document.write( "Question 1110462: In what bases, b, does (b+7) divide into (9b+7) without any remainder? \n" ); document.write( "

Algebra.Com's Answer #725456 by greenestamps(13203) $\"\"$ $\"About$

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\n" ); document.write( "We have to have $\"%289b%2B7%29%2F%28b%2B7%29\"$ equal to a whole number; and presumably the value(s) of b we are looking for are integers greater than or equal to 2.

\n" ); document.write( " $\"%289b%2B7%29%2F%28b%2B7%29+=+%28%289b%2B63%29-56%29%2F%28b%2B7%29+=+9+-+56%2F%28b%2B7%29\"$

\n" ); document.write( "Since 9 is a whole number, $\"56%2F%28b%2B7%29\"$ has to be a whole number also.

\n" ); document.write( "So we know b has to be a whole number greater than or equal to 2; and we know 56 has to be divisible by (b+7). The only divisors of 56 that are 9 or greater are 14, 28, and 56.

\n" ); document.write( "Since those divisors are the values of (b+7), the three bases for which (b+7) divides into (9b+7) without any remainder are 7, 21, and 49.

\n" ); document.write( "Check:
\n" ); document.write( "b=7: (9b+7)(b+7) = 70/14 = 5
\n" ); document.write( "b=21: (9b+7)/b+7) = 196/28 = 7
\n" ); document.write( "b=49: (9b+7)/b+7) = 448/56 = 8 \n" ); document.write( "

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