Number Theory or Algebra? Part 3

$149 \equiv -1\pmod5$

$149^{3} \equiv (-1)^{3}\pmod5$

$149^{3} \equiv -1\pmod5$

$\boxed{149^{3} \equiv 4\pmod5}$

$152 \equiv 2\pmod5$

$152^{2} \equiv 2^{2}\pmod5$

$\boxed{152^{2} \equiv 4\pmod5}$

$\therefore (149^{3})(152^{2}) \equiv 16 \pmod5$

$\therefore \boxed{(149^{3})(152^{2}) \equiv \boxed{1} \pmod5}$

Without modulo, simply multiply all the last digits together. With that, we get $9\times9\times9\times2\times2$ . Considering only the last digit, obviously that multiplication results in a last digit of $6$ . Hence, the remainder when divided by $5$ is $\boxed{1}$ .

9^3=9 as a unit place and 2^2=4 as a unit place. so, 9*4=36 and divided by 5 is 1

Level 0: Bashing

$149^3 \times 152^2 = 76426853696 \equiv 1 \pmod 5$

• Advantage: no brain needed
• Disadvantage: get one digit wrong and you're finished

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Level 1: Clever bashing

Let $x=150$ .

$149^3 \times 152^2 = (x-1)^3(x+2)^2 = (x^3-3x^2+3x-1)(x^2+4x+4) = x^5+x^4-5x^3-x^2+8x-4 \equiv -4 \equiv 1 \pmod 5$

• Advantage: seems pro because of $x$
• Disadvantage: too complicated

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Level 2: Pro

$149^3 \times 152^2 \equiv 4^3 \times 2^2 = 64 \times 4 \equiv 4 \times 4 = 16 \equiv 1 \pmod 5$

• Advantage: quick
• Disadvantage: see next level

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Level 3: Master

$149^3 \times 152^2 \equiv (-1)^3 \times 2^2 = -4 \equiv 1 \pmod 5$

• Advantage: you can actually mentally calculate this (I did), using negative numbers like a master
• Disadvantage: none

$Assume\quad \rho \quad to\quad be\quad the\quad right\quad answer;\quad therefore,\quad \rho .\\ QED$