Math CentralQuandaries & Queries


Question from Cheryl, a parent:

Sunshine Math - 4, Jupiter 1, question 4 which is (a, b, c & d). I am wanting to know question 4d. Think about the following list of number pairs. Three is the first number of a pair, and 8 is the second.

3 - 8
4 - 11
5 - 14
6 - 17 and so forth

Question d. - If a number "n" is the first number, what is the second number?

I don't know how my son should answer this question. It sounds like it should be a formula but how does he get to it and how do I explain it to him.

Looking forward to your help.

Hi Cheryl,

Sometimes before we can find a formula, we need to find the pattern from getting to the second number given the first. Let me guide you through my though process as to how I came up with the answer:

First, I try to see what the difference (subtraction) or quotient (division) is between the numbers:

Difference Quotient
8-3 = 5 8/3 = 2 2/3
11-4 =7 11/4 = 2 3/4
14-5 = 9 14/5 = 2 4/5
17-6=11 17/6 = 2 5/6

In this case, the quotients don't help me so I'll ignore them. I do notice that an odd number is added to each of the first numbers to yield the second number (e.g. 3+5=8). So I need to think: "How can I make the odd number using the first number?"

If multiply 3 by 2, it will yield 6 which is close to 5. If I subtract 6 by 1 then I will have 5. So to put this in formula terms, if my first number is n then I would be adding 2n-1 to yield the second number:


Now let's check to make sure that this formula works for all of the first numbers:

first number: n second number: 3n-1
3 3(3)-1=9-1=8
4 3(4)-1=12-1=11
5 3(5)-1=15-1=14
6 3(6)-1=18-1=17

The formula appears to work so we can assume it will work for all numbers. There are many ways that you could have arrived at the result but sometimes if you cannot see the pattern you just have to "play" with the numbers a bit until you find something that works.

Hope this helps,

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