How many two digit numbers have both digits odd, and add to 10? How many three digit number have all odd digits, and add to 10? What about four digit numbers? Ask your students to find out how many altogether have this property.
When 37 is multiplied by the sum of its digits, we obtain 370 which is the sum of its digits cubed. That is,
37 (3 + 7) = [3.sup.3] + [7.sup.3]
There is only one other two-digit number with this property. See if your students can discover it.
While we are referring to cubes it follows that:
370 = [3.sup.3] + [7.sup.3] + [0.sup.3]
371 = [3.sup.3] + [7.sup.3] + [1.sup.3]
See if your students can find two other three-digit numbers with this property.
An easier task is to find out what happens when 37 is multiplied in turn by 3, 6, 9, 12, 15, 18, 21, 24 and 27, respectively. Now find the sum of the digits in each product obtained. The first two products and sums yield:
37 x 3 = 111
1 + 1 + 1 = 3
37 x 6 = 222
2 + 2 + 2 = 6
Note what happens when the multipliers are further multiples of 3 such as 30, 33, 36 ...
Now when 37 is multiplied by other numbers less than 27, which are not multiples of 3, the products can be arranged in groups of three where the multiples differ by 9 and 18. For example, two of these groups are:
37 x 1 = 037
37 x 2 = 074
37 x 10 = 370
37 x 11 = 407
37 x 19 = 703
37 x 20 = 740
Let your students discover the other groups. Although the sum of the digits in each group's products are the same, what else can be said about repeated additions of these sums?
Some scientists in Kazakhstan have recently devised a hypothetical mathematical approach to look for patterns in the genetic codes that create proteins. Atomic masses that appear in some protein patterns include 74, 703 and 1665 which are all multiples of 37. The jury is still out on whether this is significant or not (Kemp, 2014).
The fourth hexagonal number is 37 (= 1 + 6 + 12 + 18), obtained by substituting n = 4 in the general expression 3n (n-1) + 1.
See if your students can generate this answer from the expression A[n.sup.2] + Bn + C using the first three hexagonal numbers 1, 7 (= 1 + 6), 19 (= 1 + 6 + 12).
Consider a circular pizza. Ask your students to show that the maximum number of pieces obtained after 8 straight cuts right across is 37. Investigate the formula for the maximum number of pieces after 'n' straight cuts.
Finally, a problem that does not involve 37, but shows the power of mathematics to simplify what at first appears to be a major task. The digits 1 to 9 are to be arranged to form a nine-digit number, so that the leading number is divisible by 1, the leading two numbers are divisible by 2, the leading three numbers are divisible by 3, and so on to the whole number being divisible by 9. Now the number of ways to rearrange the 9 digits is:
9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 9! = 362 880 Therefore, attempting to solve this problem by trying every possible combination of the nine digits would be a very long task. But if your students can realise that the fifth digit from the front must be 5, and the digits in the even places can only be 2, 4, 6 or 8, the number of possibilities to be investigated is reduced to 4! x 4! = 576, which is a much smaller number to investigate. When the rules for division by 3 and 9 are also used, the workload is even less. See if your students can find the only solution.
Kemp, C. (2014). Is the answer to life, the universe and everything 37? New Scientist (3000/3001), 61-63.
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|Title Annotation:||discovery with Neville de Mestre|
|Author:||de Mestre, Neville|
|Publication:||Australian Mathematics Teacher|
|Date:||Dec 22, 2015|
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