Pow: the Broken Eggs
Group 7
Hearts 09-17-10 Period 5
The Broken Eggs
POW #1 1. Problem Statement: A farmer has some eggs in a cart, and is going to market them. She accidently breaks every egg. She doesn’t remember how many she had, but she remembers some things. She knows that when she put them in groups of 2, 3, 4, 5 and 6, there was one egg left over. When she put them in groups of 13 no eggs were left over. You need to find out how many eggs there are in total.
2. Process:
I first thought about a common multiple of 2, 3, 4, 5, and 6. The least common multiple of those numbers is 60. I know the amount of eggs the farmer has must be a multiple of 60 plus 1. It has to be a multiple of 60 plus 1, because any multiple of 60 plus 1 divided by 2, 3, 4, 5, and 6 will result with a remainder of 1. Then I wrote some multiples of 60 and added 1.
60 | + | 1 | = | 61 | 120 | + | 1 | = | 121 | 180 | + | 1 | = | 181 | 240 | + | 1 | = | 241 | 300 | + | 1 | = | 301 | 360 | + | 1 | = | 361 | 420 | + | 1 | = | 421 | 480 | + | 1 | = | 481 | 540 | + | 1 | = | 541 | 600 | + | 1 | = | 601 | 660 | + | 1 | = | 661 |
Those numbers are possible solutions, but now we have to see if they are divisible by 13. I noticed they all end in 1; therefore the answer must end in 1. I then highlighted the multiples of 13 that end with a 1. 13 | 26 | 39 | 52 | 65 | 78 | 91 | 104 | 117 | 130 | 143 | 156 | 169 | 182 | 195 | 208 | 221 | 234 | 247 | 260 | 273 | 286 | 299 | 312 | 325 | 338 | 351 | 364 | 377 | 390 | 403 | 416 | 429 | 442 | 455 | 468 | 481 | 494 | 507 | 520 | 533 | 564 | 559 | 572 | 585 | 598 | 611 | 624 | 637 | 650 | 663 | 676 | 689 | 702 | | | | | | | | | |
I checked to
Hearts 09-17-10 Period 5
The Broken Eggs
POW #1 1. Problem Statement: A farmer has some eggs in a cart, and is going to market them. She accidently breaks every egg. She doesn’t remember how many she had, but she remembers some things. She knows that when she put them in groups of 2, 3, 4, 5 and 6, there was one egg left over. When she put them in groups of 13 no eggs were left over. You need to find out how many eggs there are in total.
2. Process:
I first thought about a common multiple of 2, 3, 4, 5, and 6. The least common multiple of those numbers is 60. I know the amount of eggs the farmer has must be a multiple of 60 plus 1. It has to be a multiple of 60 plus 1, because any multiple of 60 plus 1 divided by 2, 3, 4, 5, and 6 will result with a remainder of 1. Then I wrote some multiples of 60 and added 1.
60 | + | 1 | = | 61 | 120 | + | 1 | = | 121 | 180 | + | 1 | = | 181 | 240 | + | 1 | = | 241 | 300 | + | 1 | = | 301 | 360 | + | 1 | = | 361 | 420 | + | 1 | = | 421 | 480 | + | 1 | = | 481 | 540 | + | 1 | = | 541 | 600 | + | 1 | = | 601 | 660 | + | 1 | = | 661 |
Those numbers are possible solutions, but now we have to see if they are divisible by 13. I noticed they all end in 1; therefore the answer must end in 1. I then highlighted the multiples of 13 that end with a 1. 13 | 26 | 39 | 52 | 65 | 78 | 91 | 104 | 117 | 130 | 143 | 156 | 169 | 182 | 195 | 208 | 221 | 234 | 247 | 260 | 273 | 286 | 299 | 312 | 325 | 338 | 351 | 364 | 377 | 390 | 403 | 416 | 429 | 442 | 455 | 468 | 481 | 494 | 507 | 520 | 533 | 564 | 559 | 572 | 585 | 598 | 611 | 624 | 637 | 650 | 663 | 676 | 689 | 702 | | | | | | | | | |
I checked to