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

A palindrome, such as 12321, is a number that remains the same when its digits are reversed. How many 4-digit palindromes are divisible by 4?

A. 10
B. 20
C. 25
D. 28
E. 30

If a number is divisible by 4, then the number created by the last 2 digits is divisible by 4.
For example, we know that 76512, 311,244 and 2128 are divisible by 4 because 12, 44, and 28 are divisible by 4

So, the last 2 digits of the 4-digit palindromes must be 00, 04, 08, 12, . . . , 92 or 96
There are 25 such values

KEY CONCEPT: Once we've selected the LAST 2 digits, we automatically know what the FIRST 2 digits must be, since we need to create a palindrome.
For example, if the last two digits are 12, then the first two digits are 21 to get the palindrome: 2112
Likewise, if the last two digits are 36, then the first two digits are 63 to get the palindrome: 6336
And, if the last two digits are 80, then the first two digits are 08 to get the palindrome: 0880
STOP
0880 is NOT a 4-digit number.
In fact, any time the units digit is ZERO, the thousands digit of the 4-digit number is ZERO), which means we DON'T have a 4-digit number.
So, we must SUBTRACT from 25 all of those instances in which units digit is ZERO

Let's count all of those instances: 00, 20, 40, 60 and 80
There are 5 such instances.

So, the TOTAL number of 4-digit palindromes divisible by 4 = 25 - 5 = 20

Answer: B

Cheers,
Brent