Recently at an expo where Residex had a stall, our budding sales rep Jono Winfield went there to give away promotional Residex hacky-sacks.
Being an orderly sort of person (Army background), he attempted stacking them in piles of three each. However, much to his chagrin (and annoyance), he had one left over.
So he tried stacking them in piles of 5. Again, one left over.
He tried twice more, this time in piles of 7 and piles of 9. Again, however, there was one elusive Residex hacky-sack left over.
Finally, he tried one last time in piles of 11. This time, the piles worked out beautifully.
What was the minimum number of hacky-sacks that Jono would have had?
2 comments:
946.
Lets say No of hacky sacks=n. Then (n-1) is a multiple of 3,5,7,9. Since 3,5,7 are all prime numbers, and divisible by 9 implies divisible by 3, (n-1) is a multiple of (5*7*9)=315. So (n-1) could be 315, 630,945, 1260.. Of these, the first n (add 1 to above numbers) divisible by 11 is 946.
Say what? Did the fellow really arrange close to 1000 sacks!?!!
You were quite correct in your analysis. Here is the solution put in analytic form (or another presentation of same philosophy)
Assume the number of hacky-sacks is H. Because there's always one left over, we know that H-1 is divisible by 3, 5, 7 and 9. So, H-1 is a multiple of 5×7×9 = 315 (Note: 9 is also a multiple of 3, so 3 must not be included!). So we are looking for the smallest value of N where 315×N + 1 is divisible by 11.
After some trying it turns out that the smallest N for which this holds is N = 3. This means that Jono has at least 946 hacky-sacks.
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