The people at a party tried to form teams with the same number of people on each team, but when they tried to split up into teams of 2, 3, 5, or 7, exactly one person was left without a team. what is the smallest amoutn of people who could have been at the party?

211 people

Step-by-step explanation:

1.  This is a least common multiple question, otherwise known as an LCM question. We know this because if the one extra person had not shown up to the party, all groups would have been formed evenly. This means the amount of people at the party is 1 more than a number 2, 3, 5, and 7 can go into.

2. We also know this is an LCM question because we are being asked for the smallest amount of people who could have possibly attended the party.

3.  From this we know the amount of people at the party must be an odd number. If the number were even, there would have been no left over when groups of 2 were formed.

4. The amount of people at the party must end in a 1. This is because all multiples of 5 always end in a 0 or 5. Because there is on extra person, we must add 1 to all multiples of 5 we check. However, 5 + 1 is 6. This is a problem because 6 is an even number, and as we already established, the amount of people at the party must end in an odd number. So, we now know the smallest amount of people at the party will end in a 1.

5. Because we know the largest teams attempted to be formed with 1 left over is teams of 7 people, we only need to check multiples of 7. It is the largest number, so doing this will save us time.

6. Since we know the amount of people at the party must end in a 1, and we are only checking multiples of 7, we only need to check multiples of 7 that end in a 0. This is because any multiple of 7, 7 will go into evenly without a remainder. So we must add 1 to every multiple we check in order to make a remainder of 1. The only number we can add 1 to in order to get 1 is 0, so we only need to check multiples of 7 that end in 0. The only multiples of 7 that end in a 0 are when 7 is multiplied by a ten, ex: 10, 20, 30, 40, 50, ect.

7. Only searching for odd numbers, numbers that end in 1, and multiples of 7 means we only have to check if any possible answer when divided by 3 has a remainder of 1. We only have to check by the number 3 because any number ending in 1 will automatically have a remainder of one for 2 and 5, and because we are using multiple of 7 we don't need to check through 7.

8. Now that we know all our rules, all we need to do is list multiples of 7 that end in 0. Then we will add 1 to them and check to see if they have a remainder of 1 when divided by 3.

Using these rules will narrow our search drastically.

Applicable multiples of 7

(7 × 10) = 70 → 70 + 1 = 71 → 71/3 = 23 R2 → not possible

(7 × 20) = 140 → 140 + 1 = 141 → 141/3 = 47 → not possible

(7 × 30) = 210 → 210 +1 = 211 → 211/3 = 70 R1 → possible

The smallest possible amount of people at the party is 211.

i leave the calculator work to you.

7. diameter 4 miles, circumference 4π miles

8. radius 0.2 miles, circumference 0.4π miles

[this is a bad problem because the actual radius of a car on the track is less than 0.2 miles so the answer is less than 0.4π miles]

9. radius 6 feet, circumference 12π feet

10. diameter 14 cm, circumference 14π cm

11. diameter 4.8 feet, circumference 4.8π feet