Let
P=1*2*3*4* … *2022*2023
What is the lowest quantity of numbers out of the above 2023 that erasing them will cause the remaining result to be terminated by the digit 1 ?
First, we must erase all of the even number and multiples of 5, ie all numbers ending in 0,2,4,5,6 and 8.
This removes 6*202 = 1212 numbers through 2020
Plus removing 2022 means that 1213 have been removed.
This leaves:
203 ending in 1
203 ending in 3
202 ending in 7
202 ending in 9
1 ends in 1 and 9*9 ends in 1 and 7*3 ends in 1, so all we need to do is erase one number ending in 3.
Final answer: 1214 removed.
Checking:
This leaves:
203 ending in 1
202 ending in 3
202 ending in 7
202 ending in 9
Count of numbers remaining = 809, so we have erased 2023 - 809 = 1214
More counting (spoiler)
