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?

Answer 1

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

Answer 2

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