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4vir4ik [10]
2 years ago
14

how many 3 digit positive numbers have their hundreds, tens, and units digits in strictly ascending order

Mathematics
1 answer:
zaharov [31]2 years ago
6 0

There are seven 3 digit positive numbers that have their hundreds, tens, and units digits in strictly ascending order

Since the hundreds, tens, and units digits are in strictly ascending order.

Let the digit in hundreds place be n, since its is strictly ascending order, the digit in the tens place is n + 1 and the digit in the units place is n + 2.

So, the digits are n, n + 1, n + 2.

<h3>The positive 3 digit numbers</h3>

So, starting with n = 1, we find the number of 3 digit positive integers in ascending order.

<h3>So, when n = 1,</h3>

n, n + 1, n + 2.

1, 1 + 1, 1 + 2.

1, 2, 3

So, the first number is 123

<h3>So, when n = 2,</h3>

n, n + 1, n + 2.

2, 2 + 1, 2 + 2.

2, 3, 4

So, the second number is 234

<h3>So, when n = 3,</h3>

n, n + 1, n + 2.

3, 3 + 1, 3 + 2.

3, 4, 5

So, the third number is 345

<h3>So, when n = 4,</h3>

n, n + 1, n + 2.

4, 4 + 1, 4 + 2.

4, 5, 6

So, the fourth number is 456

<h3>So, when n = 5,</h3>

n, n + 1, n + 2.

5, 5 + 1, 5 + 2.

5, 6, 7

So, the fifth number is 567

<h3>So, when n = 6,</h3>

n, n + 1, n + 2.

6, 6 + 1, 6 + 2.

6, 7, 8

So, the sixth number is 678

<h3>When n = 7,</h3>

n, n + 1, n + 2.

7, 7 + 1, 7 + 2.

7, 8, 9

So, the seventh number is 789

<h3>When n = 8,</h3>

n, n + 1, n + 2.

8, 8 + 1, 8 + 2.

8, 9, 10

Since we cannot have 10 in the last digit, we ignore this number.

<h3>The 3 digit positive numbers</h3>

So, the numbers are 123, 234, 345, 456, 567, 678, and 789

So, there are seven 3 digit positive numbers that have their hundreds, tens, and units digits in strictly ascending order

Learn more about positive numbers here:

brainly.com/question/23270521

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