How many 4-digit integers have the form abcd, where b is eve

Hello Max@Math Revolution.

I would like to try to solve this PS question.

4-digit integers have the form abcd, where b is even and d >= 2b

- The number of options that digit "a" has is 9 (any number from 1 to 9).
- The number of options that digit "c" has is 10 (any number from 0 to 9).
- The number of options that digits "b" and "d" depends on each case.

I will start by fixing a number for "b":

* This implies that "d" can be any number from 0 to 9 (10 options).

Hence, there number of 4 digit numbers, in this case, is equal to $$b_0=9\cdot10\cdot10\cdot1\ =900.$$

* This implies that "d" can be any number from 2 to 9 (8 options).

Hence, there number of 4 digit numbers, in this case, is equal to $$b_1=9\cdot8\cdot10\cdot1\ =720.$$

* This implies that "d" can be any number from 4 to 9 (6 options).

Hence, there number of 4 digit numbers, in this case, is equal to $$b_2=9\cdot6\cdot10\cdot1\ =540.$$

* This implies that "d" can be any number from 6 to 9 (4 options).

Hence, there number of 4 digit numbers, in this case, is equal to $$b_3=9\cdot4\cdot10\cdot1\ =360.$$

* This implies that "d" can be any number from 8 to 9 (2 options).

Hence, there number of 4 digit numbers, in this case, is equal to $$b_4=9\cdot2\cdot10\cdot1\ =180.$$

Now, if b>=5 then there are no 4-digit number that holds the condition.

Hence, the number of 4-digits integers that holds the given condition is $$b_0+b_1+b_2+b_3+b_4=900+720+540+360+180=2700.$$ This implies that the correct answer is the option E.

I don't know if there is a shorter way. I would like to see it.

Max@Math Revolution wrote:[GMAT math practice question]

How many 4-digit integers have the form abcd, where b is even and d >= 2b?

A. 900
B. 1200
C. 1520
D. 2400
E. 2700

Case 1: b=0, implying that d ≥ 0
Number of options for d = 10. (Any of the 10 digits.)
Number of options for a = 9. (Any digit 1-9.)
Number of options for c = 10. (Any of the 10 digits.)
To combine the options above, we multiply:
10*9*10 = 900.

Case 2: b=2, implying that d ≥ 4
Number of options for d = 6. (Any digit 4-9.)
Number of options for a = 9. (Any digit 1-9.)
Number of options for c = 10. (Any of the 10 digits.)
To combine the options above, we multiply:
6*9*10 = 540.

Case 3: b=4, implying that d ≥ 8
Number of options for d = 2. (Must be 8 or 9.)
Number of options for a = 9. (Any digit 1-9.)
Number of options for c = 10. (Any of the 10 digits.)
To combine the options above, we multiply:
2*9*10 = 180.

Total ways = Case 1 + Case 2 + Case 3 = 900 + 540 + 180 = 1620.

The correct answer is not listed.

=>

Suppose abcd is a 4-digit number.
There are 9 possible values for a: a = 1, 2, …, 9.
There are 10 possible values of c: c = 0,1,2,…,9.
Since b is even, b can take on the values 0,2,4,6 and 8.
However, the condition d >= 2b limits the possible values of b to 0, 2 and 4.

Case 1 : b = 0 => d = 0, 1, … , 9
The number of possible values of d is 10.
There are 10 * 9 * 10 = 900 4-digit integers with b = 0.

Case 2: b = 2 => d = 4, 5, …, 9
The number of possible values of d is 6.
There are 6 * 9 * 10 = 540 4-digit integers with b = 2.

Case 3: b = 4 => d = 8, 9
The number of possible values of d is 2.
There are 2 * 9 * 10 = 180 4-digit integers with b = 4.
In total, there are 900 + 540 + 180 = 1620 possible 4-digit integers of this form.

Therefore, the answer is C.

Answer: C

In total, there are 900 + 540 + 180 = 1520 possible 4-digit integers of this form.

The sum in red is incorrect.
Correct:
900+540+180 = 1620.
Could you please correct the typo in your post?