Question 609908
<p>The other poster's previously stated solution is NOT CORRECT. Here is the proper math:</p>
<p>&nbsp;453<br />&nbsp; x 4<br />&nbsp;-----<br />&nbsp; &nbsp;&nbsp; 5&nbsp; (3x4 is 12 which is ONE 7 with remainder of 5, so carry the ONE)</p>
<p>&nbsp;&nbsp; 1<br />&nbsp;453<br />&nbsp; x 4<br />&nbsp;-----<br />&nbsp;&nbsp; 05 (5x4 is 20, then plus 1 is 21 which is THREE 7s with 0 left over, so carry the THREE)</p>
<p>&nbsp;31<br />&nbsp;453<br />&nbsp; x 4<br />&nbsp;-----<br />&nbsp; 505 (4x4 is 16, then plus 3 equals 19 which is TWO 7s with 5 left over, so carry the TWO)</p>
<p>231<br />&nbsp; 453<br />&nbsp;&nbsp; x 4<br />&nbsp; -----<br />&nbsp; 2505 (you can imagine a zero in front of the "453" so 4x0 is 0 plus 2 is 2 which is ZERO 7s with 2 left over)</p>
<p>Notice the pattern? The process of multiplication is no different regardless of base. The only thing different is in what is carried. In this case it is sets of 7 instead of the sets of 10 you are used to.</p>
<p>The correct answer is 2505 in base 7.</p>
<p>This can be checked by converting the numbers to base 10, then multiplying, and converting back to base 7:</p>
<p>453 base 7 = (4 sets of 7^2) + (5 sets of 7^1) + (3 sets of 7^0)<br />&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; =&nbsp;&nbsp;&nbsp;&nbsp; (4*49)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; +&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (5*7)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; +&nbsp;&nbsp;&nbsp;&nbsp; (3*1)<br />&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp; &nbsp; &nbsp;&nbsp; &nbsp; =&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 196&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp; +&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 35&nbsp;&nbsp;&nbsp; &nbsp;&nbsp; &nbsp; +&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 3<br />&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp; &nbsp; &nbsp;&nbsp; = 234</p>
<p>4 base 7 = (4 sets of 7^0)<br />&nbsp;&nbsp;&nbsp;&nbsp; &nbsp; &nbsp;&nbsp; &nbsp;&nbsp; =&nbsp;&nbsp;&nbsp;&nbsp; (4*1)<br />&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; = 4</p>
<p>234 x 4 = 936 and that converted to base 7 is:</p>
<p>936/7=133 with remainder of 5 so put 5 in rightmost position and<br />133/7=19 with remainder of 0 so put 0 in position to left of the 5.... 05 and<br />19/7=2 with remainder of 5 so put 5 in position to left of the 05... 505 and<br />2 is less than 7 so put the 2 in position to left of the 505... 2505</p>
<p>So to sum up (carried numbers in both cases are in the top row):</p>
<p>&nbsp;&nbsp; Base 7&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; Base 10<br />-----------------------------<br />&nbsp; &nbsp; 231&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp; &nbsp; 11<br />&nbsp; &nbsp;&nbsp; 453&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 234<br />&nbsp; &nbsp; &nbsp; x 4&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp; &nbsp; x 4<br />&nbsp;&nbsp; ----------------------<br />&nbsp;&nbsp; &nbsp; 2505&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 936</p>
<p>I hope that helps :)</p>