4 ms·
Well, if either number ends in a 0, the product will end in a 0, so that's almost 20% of the cases right there. And if you get any even number and a number endi
by lucasmullens 5y ago
Well, if either number ends in a 0, the product will end in a 0, so that's almost 20% of the cases right there. And if you get any even number and a number ending in 5 you'll get a product ending with 0 also, so 24-31% total sounds about right.
- function_seven 5y agoSuch a simple answer. Thanks. I just did a 10x10 times table and got the following prevalences: [27, 4, 12, 4, 12, 9, 12, 4, 12, 4] So '0' has 27% probability, '5' has 9%, and the rest of the evens 12%, the rest of the odds 4%
- inimino 5y agoAnother way to explain it is consider the prime factorization of both numbers. The product will contain all of the prime factors from both, so if there is at least one 2 and at least one 5 in then the result will be divisible by 10.