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 ?
My first thought was: eliminate all numbers ending in 0 or 5 to get rid of the trailing zeros.
But the only single digit multipliers yielding a terminal 1 are 1*1, 9*9, 3*7.
So eliminate everything except factors that end in 1, 3, 7, or 9.
This will result in:
203 1s
203 3s
202 7s and
202 9s
We can keep all these factors except for one of those ending in 3, because there must be an equal number of 3s and 7s (and also an even number of 9s).
203 + 203 + 202 + 202 - 1 = 809 factors
So we erase 2023 - 809 = 1214.
Erase 1214 of the numbers.
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Posted by Larry
on 2023-05-11 08:04:26 |
solution
