Find the 6-digit number beginning and ending in the digit 2 that is the product of three consecutive even integers.
Since the 6-digit number ends in 2 consider the last digit of the 3 consecutive integers and the last digit of their product:
(0,2,4) --> 0
(2,4,6) --> 8
(4,6,8) --> 2 *** (4,6,8) are the last digits
(6,8,0) --> 0
(8,0,2) --> 0
Since the 6-digit number begins with 2, the even integers must be, ballpark, between the cube root of 200,000 and 300,000, so between about 58 and 70.
The only number in that range ending in 6 is 66.
product(64,66,68) = 287,232 Bob's your uncle.
|
|
Posted by Larry
on 2025-05-09 12:32:58 |
p&p solution
