Divisible by 3,4 but not 5

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0 replies

jlacosta

Mar 2, 2025

0 replies

Inequalities

sqing 4

N an hour ago by sqing

Let $a,b,c\ge \frac{1}{2}$ and $\left(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}\right)\left(\frac{1}{a}-\frac{1}{b}+\frac{1}{c}\right)\le 1.$ Prove that
$a+b+c\geq 2$ Let $a,b,c\ge \frac{1}{2}$ and $\left(a+\frac{1}{a}+\frac{1}{b}-\frac{1}{c}\right)\left(a+\frac{1}{a}-\frac{1}{b}+\frac{1}{c}\right)\le \frac{9}{2}.$ Prove that
$a^2+b^2+c^2\geq 1$ Let $a,b\ge \frac{1}{2}$ and $\left( \frac{1}{a}-\frac{1}{b}+2\right)\left( \frac{1}{b}-\frac{1}{a}+2\right) \le \frac{20}{9}.$ Prove that
$a+b\geq 2$ Let $a,b\ge \frac{1}{2}$ and $a^2+b^2=1.$ Prove that
$\left(\frac{2}{a}+\frac{1}{b}-1\right)\left(\frac{2}{a}-\frac{1}{b}+1\right)\ge \frac{13}{3}$

4 replies

sqing

Mar 15, 2025

sqing

an hour ago

Binomial Summation

Saucepan_man02 0

3 hours ago

Given that: $l_n = \sum_{r=1}^{n-1} \frac{2r^2-r(n-2)+1}{(n-r) \binom{n}{r}}$ , find $l_{15}$ .

0 replies

Saucepan_man02

3 hours ago

0 replies

Inequality with integers and indices

Michael Niland 5

N 3 hours ago by ohiorizzler1434

Without using a calculator, show that $2^{\frac{1}{2}}$ $< 3^{\frac{1}{3}}$ , and for a positive integer $n\geq3$ , prove that

$n^{\frac{1}{n}}$ $>(n+1)^{\frac{1}{n+1}}$ .

5 replies

Michael Niland

Yesterday at 6:52 AM

ohiorizzler1434

3 hours ago

Quadratic Equation Problems

Saucepan_man02 4

N 3 hours ago by Saucepan_man02

4 replies

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Mar 3, 2025

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3 hours ago

No more topics!

Divisible by 3,4 but not 5

Silverfalcon 4

N Dec 4, 2005 by nonie

How many positive integers not exceeding 2001 are multiple of 3 or 4 but not 5?

$\textbf{(A)} \ 768 \qquad \textbf{(B)} \ 801 \qquad \textbf{(C)} \ 934 \qquad \textbf{(D)} \ 1067 \qquad \textbf{(E)} \ 1167$

4 replies

Silverfalcon

Dec 3, 2005

nonie

Dec 4, 2005

Divisible by 3,4 but not 5

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Silverfalcon

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Silverfalcon

#1 h Dec 3, 2005, 2:26 AM • 2 Y

Y by Adventure10, Mango247

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dangvh

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dangvh

#2 h Dec 3, 2005, 2:36 AM • 2 Y

Y by Adventure10, Mango247

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By dividing $2001$ by $3$ and $4$ and taking the integer part of the quotient we get $667$ and $500$ respectively. By the principle of inclusion and exclusion, we have to subtract the quotients when $2001$ is divided by $12$ , $15$ , and $20$ . Then add the quotient when $2001$ is divided by $60$ . Thus we get $667+500-166-133-100+33=801$ or $\boxed{B}$

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236factorial

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236factorial

#3 h Dec 3, 2005, 3:16 AM • 1 Y

Y by Adventure10

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There are 667 multiples of 3 and 500 multiples of 4 from 0 - 2001, or 1167 total. The multiples of both three and four (or multiples of 12) are counted twice, so 1167 must be reduced by the number of multiples of 12 from 0 - 2001 or 166 numbers, leaving 1001 numbers. The problem states that multiples of 5 must be excluded, so we must eliminate the numbers that are multiples of both 3 and 5 (or multiples of 15), and numbers that are multiples of both 4 and 5 (or multiples of 20). After subtracting, there are 1001 - 133 - 100 = 768 numbers. We also eliminated the numbers that are multiples of 3, 4, and 5 (or multiples of 60) twice. We must add the number of multiples of 60 from 0 - 2001. There are 33, and adding this to 768 gives us 801 numbers.

This post has been edited 1 time. Last edited by 236factorial, Dec 4, 2005, 3:39 AM

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JesusFreak197

1939 posts

JesusFreak197

#4 h Dec 3, 2005, 5:26 AM • 2 Y

Y by Adventure10, Mango247

Answer

We take multiples of $3$ and multiples of $4$ , and subtract the overlap, so we have $667+500-166=1001$ . However, we must remove multiples of $5$ . Every $60$ , we have $15$ , $20$ , $30$ , $40$ , $45$ , $60$ that overlap, so we remove $6$ numbers per $60$ . There are 33 $60$ 's before $2001$ , so we subtract $198$ , and finally the $15$ and $20$ of the next $60$ , so we have $1001-200=801\Rightarrow \boxed{B}$ .

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nonie

834 posts

nonie

#5 h Dec 4, 2005, 12:54 AM • 1 Y

Y by Adventure10

Silverfalcon wrote:

How many positive integers not exceeding 2001 are multiple of 3 or 4 but not 5?

$\text{(A)} \ 768 \qquad \text{(B)} \ 801 \qquad \text{(C)} \ 934 \qquad \text{(D)} \ 1067 \qquad \text{(E)} \ 1167$

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Take the # of multiples of 4 and add the # of multiples of 3 subtract the # of multiples of 3 and 4 (12) and you get the 3 or 4 part to get the non 5's you subtract the # of multiples of 3 and 5 (15) and the multiples of 4 and 5 (20) and finally add the # of multiples of all 3 numbers (60) (because we got rid of them twice).

So then you get .....

500+667-166-133-100+33=801.
SO the answer is B 801
Sometimes using a venn diagram helps me when I do these kind of problems.

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