Pied Piper piled rats

Just find the LCM of 2,3,4,5,6,7 and you will get the answer, i.e., 420

_ We need to find a number such that all the elements of the set $\{2,3,4,5,6,7\}$ divides it. So named and defined that number as $n=7!=1\times 2 \times 3 \times 4 \times 5\times 6 \times 7.$ Clearly this number works, but is not the smallest, all we have to do is to erase the numbers 2 and 6, So $n=3 \times 4 \times 5 \times 7$ . Why? -Becuase 2 and 4 divides 4 but we don't need $2\times4$ is enough with 4 and 2 and 3 divides 6 so we don't need 6. Hence $n=3 \times 4 \times 5 \times 7=\boxed{420}$ . _

As the minimum number of rats cannot be zero,

and the number of rats is divisible by 2, 3,4,5,6,7 because we can divide the number of rats into equal parts of 2,3,4,5,6,7

so minimum number will be LCM of 2,3,4,5,6,7, which is equal to 420.

so the minimum number of rats are 420 rats.

$\text{LCM}(2,3,4,5,6,7)=\boxed{420}$

It is given that the result should be perfectly divisible by 2, 3, 4, 5, 6 and 7.

Hence, the result is the L.C.M. of {2, 3, 4, 5, 6, 7}.

Now, To obtain L.C.M. -

Divide the set by 2 , the set becomes

{1, 3, 2, 5, 3, 7}

Divide the set by 3 , the set becomes

{1, 1, 2, 5, 1, 7}

Divide the set by 2 , the set becomes

{1, 1, 1, 5, 1, 7}

Divide the set by 5 , the set becomes

{1, 1, 1, 1, 1, 7}

Divide the set by 7 , the set becomes

{1, 1, 1, 1, 1, 1}

Since all the elements in set is 1, L.C.M of the numbers is 2 x 3 x 2 x 5 x 7 = $\boxed{420}$

That's the result.

The problem is asking for the least number that is divisible by 2, 3, 4, 5, 6 and 7.

This number is obtained by $2 \times 3 \times 4 \times 5 \times 6 \times 7$ .

However, all numbers divisible by 6 is automatically divisible by 2 and 3. 4 and 6 have a common factor of 2.

Therefore, to get the Least Common Multiple of the above mentioned numbers, $2 \times 5 \times 6 \times 7$ which is $\boxed{420}$

SInce they can be divided into 2, 3, 4, 5, 6 and 7 equal piles, we have to get the least common multiple which is LCM = 2 3 2 5 7=420

(4 has factors 2 and 2, so that we will not include it, and the same for 6, since 6=2*3)

write the following python code to get the o/p:

for i in range(1,1000): if i%2==0 and i%3==0 and i%5==0 and i%7==0 and i%4==0 and i%6==0: print(i)

in the end you will get 2 answers, select the one which is minimum.

If $N =$ the number of rats he has, then $2, 3, 4, 5, 6$ and $7$ must all be factors of $N$ . Therefore, prime factors of $N$ must include $2, 3, 5$ and $7$ . If $N$ was $2\times3\times5\times7$ , then he could divide his rats into 2, 3, 5, 6 or 7 piles. However, he could not divide them into 4 equal piles. For him to be able to do so, $N$ must have $2^2$ as a prime factor; therefore, $N=2^2\times3\times5\times7 = 420$ .

For this, we need the LCM of 2,3,4,5,6 & 7. As their LCM is 420, the minimum number of rats the Pied Piper of Hamelin has is 420.

the minimum number of rats = least common multiple (LCM) of 2,3,4,5,6,7 The LCM of 2,3,4,5,6,7 is 420.

Therefore the answer is \boxed{420}

we have to find least common number of rats that can be divided by 2, 3, 4, 5, 6, 7.

for this we will find out the LCM of 2, 3, 4, 5, 6, 7 = 2 * 2 * 3 * 5 * 7 = 420

all you've got to do is get the least common multiple of 2, 3, 4, 5, 6 & 7. to make it easier, divide all possible common factors:

1. divide 2, 4 & 6 by 2 making the numbers 1, 3, 2, 5, 3 & 7
2. divide the two 3's by 3 to make 1, 1, 2, 5, 1 & 7
3. multiply all numbers and the factors/divisors: (1)(1)(2)(5)(1)(7)(2)(3) = 420

Take L.C.M of all. That gives the least number of rats So, answer is 420

If any no. is divisible by any set of numbers than the no. itself is a multiple of LCM of that set of numbers. For minimum no., the number should be exactly equal to the LCM of the set of the numbers.

So, LCM (2,3,4,5,6,7) = 420

6*7=42 and it can be divided by 3,4,6,7 now if u multiple 5 it would not be divided by 3,4,6,7 so if u multiple 10 it would be divided by all the numbers mentioned above

LCM of 2,3,4,5,6,7 then think a prime #s that can be divide in that values 2[2,3,4,5,6,7 2[1,3,2,5,3,7 3[1,3,1,5,3,7 5[1,1,1,5,1,7 7[1,1,1,1,1,7 [1,1,1,1,1,1

then multiply all the prime numbers 2x2x3x5x7

ANSWER: 420

check for a number which is a multiple for for 2,3,4,5,6&7i.e., 420 satisfies the division of 2,3,4,5,6&7 with remainder zero