My son and his soccer had to fundraise money for the team Th

My son and his soccer had to fundraise money for the team. They sold chocolates for $1.00. After collecting all of his teammates\'s $1 bills (assuming buyers only paid using $1 bills), he tried to put the $1 bills into 2 equal piles and found 1 left over at the end. He then tried to put the bills in 3 equal piles and also ended up one extra bill. Frustrated, he tried 4, 5, and 6 equal piles and each time had $1 left over. Finally, he put all the bills into 7 equal piles and none were left over.

Based on this information, the soccer team is sure to have enough money for all the team\'s needs.

a) What is the least amount of money they could have raised? Show how you know.

b) It turns out that they raised more than $500. Knowing this, what is the least amount of money they could have raised?

****PLEASE HELP, URGENT***

PLEASE HELP emegenerat

Solution

Let N be the number. Since it leaves a remainder of 1 on division by 6, it must be of the form N = 6*M + 1, where M is an integer. (We willassume that all variables below are integers, too.) This also takes care of division by 2 and 3, since N = 6*M + 1 = 2*(3*M) + 1 = 3*(2*M) + 1. Since N leaves a remainder of 1 on division by 5, it must be of the form N = 5*L + 1 = 6*M + 1, so 5*L = 6*M, so L = 6*K and M = 5*K. Thus N = 30*K + 1. (This uses the fact that 5 and 6 have no common factor.) Furthermore, N leaves a remainder of 1 on division by 4, so N = 4*J + 1 = 30*K + 1, so 4*J = 30*K, or 2*J = 15*K. This means that J = 15*I and K = 2*I, so N = 60*I + 1. (This uses the fact that 2 and 15 have no common factor.) At this point, you could try values of I = 0, 1, ..., until you found a solution. Alternatively, you could continue in the following vein. Finally N leaves a remainder of 0 on division by 7, so N = 7*H = 60*I + 1 = (7*8 + 4)*I + 1, 7*(H - 8*I) = 4*I + 1, 7*(2*H - 16*I) = 8*I + 2, 7*(2*H - 17*I) - 2 = I, so put 2*H - 17*I = G, and I = 7*G - 2, so H = 60*G - 17. Then N = 420*G - 119. a) A number is of the above form if and only if it satisfies the divisibility conditions of the problem. The smallest positive answer occurs when G = 1, N = 301. b) For smallest positive answer and the number to be greater than 500, put G = 2. N = 420x2-119 = 721