Question 388549
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



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Jim