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Difficult Math Problem #79 - Number Theory

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by aim-wsc » Sat Dec 23, 2006 5:10 am
5000/15 = 333 quotient & 5 remainder

5000/21 = 238 quotient & 2 remainder

LCM for 15 & 21 =105

5000/105 = 47 quotient

5000-333-238+47 = 4476

ANSWER :
4475

C'mon 800guy i dont think they will ask such questions...
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by thankont » Mon Dec 25, 2006 12:42 am
agree with aim-wsc
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OA

by 800guy » Mon Dec 25, 2006 9:31 am
here's the OA. merry christmas!

We first determine the number of integers less than 5,000 that are evenly divisible by 15. This can be found by dividing 4,999 by 15:

= 4,999/15
= 333 integers

Now we will determine the number of integers evenly divisible by 21:

= 4,999/21
= 238 integers

some numbers will be evenly divisible by BOTH 15 and 21. The least common multiple of 15 and 21 is 105. This means that every number that is evenly divisible by 105 will be divisible by BOTH 15 and 21. Now we will determine the number of integers evenly divisible by 105:

= 4,999/105
= 47 integers

Therefore the positive integers less than 5000 that are not evenly divisible by 15 or 21 are 4999-(333+238-47)=4475
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by 800guy » Mon Dec 25, 2006 9:32 am
aim-wsc wrote:5000/15 = 333 quotient & 5 remainder

5000/21 = 238 quotient & 2 remainder

LCM for 15 & 21 =105

5000/105 = 47 quotient

5000-333-238+47 = 4476

ANSWER :
4475

C'mon 800guy i dont think they will ask such questions...
you're probably right, but it always helps to prepare for the worst
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