Word prob: If you divide by 2, 3, 4, 5, or 6, the remainder

[penny]

I'm just returning to school after many years and need help in math. Like others, word problems seem to be my biggest challenge so far. How do you even know where to begin to solve them?

For example (taken from my textbook):
A passenger in a taxicab asks for the driver's company number. The driver says abruptly, "Sure, you can have my number. Work it out:

If you divide it by 2,3,4, 5, or 6 you will get a remainder of 1. If you divide it by 11, the remainder will be 0 and no driver has a company number that meets these requirements and is smaller than this one."

Determine the number.

 

[stapel]

Re: Word prob: If you divide by 2, 3, 4, 5, or 6, the remain

...word problems.... How do you even know where to begin to solve them?

You learn how to work with them by working with them. It's very much a matter of experience, learning how to recognize patterns, and learning useful techniques. There is no "master formula" that will solve them all. There is no "royal road".

A passenger in a taxicab asks for the driver's company number. The driver says abruptly, "Sure, you can have my number. Work it out:

If you divide it by 2,3,4, 5, or 6 you will get a remainder of 1. If you divide it by 11, the remainder will be 0 and no driver has a company number that meets these requirements and is smaller than this one."

Determine the number.

If the number is evenly divisible by 11, then it must be a multiple of 11. So start working with multiples of 11. Either solve by "brute force" (trying every multiple of 11 until you find one that works) or else try to find a pattern (or a rule that they gave you in class, if applicable).

Note: The exercise does not state in which country this interchange took place. Knowing that would have been helpful in narrowing your search. For instance, had the exercise been placed within the USA, the number would have had seven digits (123-4567) or, if the area code is included, ten digits (123-456-7890).

Eliz.

 

[soroban]

Re: Word prob: If you divide by 2, 3, 4, 5, or 6, the remain

Hello, Penny!

This problem looks like trial-and-error, but there is a logical solution.

A passenger in a taxicab asks for the driver's company number.

The driver says abruptly, "Sure, you can have my number. Work it out:
. . If you divide it by 2,3,4, 5, or 6 you will get a remainder of 1.
. . If you divide it by 11, the remainder will be 0.
And no driver has a number that meets these requirements and is smaller than this one."

Determine the number.

Let \(\displaystyle N\) be the driver's company number.

When dividing by 2, 3, 4, 5, or 6, there is a remainder of 1.
. . Then \(\displaystyle N\) is of the form: \(\displaystyle \,60a\,+\,1\,\) for some positive integer \(\displaystyle a.\)
Note: 60 is the LCM of 2,3,4,5,6.

But \(\displaystyle N\) is evenly divisible by 11: \(\displaystyle \:N \:=\:60a\,+\,1\:=\:11b\,\) for some positive integer \(\displaystyle b.\;\) [1]

Solve for \(\displaystyle b:\;\;b\:=\:\frac{60a\,+\,1}{11}\;\;\Rightarrow\;\;b \:=\:5a\,+\, \frac{5a\,+\,1}{11}\;\) [2]

Since \(\displaystyle b\) is an integer, \(\displaystyle 5a\,+\,1\) must be divisible by 11.
. . The smallest case is when: \(\displaystyle a\,=\,2\)

Substitute into [2]: \(\displaystyle \:b\:=\:5(2)\,+\,\frac{5(2)\,+\,1}{11}\;\;\Rightarrow\;\;b\:=\:11\)

Substitute into [1]: \(\displaystyle \:N \:=\:11(11)\)


Therefore: \(\displaystyle \L\:N\:=\:121\;\) . . . There!