How many three-digit numbers can be expressed as an integral power of 2?

The correct option is (c) - 3.

Solution

The question asks for the number of three-digit numbers that can be expressed as an integral power of 2. An integral power of 2 is any number in the form $2^{n}$2n, where n is an integer (0, 1, 2, 3, ...).

Let's list the powers of 2 and identify those that fall within the three-digit range (100 to 999):
$2^{0} = 1$20=1 (1 digit)
$2^{1} = 2$21=2 (1 digit)
$2^{2} = 4$22=4 (1 digit)
$2^{3} = 8$23=8 (1 digit)
$2^{4} = 16$24=16 (2 digits)
$2^{5} = 32$25=32 (2 digits)
$2^{6} = 64$26=64 (2 digits)
$2^{7} = 128$27=128 (3 digits)
$2^{8} = 256$28=256 (3 digits)
$2^{9} = 512$29=512 (3 digits)
$2^{10} = 1024$210=1024 (4 digits)

The numbers that satisfy the condition are 128, 256, and 512. Counting these, we find there are exactly 3 such numbers.

Why the other options are incorrect

• Option (a) - 1: This is incorrect because it accounts for only one power (perhaps only $2^{7}$27 or $2^{9}$29), ignoring the other two valid three-digit powers of 2.
• Option (b) - 2: This is incorrect as it misses one of the three valid values (128, 256, or 512) in the sequence of powers.
• Option (d) - 4: This is incorrect because the next power of 2 after 512 is 1024, which is a four-digit number, and the power before 128 is 64, which is a two-digit number.

Key Concept

The question tests the ability to identify values within a specific range (100-999) using a geometric progression with a common ratio of 2.