Let n be a positive integer such that n= ABC_7 and n=CBA_11. Find the largest possible value of n in base 10.

nerdboy Oct 1, 2024

#1**0 **

To solve this problem, let's first understand the given information:

n = ABC_7: This means n is a 3-digit number in base 7, where A, B, and C are digits from 0 to 6.

n = CBA_11: This means n is a 3-digit number in base 11, where C, B, and A are digits from 0 to 10.

We need to find the largest possible value of n in base 10.

Approach:

Convert both representations to base 10:

n = ABC_7 = 7^2A + 7B + C

n = CBA_11 = 11^2C + 11B + A

Set the two expressions equal to each other:

7^2A + 7B + C = 11^2C + 11B + A

Rearrange the equation:

48A - 4B - 120*C = 0

Divide the equation by 4:

12A - B - 30C = 0

Analyze the equation:

We want to maximize n, so we want to maximize A and minimize C.

Since A, B, and C are digits, A must be at least 1 and C must be at most 5 (otherwise, A would have to be greater than 6).

Trying different values:

A = 6, C = 1:

126 - B - 301 = 0

B = 36

A = 6, C = 2:

126 - B - 302 = 0

B = 24

A = 6, C = 3:

126 - B - 303 = 0

B = 12

A = 6, C = 4:

126 - B - 304 = 0

B = 0

A = 6, C = 5:

126 - B - 305 = 0

B = -12 (not possible)

The largest possible value of n in base 10 is when A = 6, B = 36, and C = 1:

n = 7^26 + 736 + 1 = 49*6 + 252 + 1 = 294 + 252 + 1 = 547

Therefore, the largest possible value of n in base 10 is 547.

AnswerscorrectIy Oct 1, 2024