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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.

 Oct 1, 2024
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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.

 Oct 1, 2024

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