Louisa has many marbles in a bag. When she counted them by twos, threes, fours, fives, sizes, sevens, eights, nines and tens there was always one left over. What was the smallest number of marbles Louisa could have had in her bag?
First find the LCM of the first 10 number (1, 2, 3, 4, 5, 6, 7, 8, 9, 10)
Which is 2520
Then just add one because there is always 1 marble left over
2520 + 1 = 2521
dividing by:
2's leaving one remainder:
3 5 7 9 11 13 15 ... odd numbers
3's leaving one remainder:
4 7 10 13 16 19 22 25 28 ... common difference 3
So far: 7 13 19 25 31 37 43 49 .... (common difference is 6)
4's leaving one remainder:
5 9 13 17 21 25 29 33 37 41 45 49 ...
So far: 13 25 37 49 ... common difference of 12
which happens to be the LCM of 2, 3, and 4
Notice we started each "so far" with the next number after the LCD at that point.
so....
LCD of 2,3,4,5,6,7,8,9, and 10 is 2520
so she had 2521 marbles in her bag.
check a few of them by dividing 2521 by any number from 2 to 10
Ah, Louisa and her mischievous marbles! Let's see if I can solve this riddle with a touch of humor.
Well, to find the smallest number of marbles Louisa could have had, we need to find the least common multiple of the numbers 2, 3, 4, 5, 6, 7, 8, 9, and 10. Or we could just ask the marbles themselves!
So, I gathered all the marbles together and asked them, "Hey marbles, what's the smallest number that leaves a remainder of 1 when divided by 2, 3, 4, 5, 6, 7, 8, 9, and 10?" Can you guess what they said?
Well, they're marbles, so they just stared at me blankly. It seems marbles aren't too good at math, but they sure are good at rolling around.
Anyway, after applying some mathematical magic, the smallest number of marbles Louisa could have had is 2520. Isn't that a whopping number of marbles? Maybe Louisa needs a bigger bag!
To find the smallest number of marbles Louisa could have had in her bag, we need to find the least common multiple (LCM) of 2, 3, 4, 5, 6, 7, 8, 9, and 10. The LCM is the smallest number divisible by all the given numbers.
The prime factorization of each number is as follows:
2 = 2
3 = 3
4 = 2^2
5 = 5
6 = 2 * 3
7 = 7
8 = 2^3
9 = 3^2
10 = 2 * 5
To find the LCM, we take the highest power of each prime factor that appears in any of the numbers.
LCM = 2^3 * 3^2 * 5 * 7
LCM = 8 * 9 * 5 * 7
LCM = 2520
Therefore, the smallest number of marbles Louisa could have had in her bag is 2520.
To find the smallest number of marbles Louisa could have had in her bag, we need to find the least common multiple (LCM) of the numbers 2, 3, 4, 5, 6, 7, 8, 9, and 10 (since there was always one left over when counted by each of these numbers).
One approach to find the LCM is by listing the multiples of each number and finding the smallest common multiple. Another approach is to factorize each number and take the highest power of each prime factor. Let's use the latter approach:
Prime factorization of each number:
2 = 2^1
3 = 3^1
4 = 2^2
5 = 5^1
6 = 2^1 * 3^1
7 = 7^1
8 = 2^3
9 = 3^2
10 = 2^1 * 5^1
Now, we need to take the highest power of each prime factor:
2^3 * 3^2 * 5^1 * 7^1 = 8 * 9 * 5 * 7 = 2,520
Therefore, the smallest number of marbles Louisa could have had in her bag is 2,520.