Problema Solution
Lee has a collection of records. When he puts them in piles of two, he has one left over. He also has one left over when he puts them in piles of 3 or piles of 4. He has none left over when he puts them in piles of 7. What is the least number of records Lee can have?
Answer provided by our tutors
let 'x' be the number of records that Lee has
When he puts them in piles of two, he has one left over:
x - 1 is dividable by 2
He also has one left over when he puts them in piles of 3 or piles of 4
x - 1 is dividable by 3
x - 1 is dividable by 4
LCM(2, 3, 4) = 12 => x-1 = 12s, where s is integer
x = 12s + 1
He has none left over when he puts them in piles of 7
x is dividable by 7 => x = 7k, k is integer
12s + 1 = 7k
with tries we find
s = 4
k = 7
x = 7*7 = 49
Lee had 49 records.