....3,2,6,4,5,1.... and we know that 5555 mod 6= 5....... so 2222^5555 mod 6= the fifth term in the sequence, i.e, 5....

Fermat little a^p = mod ( p) & we have 3 ^ 6 = 1 mod (p) # ( 2222 ) ^ ( 5555 ) = 2222/ 7 = (317 * 7 + 3 ) ^ 5555 = 3^ 5555 = 3 ^ ( 6) (925) * (3) ^ ( 5 ) = ( 3 ) ^ ( 5 ) = 5 mod ( 7)

Ebrahim ELromancy - 6 years, 2 months ago

Twos and Fives

4 solutions

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