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get the prime factors of 180 using continuous divition​

Sagot :

Starting with 180, divide it by the lowest prime number 2 , repeat until it cannot be divided by 2 and move on to next prime number 3 and so on. When the remainder is 1 stop-we have obtained our prime factors.

∙ 180 ÷ 2 = 90

∙ 90 ÷ 2

= 45 which cannot be divided by 2 → 3

∙ 45 ÷ 3 = 15

∙ 15 ÷ 3

= 5 which cannot be divided by 3 → 5

∙ 5÷5

= 1 having reached 1, we are there

⇒180 = 2 × 2 × 3 × 3 × 5

= 2² × 3² × 5

[tex] \LARGE\sf\color{orange}\overline\color{orange}{DIRECTIONS :}[/tex]

[tex] \color{orange}•••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••[/tex]

  • ➝ 180 ÷ 2 = 90
  • ➝ 90 ÷ 2
  • ➝ 45 ÷ 3 = 15
  • ➝ 15 ÷ 3
  • ➝ 5 ÷ 5
  • ➝ 180 = 2 • 2 • 3 • 3 • 5
  • ➝ 2² × 3² × 5

[tex] \color{orange}•••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••[/tex]

[tex]\large\sf\color{orange}{- \: CarryOnLearning}[/tex]