Bobby found the sum of 3 consecutive integers
and then the sum of the next 3.
Multiplying both results he got 111111111.
Is it possible?
Source Russian Kвант (Quant)
The prime factors of 111111111 are:
3, 3, 37, 333667
Each of the two sums must have had one of the factors that are 3, as the sum of three consecutive numbers is 3 times the middle number. Given the size of the largest factor, none of the integers could have been negative, to allow violation of that rule.
The only pair of triplets could have been 36, 37, 38; and 333666, 333667, 333668 which are not consecutive.
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Posted by Charlie
on 2022-04-02 14:15:44 |
another route to the solution
