Question 1) How many positive five-digit integers contain the digit grouping "24" at least once? For instance, 24380 and 24440 are two such integers to include, but 23480 and 42034 do not meet the restrictions.

Question 2) For how many positive values of n are both and 3n four-digit base 9 integers?

timmynotjimmy Apr 11, 2024

#2**0 **

For question 1:

Here's how to find the number of positive five-digit integers containing the digit grouping "24" at least once:

Total Possible Numbers:

There are 9 choices (excluding 0) for each digit in a five-digit number (1, 2, 3, 4, 5, 6, 7, 8, 9).

So, the total number of possible five-digit positive integers is 9 * 9 * 9 * 9 * 9 = 59,049.

Unwanted Scenarios (Numbers without "24"):

No "24" group: There are 8 choices for the digit in the place where "24" could be (all digits except 2 and 4). For the remaining 4 digits, there are 9 choices each. So, there are 8 * 9 * 9 * 9 = 5832 cases where "24" doesn't appear in any of the five digits.

"24" only appears once, but not next to each other:

Choose the digit that will be between "2" and "4": There are 7 choices (all digits except 0, 2, and 4).

Choose the positions for "2" and "4" (not next to each other): There are 4 choices for the first position (remaining slots after fixing one digit). There are 3 choices for the second position (remaining slots after fixing the first and the middle digit). So, there are 4 * 3 = 12 ways to arrange "2" and "4" with another digit in between.

Choose the remaining 2 digits: There are 9 choices each.

Total unwanted cases with "24" not next to each other: 7 * 12 * 9 * 9 = 5832.

Total Numbers with "24" at least once:

We want to find the number with "24" at least once. So, subtract the unwanted cases from the total:

Total numbers - Unwanted cases = Numbers with "24"

59,049 - (5832 + 5832) = 47,385

Therefore, there are 47,385 positive five-digit integers containing the digit grouping "24" at least once.

parmen Apr 14, 2024

#3**0 **

For question 2:

Absolutely, we can find the positive values of n for which both n and 3n are four-digit base-9 integers.

Understanding the Problem:

Base-9 Integers: Numbers are represented in base-9 using digits 0 to 8, just like base-10 uses 0 to 9.

Four-digit Integers: We're interested in n such that both n and 3n have four digits in base-9.

Conditions for Four-digit Numbers:

For a base-b number to be four-digit, it must satisfy:

b^3 ≤ n < b^4

In this case, the base is b = 9. Therefore:

9^3 ≤ n < 9^4

729 ≤ n < 6561

Constraint for 3n:

We also need 3n to be a four-digit base-9 integer. This translates to:

9^3 ≤ 3n < 9^4

243 ≤ 3n < 6561

Dividing the second inequality by 3, we get:

81 ≤ n < 2187

Combining Conditions:

To satisfy both conditions, n must lie between 729 and 2187 (inclusive):

729 ≤ n ≤ 2187

Counting Possible Values of n:

Within this range, n must be a multiple of 3 (since 3n is a multiple of 3 and needs to be a four-digit integer).

The first multiple of 3 within the range is 732, and the last multiple of 3 is 2187.

To count the number of multiples of 3 in this range, we can divide the last multiple by the first multiple and subtract 1 (since we only consider integers from 732 to 2187, inclusive):

(2187 / 732) - 1 = 3

Answer:

Therefore, there are 3 positive values of n (732, 735, and 738) for which both n and 3n are four-digit base-9 integers.

parmen Apr 14, 2024