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If all multiples of 3 and all multiples of 4 are removed from the list of whole numbers 1 through 1200, then how many whole numbers are left?

 Jun 4, 2021
 #1
avatar+1893 
+1

2/3 of the numbers will not be multiples of 3, and 3/4 of the numbers won't be multiples of 4. 

1200 * 2/3 * 3/4 = 600

 

=^._.^=

 Jun 4, 2021
 #3
avatar+483 
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Wow that's slick

 #4
avatar+1893 
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Thank you. :))

I learnt it from an amc problem. 

 

=^._.^=

catmg  Jun 4, 2021
 #2
avatar+483 
+1

We try a complementary approach

Multiples of 3 from 1 - 1200: $\left \lfloor{\frac{1200 - 1 + 1}{3}}\right \rfloor = \left \lfloor{\frac{1200}{3}} \right \rfloor = 400$

Multiples of 4 from 1 - 1200: $\left \lfloor{\frac{1200 - 1 + 1}{4}}\right \rfloor = \left \lfloor{\frac{1200}{4}} \right \rfloor = 300$

Multiples of $4 \cdot 3 = 12$ from 1 - 1200: $\left \lfloor{\frac{1200 - 1 + 1}{12}}\right \rfloor = \left \lfloor{\frac{1200}{12}} \right \rfloor = 100$

 

$1200 - (400 + 300 - 100) = 1200 - 600 = \boxed{600}$

edited by MathProblemSolver101  Jun 4, 2021

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