This is a COMBINATORICS problem.
If we're looking for a prime tens and units digit, we're counting single-digit prime numbers: 2, 3, 5, and 7.
Now we need to count the possibilities for each digit. Since we have the constraint that we're looking for all #s 0-1570, we have to be careful, since we can't always set constraints unilaterally on a given digit. We can set a unilateral constraint on the thousands digit (must be a 0 or 1), but not on the hundreds digit. E.g. the digit 9 is a possibility for the hundreds digit... but only if the thousands digit is 0, or we'd have 1900+.
So it's easiest to count in 2 parts:
1. The number of 4-digit numbers less than 1570 with prime tens and units digits:
thousands digit: 1 possibility (thousands digit must be 1)
hundreds digit: 6 possibilities (0, 1, 2, 3, 4, or 5)
tens digit: 4 possibilities (2, 3, 5, 7)
units digit: 4 possibilities (2, 3, 5, 7)
Now we multiply the # of possibilities for each digit:
$$1\cdot6\cdot4\cdot4=96$$
But... this number includes a few possibilities that don't work because they're greater than 1570: 1577, 1575, 1573, 1572. We simply need to exclude these 4 possibilities by subtracting them from the total:
96 - 4 = 92.
(Note: we wouldn't want to exclude 7 unilaterally as a possibility for the tens digit, since 1477 or 1372 would have worked, etc).
2. Now, count the number of 2- and 3-digit numbers with prime tens and units digits:
hundreds digit: 10 possibilities (any digit 0-9, giving us a 2- or 3-digit number)
tens digit: 4 possibilities (2, 3, 5, 7)
units digit: 4 possibilities (2, 3, 5, 7)
$$10\cdot4\cdot4=160$$
Since all of these possibilities are between 0 and 1570, we don't need to exclude any.
Now simply add the two parts together:
92 + 160 = 252
The answer is B.
Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education