How many positive 3-digit numbers have property that the first digit is at least three times the second digit?

1 answer:

3 0

The number of positive 3-digit numbers have a property that the first digit is at least three times the second digit is 285.

<h3>How to get the number?</h3>

Since restriction is set to the first digit, all you have to think is about the last two digits e.g if the first digit is one, the last two digits can only be zeros.

Thus, by the principle of multiplication, 1×1=1 way. If the first digit is 2, then the last digits can be either 1 or 0. You have 2×2=4 ways

For first digit 3, the last two digits can be 0,1 or 2. 3x3=9, there is a succinct pattern. Therefore, 1+4+9+16+25+36+49+64+81=285.

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Sphere and a cylinder have the same radius and height. The volume of the cylinder is 64 meters cubed.

Mrrafil [7]

Answer:

C. 128/3 meters cubed

Step-by-step explanation:

The volume of a cylinder is denoted by: $V=\pi r^2h$ , where r is the radius and h is the height. We know it's equal to 64, so we can set that equal to V:

$V=\pi r^2h$

$64=\pi r^2h$

We know that the sphere and cylinder have the same height and radius. However, the "height" of a sphere is actually the same as its diameter, which is twice its radius. Then, we can replace h in the above equation with 2r:

$64=\pi r^2h$

$64=\pi r^2*2r=2\pi r^3$

$\pi r^3=64/2=32$

Now, the volume of a sphere is denoted by: $V=\frac{4}{3} \pi r^3$ , where r is the radius. From above, we know that $\pi r^3=32$ , so we can plug this into the equation: