Explanation By Claudio Hurtado Coach GMAT QUANT, GRE QUANT, SAT QUANT: clasesgmatchile@gmail.com , +56 945 517 215, gmatchile.cl or clasesgmat.es

Here it is convenient to separate the problem into 3 parts:

1st part three-digit number, even, all digits other than zero and whose hundreds digit is 9:

Possibilities for hundreds: 1 (only 9)
Possibilities for tens: All digits except 9, except an even number (the one that occupies the units figure, remember that all three-digit numbers must be even), except zero (remember that all numbers must be different from zero), that is, 7 possibilities.

Possibilities for units: We know that a number is even if the units figure is even, among the digits there are 5 pairs (0, 2, 4, 6 and 8), from these we must discount the 0, since the figures do not They can be zero, so for units we have 4 possibilities.

Then all the three-digit numbers, which meet the stated condition and with a hundred digit 9, are:

1x4x7 = 28

2nd part, 3-digit number, where the hundred digit is 8:

It is similar to the 1st part, except that since the hundreds figure (8) is an even number, then from the possibilities of the units figure, which must be even, we must subtract one more than in the first part, this is left over:

1x7x3=21

3rd part, analysis similar to the first part:

1x7x4=28

Thus adding the three parts, since it is the 1st part or second part or third part, we have:

28 + 21 + 28= 77

Answer E.