In 1962, I gave a version of the following problem to 25 Third-Grader Students (13 Native Puerto Ricans, 12 from The States) at our Campus School, Inter American University, San Germán, Puerto Rico. Each child received a DIFFERENT set of 3 numbers -- so copying from the next kid wouldn't help. Using a number trick and colored charts, each child answered ALL QUESTIONS CORRECTLY. Can you? (See an associated file, COLORED MULTIPLICATION TABLES, for methodology.)Is each of these numbers (exactly) divisible by 144? And, if not, why not? (To make a CHOICE, click in the associated eyelet. To remove the REPLY, click on "OK".)
Kid, did you do as well as our Campus School Kids? Please see below for Solution and Explanation.
Is each of these numbers (exactly) divisible by 144? And, if not, why not?
- (A) 1729306448
- (B) 1825906446
- (C) 1724906448
144 = 16 X 9. And there is an Algorithm (described in an associated file) for Factor 16, with another Algorithm for Factor 9. As I show there:
- A ("decimal") number has 16 as a factor if, and only if, the last 4 digits of the number forms a multiple of 16 -- the number preceding these don't matter. (Kids discover this by coloring "the 16-pattern": the multiples of 16 -- obtaining a pattern which repeats every 10000. Why? Because 16 = 2 X 2 X 2 X 2, which is a subpattern of 10 X 10 X 10 X 10.)
- A ("decimal") number has 9 as a factor if, and only if, the SUM OF ITS DIGTS FORMS A MULTIPLE OF 9 Thus, 1 + 4 + 4 = 9, so 144 has a factor 9. Keep adding the digits until a single digit (9) is obtained. You can also test by "casting out the nines" formed by the digits.
- A ("decimal") number has 144 as a factor if, and only if, it passes both the above tests. Now, consider, 1729306448. Its last 4 digits, 6448 forms a multiple of 16: 6448 = 16 x 403. So it passes the first test. But, 1 + 7 + 2 + 9 + 3 + 0 + 6 + 4 + 4 + 8 = 44 and 4 + 3 = 7. So this number, when divided by 9, would yield remainder 8. (You can also test by casting out the nines. Cast out the 0 and 9 digits in the number; cast out 7 + 2 = 9; cast out 6 + 3 = 9; cast out 1 + 8 = 9. This leaves the digits 4 + 4 = 8, yielding remainder 8, when divided by 9.) Hence, it fails both tests, clearly not a multiple of 144.
Next, consider number 1825906446. It passes the fails test (the 16-test) by way of the last 2 digits. (If 6448 passes in the first number, 6446 will fail by 2 in this second number.) What about its 9-ness? 1 + 8 + 2 + 5 + 9 + 0 + 6 + 4 + 4 + 6 = 45, and 4 + 5 = 9, so the number is divisible by 9. (You can also cast the nines: 1 + 8 = 9; 9 + 0 = 9; 5 + 4 = 9; leaving 2 + 6 + 4 + 6 = 18, and 1 + 8 = 9, so 9's cast out.) So, it passes the second test (9-test). But th 16-failure indicates it is not a multiple of 144.
Finally, consider 1724906448. It has the same last 4 digits (6448) as the previous number and we established that this passes the first test (the 16-test). What about its 9-ness? 1 + 7 + 2 + 4 + 9 + 0 + 6 + 4 + 4 + 8 = 45, and 4 + 5 = 9. (You can also cast out the nines: 9 and 0; 7 + 2 = 9; 1 + 8 = 9; leaving 4 + 6 + 4 + 4 = 18, and 1 + 8 = 9.) It passes the second test (9-test). Having passed both tests, you find its a multiple of 16 X 9 = 144. In fact, 1724906448 = 144 X 11978517.
Charlotte?
This is neither a trick problem nor a trivial problem. Students need to know factoring methods to do ordinary arithmetic. And this prepares them for algorithms in general mathematics and for conservation laws in physics.