(I) Consider two 2-digit base ten positive integers, distinct or otherwise, each having the last digit as 5.
Devise an algorithm for mentally multiplying the two numbers.
(II) Derive the 2-digit duodecimal analogue to (I) that utilizes a very very similar algorithm.
Extra Challenge: Generalize (I) and (II) to all positive integer bases less than 37.
Clarification: It may NOT be very facile to effect the mental multiplication in part (II), but with the algorithm, it would be fairly easy to derive the product very quickly by way of p&p.
Writing the numbers as (10a+5)(10b+5)
the product 100ab+50a+50b+25
can be written as 100(ab +(a+b)/2)+25
So you just need to add ab with half of a+b.
If a+b is odd, ignore the remainder and the number ends in 75 instead of 25.
Example 45*75
4*7=28, (4+7)/2=5r1, 28+5=33
Answer 3375
In duodecimal the trick for numbers ending in 6 (half the base) would look similar
(10a+6)(10b+6)
=100ab+60a+60b+30
=100(ab+(a+b)/2)+30
Where the last two digits are 90 if a+b is odd.
Example 46*76 (remember this is all base 12)
4*7=24, (4+7)/2=5r1, 24+5=29
Answer 2990
A similar trick should hold in and EVEN base 2n where both numbers end in n. But you'd need to know your times tables in that base.
For ODD bases, I'm not sure how you'd want to generalize, since the last digit can't be half the base.
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Posted by Jer
on 2022-09-26 15:48:00 |
Solution for (I) and (II)
