Singapore Primary Math Explained scmath
Updated: 1 March, 2007
The question below was raised by a member of the SingaporeMath group at Yahoo sometime back. I had then created a web page to show my solution to the problem. I've decided to put the question here (with some amendments for clarity) for those who are interested.
What is the 2001st number in the following number sequence?
1 – 1,  2 – 1,  1 – 2,  3 – 1,  2 – 2,  1 – 3,  4 – 1,  3 – 2,  2 – 3,  1 – 4,  5 – 1,  4 – 2,  3 – 3,  .  .  . 
I would classify this type of math problem as "investigative math". I think that such "investigative math" problems are common for students in the primary Gifted Education Programme (GEP).
Look for patterns
We are asked to find the value of the 2001st number in the above number sequence. In such a situation, I usually examine the number sequence to look for patterns or relationships to help me predict the number at the required position.
Looking at the numerator and denominator as separate number sequences, we can see that the numbers in each number sequence can be grouped into subseries. The first 6 subseries of the numerator and denominator number sequence are shown below.
Numerator:
1  2 1  3 2 1  4 3 2 1  5 4 3 2 1  6 5 4 3 2 1
Denominator:
1  1 2  1 2 3  1 2 3 4  1 2 3 4 5  1 2 3 4 5 6
Can you spot any more patterns or relationships?
(I'll use the term "rank" to refer to the location of a subseries within the number sequence. The first subseries is ranked 1, the second subseries is ranked 2, the third subseries is ranked 3 and so on. The term "sequence position" refers to the position of a number within the number sequence.)
Note that the largest number in each subseries is the same as it's rank in the number sequence. The number of numbers in a subseries is also the same as it's rank or it's largest number. As an example, the largest number in the 5th subseries ("5 4 3 2 1" for the numerator or "1 2 3 4 5" for the denominator) is 5. There are also 5 numbers in the 5th subseries.
To find the value of the 2001st number in the number sequence, we need to find the rank of the subseries within which the 2001st number occurs and the position of the 2001st number within that subseries.
Make a systematic list
To do this, we'll tabulate the first six subseries in such a way as to show the relationship between the sequence position of a number in the number sequence and the subseries within which it falls. I've included the subseries itself beside the tabulation for clarity.
Rank of sub−series  Sequence position of number in number sequence  Sub−series (Denominator)  

1  1  1  
2  2  3  1  2  
3  4  5  6  1  2  3  
4  7  8  9  10  1  2  3  4  
5  11  12  13  14  15  1  2  3  4  5  
6  16  17  18  19  20  21  1  2  3  4  5  6 
(To check that my tabulation is clear to you, it shows that the 3rd number of the number sequence falls within the 2nd subseries and the 12th number of the number sequence falls within the 5th subseries. The sequence position of the 2nd number of the 6th subseries is the 17.)
Look for patterns
Can you spot any patterns or relationships between the sequence position of a number and the rank of the subseries?
Note that the sequence position of the last number in each subseries (1, 3, 6, 10, 15, 21) equals the sum of the ranks (1, 2, 3, 4, 5, 6 ) of all the subseries before it and it's own rank. For example, the sequence position of the last number of the 3rd subseries is 6 which is 1+2+3 (1 and 2 being the rank of the two subseries before it and 3 being the rank of the 3rd subseries itself). Similarly, the sequence position of the last number in the 6th subseries is 21 which is 1+2+3+4+5+6.
We now know that the sequence position of the last number of a particular subseries is the value of 1+2+3+4+ ....... +N where N is the rank of the subseries. But how do we calculate the value of the series "1+2+3+.... + N"? This itself can be a problem for "investigative math" (you may wish to try it out yourself).
Anyway, to make matters short the value of "1+2+3+ .... + N" is
N × (N+1) 
2 
As an example, the value for N=4 is
4 × (4+1) 
2 
4 × 5 
2 
6 × (6+1) 
2 
Now that we know how to calculate the sequence position of the last number of a subseries, how do we find out which subseries the 2001st number belongs to?
Use guess and check
Knowing the sequence position of the last number in a subseries, we can easily find out the sequence position of it's first number. It's useful to list our results in "guess and check" to help us make better guesses. Let's start our guess with the 50th subseries (N = 50).
Guess  Rank  Last Number of subseries  First Number of subseries  

(N)  (L)=

(LN)+1  
1  50  1,275  1,226  
Check: 2001 is not within the 50th subseries. It is 726 (20011275) more than the sequence position of the last number. We know from the ranks of the subseries that follows that the sequence position of the last number of each subsequent subseries increases by 51, 52, and so on. To choose the rank for our next guess, lets assume an average increase of 55. Therefore to reach 726, we should increase the rank by 726/55 or approximately 13. Lets try 50+13 for the next value of N.  
2  63  2,016  1,954  
Check: Yes! The 2001st number is within the 63rd subseries. 
The 2001st number is within the 63rd subseries. It's position within the subseries is given by (20011954)+1 which equals to 48. For the denominator, the value of the number at the 48th position of this subseries is 48. For the numerator, the value is (63  48) + 1 which equals to 16.
Therefore, the 2001st number in the series
1 – 1,  2 – 1,  1 – 2,  3 – 1,  2 – 2,  1 – 3,  4 – 1,  3 – 2,  2 – 3,  1 – 4,  5 – 1,  4 – 2,  3 – 3,  .  .  .  is  16 –– 48 
I had mentioned earlier that students in the GEP had long been exposed to "investigative math" problems as part of their enriched curriculum. I feel that with the continuing emphasis on problem solving and thinking skills, more of such questions will filter down to the mainstream students.