This is a great puzzle to solve with your kids. Call them over!
Winter is a tough time at school. Teachers get sick frequently and principal needs to come up with some creative ideas on how to make kids happy. On one specific Monday morning two teachers called in sick. Principal asked a gym teacher to take all the kids from these two classes on a field trip to the fire station and back. Gym teacher gathered all the kids and ordered them to stand in pairs. When they finally did it, it became apparent that one child does not have a partner. OK, said the gym teacher, the sidewalk is wide, let's stand three-by-three. Again, one person was left. They tried arranging themselves in rows of four, but mysteriously someone was still remaining alone. Finally, they realized that if they will organize themselves in rows of five, they will have no one left out. They went to the fire station, climbed the fire tracks, ate lunch with the firemen, saw firemen leave for emergency, and returned back to school very satisfied. When the principal saw the gym teacher she asked him to mark in the school journal how many kids did he take on a field trip.
He was ashamed as he forgot to count the kids. He was pretty sure it was more than 30. Can you help him figure out how many kids exactly were on this trip?
(adapted from the book by C. Koval)
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7 comments:
It has to end in a 0 or 5 to be divisible evenly by 5. It can't be a 0 since that will always be divisible by 2. The digits can't add up to a number that's divisible by three.
25 leaves a 1 remainder when dividing by 2, 3, and 4 but is divisible evenly by 5. How sure was he that it was over 30?
85 is the next number that leaves a 1 remainder when dividing by 2, 3, and 4, but is divisible evenly by 5.
Robin says:
The answer is 85. He says he got this because he knew it couldn't be a multiple of 5 that ends in 0, because those numbers are also multiples of 2. He went to each multiple of 5, starting with 35, and tried to figure out which was the first one that when divided by 3, 4 or 2 left a remainder of 1. 85 was the first number that worked.
Since with rows of 2, 3, and 4, there was always one person left over, this means 1 fewer than the answer is divisible by 2, 3, and 4.
Let's say X is the total number of kids.
X-1 must be divisible 2, 3, and 4 which means it has to be a multiple of 12.
X has to be a multiple of 5.
There are infinite answers: 25, 85, 145, 205.... (+60).
Since the gym teacher "thinks" it was more than 30, and practically speaking, we'd hope the school wouldn't have that many kids in a class, agree the likely answer is 85.
You guys area amazing!
Correct answers, wonderful explanations.
I think Phil should get the puzzle point because he was first. The rest of us were too busy with the french toasts, bagels, coffee and kids activities in the morning.
Expect a new puzzle tomorrow.
LOL - My kids spilled their milk over my laptop - Im back up again! Hurray! I'm a mom that works from home with Moms With A Plan Work At Home Business Opportunity
85. Divisible by 5, but leaves a remainder of 1 when divided by 2, 3, and 4.
The LCM for 2, 3 and 4 is 12. A multiple of 12 plus 1 will yield a remainder of 1 when divided by 2, 3 and 4. Because it was a multiple of 5 the lowest multiple of 12 plus 1, divisible by 5, is 25. The PE teacher thought there were more then 30, and the next multiple of 12 plus 1, ending in 5, is 60 more than 25 - or 85. Therefore 85 is the smallest that the group could have been.
I hope for the PE yeacher's sake he had some parents and aides with him!
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