Find the last two digits of this expression:
7x19x31x43x55x........x2023
Using a computer is MASSIVE overkill. With a clever approach this problem borders on mental math.
All the numbers in the product are the sequence of integers congruent to 7 mod 12.
Note that the fifth term 55 and tenth term 105 are multiples of 5. That alone tells us the last two digits are one of 00, 25, 50, or 75. But since all the terms are odd that reduces the possibilities to 25 or 75.
Numbers 7 mod 12 are a subset of numbers 3 mod 4. They have the property if a product consists of an odd quantity of terms the product will also be 3 mod 4 but if there is an even quantity of terms the produce will be 1 mod 4. 25=1 mod 4 and 75=3 mod 4
So the arithmetic sequence from 7 to 2023 with a difference of 12 has 1+(2023-7)/12 = 169 terms, this is odd so we then conclude the last two digits of the product is 75.
Solution
