How many three digit numbers have digits whose sum is greater than 2?

896

Step-by-step explanation:

Let's talk first about how many 3 digit numbers there are. The first 3 digit number is 100 and the last is 999. So there are 999-100+1 numbers that are 3 digits long. That simplifies to 900.

Now let's find how many of those have a sum for the digits being 1, then 2 ? Then take that sum away from the 900 to see how many 3 digit numbers have the sum of their digits being more than 2.

3 digit numbers with sum of 1:

The first and only number is 100 since 1+0+0=1.

We can't include 010 or 001 because these aren't really three digits long.

3 digit numbers with sum of 2:

The first number is 101 since 1+0+1=2.

The second number is 110 since 1+1+0=2.

The third number is 200 since 2+0+0=2.

That's the last of those. We could only use 0,1, and 2 here Anything with a 3 in it would give us something larger than or equal to 3.

So there are 900-1-3 numbers who are 3 digits long and whose sum of digits is greater than 2.

This answer simplifies to 896.

i am currently in the exact same situation rn. algebra2/adv algebra is killing

step-by-step explanation:

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step-by-step explanation: