SOLUTION: Find all three-digit numbers in base 7 whose digits are reversed when converted to base 11.

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Question 1166029: Find all three-digit numbers in base 7 whose digits are reversed when
converted to base 11.
Answer by greenestamps(13206) About Me  (Show Source):
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We need to find a, b, and c such that abc (base 7) = cba (base 11).

Note that a, b, and c must be whole numbers less than 7.

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(1) We can find the solutions using algebra without any preliminary analysis. We need to have

49a%2B7b%2Bc+=+121c%2B11b%2Ba
48a+=+120c%2B4b
12a+=+30c%2Bb
b+=+12a-30c
b+=+6%282a-5c%29

So b is a multiple of 6; since it is a whole number less than 7, it can only be 0 or 6.

(a) b=0 means 2a-5c=0; again with a and c both whole numbers less than 7, we must have c=2 and a=5. That gives us the solution 502 (base 7) = 205 (base 11).

CHECK: 502 (base 7) = 5(49)+2 = 247; 205 (base 11) = 2(121)+5 = 247

(b) b=6 means 2a-5c=1; with a and c both whole numbers less than 7, the only possibility is a=3 and c=1. That gives us the solution 361 (base 7) = 163 (base 11).

CHECK: 361 (base 7) = 3(49)+6(7)+1 = 147+42+1 = 190; 163 (base 11) = 121+6(11)+3 = 190

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(2) We can use a bit of logical analysis to narrow down the possible solutions.

The largest 3-digit number in base 7 is 342; in base 11, 342 is 2??.

So we know abc (base 7) = cba (base 11) means c can only be either 1 or 2.

(a) If c=1....

49a%2B7b%2B1+=+121%2B11b%2Ba
48a+=+120%2B4b
12a+=+30%2Bb

The requirement that a and b both be whole numbers less than 7 gives only one solution to that equation: a=3 and b=6.

That gives us the solution 361 (base 7) = 163 (base 11) -- as we found by the earlier method.

(b) If c=2....

49a%2B7b%2B2+=+242%2B11b%2Ba
48a+=+240%2B4b
12a+=+60%2Bb

And the requirement that a and b both be whole numbers less than 7 again gives only one solution to that equation: a=5 and b=0.

And that gives us the other solution we found by the earlier method: 502 (base 7) = 205 (base 11).

ANSWERS:
(a) 502 (base 7) = 205 (base 11)
(b) 361 (base 7) = 163 (base 11)