Question 1166029
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We need to find a, b, and c such that abc (base 7) = cba (base 11).<br>
Note that a, b, and c must be whole numbers less than 7.<br>
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(1) We can find the solutions using algebra without any preliminary analysis.  We need to have<br>
{{{49a+7b+c = 121c+11b+a}}}
{{{48a = 120c+4b}}}
{{{12a = 30c+b}}}
{{{b = 12a-30c}}}
{{{b = 6(2a-5c)}}}<br>
So b is a multiple of 6; since it is a whole number less than 7, it can only be 0 or 6.<br>
(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).<br>
CHECK: 502 (base 7) = 5(49)+2 = 247; 205 (base 11) = 2(121)+5 = 247<br>
(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).<br>
CHECK: 361 (base 7) = 3(49)+6(7)+1 = 147+42+1 = 190; 163 (base 11) = 121+6(11)+3 = 190<br>
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(2) We can use a bit of logical analysis to narrow down the possible solutions.<br>
The largest 3-digit number in base 7 is 342; in base 11, 342 is 2??.<br>
So we know abc (base 7) = cba (base 11) means c can only be either 1 or 2.<br>
(a) If c=1....<br>
{{{49a+7b+1 = 121+11b+a}}}
{{{48a = 120+4b}}}
{{{12a = 30+b}}}<br>
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.<br>
That gives us the solution 361 (base 7) = 163 (base 11) -- as we found by the earlier method.<br>
(b) If c=2....<br>
{{{49a+7b+2 = 242+11b+a}}}
{{{48a = 240+4b}}}
{{{12a = 60+b}}}<br>
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.<br>
And that gives us the other solution we found by the earlier method: 502 (base 7) = 205 (base 11).<br>
ANSWERS:
(a) 502 (base 7) = 205 (base 11)
(b) 361 (base 7) = 163 (base 11)<br>