SOLUTION: our class planned a holiday party for disadvantaged kids. Some of us baked cookies for the party. On the day of the party,we found we could divide the cookies into packets of two,

Question 388549: our class planned a holiday party for disadvantaged kids. Some of us baked cookies for the party. On the day of the party,we found we could divide the cookies into packets of two,three,four,five,or,six and have just one cookie left over in each case. If we divided them into packets of seven, there would be no cookies left over. What is the the least number of cookies the class could have baked?

Found 2 solutions by jim_thompson5910, robertb:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
First, we know that if you divide the packets in two, you'll have one left over. So the number of packets MUST be odd. Also, since a number that's divisible by 5 must end with a 0 or a 5, this means that the number either ends with a 1 or a 6 (since one is left over). However, 6 is even, so this eliminates that number. So the number of packets MUST end with a 1.

Now list the multiples of 7:
0, 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98, 105, 112, 119, 126, 133, 140, 147, 154, 161, 168, 175, 182, 189, 196, 203, 210, 217, 224, 231, 238, 245, 252, 259, 266, 273, 280, 287, 294, 301, 308, 315, 322, 329, 336, 343, 350, 357, 364, 371, 378, 385, 392, 399, 406, 413, 420, 427, 434, 441, 448, 455, 462, 469, 476, 483, 490, 497, 504, 511, 518, 525, 532, 539, 546, 553, 560, 567, 574, 581, 588, 595, 602, 609, 616, 623, 630, 637, 644, 651, 658, 665, 672, 679, 686, 693, 700, ...

Of this list, only the numbers 21, 91, 161, 231, 301, 371, 441, 511, 581, and 651 end with 1. So it's possible that one of these numbers fits our criteria.

So let's first check 21: It's certainly odd, so it meets the first condition. However, it is divisible by 3, so if you divide it by 3, you will NOT have 1 left over. So 21 is NOT the number.

Now let's check 91. Again, it's odd, so it works for 2. Also, since 91/3 = 30 remainder 1, this means that it also works for 3. However, 91/4 = 22 remainder 3. So the number is NOT 91

Now let's check 161. Again, it's odd, so it works for 2. However, notice that 161/3 = 53 remainder 2, this means that the number is NOT 161 (since it doesn't work for 3)

Now let's check 231. Again, it's odd, so it works for 2. However, notice that 231/3 = 77 remainder 0, this means that the number is NOT 231 (since there are no leftovers when dividing by 3)

Now let's check 301. Again, it's odd, so it works for 2. Also, notice that 301/3 = 100 remainder 1 and notice that 301/4 = 75 remainder 1. Furthermore, we see that 301/5 = 60 remainder 1 and 301/6 = 50 remainder 1

So because 301 leaves a remainder of 1 when dividing by 2, 3, 4, 5, and 6, this means that 301 is the least number of cookies

If you need more help, email me at jim_thompson5910@hotmail.com

Also, feel free to check out my tutoring website

Jim
Answer by robertb(5830) (Show Source): You can put this solution on YOUR website!
The LCM of 2,3,4,5,6 is 60. What we want then is to find positive integers m and n such that 60m + 1 = 7n, or we have to find the smallest positive m that would make 60m + 1 divisible by 7. Positive integers of the form 60m+ 1 are 61, 121, 181, 241, 301, 361, etc... 301 is the least positive value of this form divisible by 7.
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