Question

the mystery marching band: students in a marching band want to line up for their performance. the problem is that when they line up in twos there is 1 left over. when they line up in threes there are 2 left over. when they line up in fours there are 3 left over. when they line up in fives there are 4 left over. when they line up in sixes there are 5 left over. when they line up in sevens there are no students left over. how many students are there? show how you figured it out.

Answer 1

119

Explanation:

"When they line up in twos there is 1 left over."

We need an odd number.

"When they line up in sevens there are no students left over."

We need a multiple of 7 that is odd.

Since there are leftovers when dividing by 2, 3, 4, 5, and 6, the number cannot be a multiple of 2, 3, 4, 5, or 6.

Let's list the first 10 odd multiples of 7:

7 × 3 = 21

7 × 5 = 35

7 × 7 = 49

7 × 9 = 63

7 × 11 = 77

7 × 13 = 91

7 × 15 = 105

7 × 17 = 119

7 × 19 = 133

7 × 21 = 147

The following multiples are eliminated because they are multiples of 3 or 5: 21, 35, 63, 105, 147.

The following odd multiples of 7 are left:

49, 77, 91, 119, 133, 147

Now we test each one by dividing by each number, 2, 3, 4, 5, 6, 7, and writing the quotient and the remainder.:

49/2 = 24 R 1

49/3 = 16 R 1

49 is not it.

77/2 = 38 R 1

77/3 = 25 R 2

77/4 = 19 R 1

77 is not it

91/2 = 45 R 1

91/3 = 30 R 1

91 is not it

119/2 = 59 R1

119/3 = 39 R 2

119/4 = 29 R 3

119/5 = 23 R 4

119/6 = 19 R 5

119/7 = 17 R 0

119 works in all the divisions.

Answer: 119