Elementary School Math Problems-2

Greatest Common Divisor of 30 and 45 is 15.

The problem can be very easily solved using the concept of greatest common divisor,but let's look at it in other way.(You'll find out that this 'other way' is just another way of using the same concept of GCD. Let n be the no. of students,x be the no. of pens and y be the no. of pencils that each of them receive. Then for no left overs,we must have, n(x+y)=45+30=75 Also, nx=30 and ny=45. Thus,x/y=30/45=2/3 Now,n=75/(x+y). To make n maximum,x+y must be made minimum and from above,we find that min(x+y)=2+3=5 because x and y must be natural numbers.This is where I've 'unknowingly' invoked the law of GCD.(Mathematics is awesome). Finally,max(n)=75/5=15

lol. max pens are 30 so the max students cant be greater than 30. Now the greatest number below 30 that can be divisor of 45 with no remainder is 15.. so 15 students

The highest common factor of 45 and 30, which is the biggest number of students equally receiving pens and pencils, is 15

The prime factorizations of $30$ and $45$ is

$30=2 \times 3 \times 5$

$45 = 3^2 \times 5$

So the Greatest Common Divisor is $3(5)=$ $\color{#D61F06}\large \boxed{15}$

It's just the GCD of 30 and 45, which is 15.

What does the asymptote tell the curve?

Don't touch me.

HCF of 30 and 45. 15 students with 2 pens and 3 pencils each

To find out the answer you have to find most common multiple between 30 and 45 and the answer is 15. I am so happy that now finally I have time to solve funny quizzes and spend time on this BRILLIANT page, however it wasn't always like that I had a lot of task connected with writing essay, etc, and this web site really helped me https://aussieessaywriter.com.au/ if you have such problem. RECOMMEND YOU!

$GCD(30,45)=\boxed{\large{15}}$

The highest common multiple of 30 and 45 is 15

G.C.D(30,45)=15.....there can be maximum of 15 students..and each student will get 2 pens and 3 pencils

X+Y=30 X-Y=45 The difference of this number is 15. Answer is 15. You can solve this way using adding metod or logic,Because only 15 students can receive the same quantity of pens and pencils.

We have 30 pens and 45 pencils. If we're to hand them out equally with none left over, we can assume that the MAXIMUM amount of students can be represented by the largest common factor of 30 and 45. This number is 15, so there can be a maximum of fifteen students.

GCD(30,45)=15

Highest common factor. All 15 students each would receive 2 pens and 3 pencils each.

Can we solve this using Combinatorial methods? I have observed that, if we were to have 30 and/or 45 students, the Pigeonhole principle would come into play.

Since $\gcd(30,45)$ is 15

Therefore the maximum number of students that he could have is 15

For maximum number of equal amounts of any two or more objects will be the HIGHEST COMMON FACTOR, which in this case is 15

Look for the greatest common factor: 30 = 5 x 3 x 2 45 = 5 x 3 x 3 GCF = 5 x 3 = 15 What's left after removing the GCF from the equation is how much of each item (ie. 2 pens and 3 pencils) each student is to receive.

GCF of 45 and 30 is 15

we find H.C.F. of 30 and 45 that is 15,so no of students is 15 ANS K.K.GARG,India

The highest common factor of 30 and 45 is 15. So, the answer is 15.

45 - 30 = 15 It is as easy as ABC

maximum diviser of 30 and 45 is 15

it's something in ratio factor to keep reminder equal Zero
30%15=0 ...... 45 %15=0

Its easy.... 15. :)

Hey who edited it and how

Greatest common divisor of 30 and 45 is 15.