I affirm that all of these statements are correct.
All prime numbers are odd with the exception of two. Hence, if we consider the sum of the first million primes, it consists of one even number combined with 999,999 odd numbers. Since the product of an odd number multiplied by another odd number results in an odd number, we conclude that the sum must be an odd number, being even + odd = odd.
It's clear that an odd number ends with an odd digit, so the only digit that can be eliminated is b (an even digit).
Honestly, I find Mrs. Garcia's method easier to perform mentally. It hinges on how familiar you are with your multiples of 5. (5*15 = 75 is a multiplication I often use)
Melissa's approach involves calculating 5*20 = 100 and 5*9 = 45, then combines the 3-digit result 100 with the 2-digit result 45, yielding 145. Adding 45 to 00 is simple and doesn’t require carrying digits, thus the arithmetic is fairly straightforward.
Mrs. Garcia's technique involves computing 5*14 = 70 and 5*15 = 75, then summing these two-digit results. Many people may not readily recall that 5*15=75, which complicates forming that product. The addition of 70 and 75 requires a carrying operation, making the math somewhat more complex. The resulting total is 145.
(The rationale behind my preference for Mrs. Garcia's method is that I can achieve the final sum by simply doubling 7 tens, followed by adding 5. The only 3-digit number to remember mentally is the ultimate total.)
_____Subtraction introduces a slight complication, yet reshaping it as $5(30 -1) = $150 - 5 = $145 is possible.
Or, you may reframe it as $5(28 +1) = $140 +5 = $145.
Dividing an even number by 2 to find the product of 5 is straightforward when you append a zero.
5*14 = 10*7 = 70
5*28 = 10*14 = 140.
