When integer a is divided by 4 . . .

[Vincen]

When integer a is divided by 4 the remainder is 2, and when integer b is divided by 5 the remainder is 1. How many integers between 20 and 29, inclusive, CANNOT be the sum a + b?

A. Zero
B. One
C. Two
D. Three
E. Four

The OA is A.

I don't know how to solve this PS question. I don't where I should start. Experts, may you give me some help here? I would be thankful.

[EconomistGMATTutor]

Hello.

Let's take a careful look at your question.

When "a" is divided by 4 the remainder is "2". Possible values for "a":
2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42, . . . .

When "b" is divided by 5 the remainder is "1". Possible values for "b":
1, 6, 11, 16, 21, 26, 31, 36, 41, . . .

The integers between 20 and 29 can be written as follows:

20 = 14 + 6
21 = 10 + 11
22 = 6 + 16
23 = 2 + 21
24 = 18 + 6
25 = 14 + 11
26 = 10 + 16
27 = 26 + 1
28 = 22 + 6
29= 18 + 11

So, ALL the integers can be the sum a+b.

This is why the correct answer is A.

I hope this explanation may help you.

Feel free to ask me if you have a doubt.

Regards.

[Jeff@TargetTestPrep]

When integer a is divided by 4 the remainder is 2, and when integer b is divided by 5 the remainder is 1. How many integers between 20 and 29, inclusive, CANNOT be the sum a + b?

A. Zero
B. One
C. Two
D. Three
E. Four

We use the "quotient remainder theorem," which states: dividend = divisor x quotient + remainder.
We can create the following equations:

a = 4Q + 2

b = 5Z + 1

So, a can be 2, 6, 10, 14, 18, 22, and 26.

So, b can be 1, 6, 11, 16, 21, and 26.

We see that a + b can be the following:

14 + 6 = 20

10 + 11 = 21

6 + 16 = 22

22 + 1 = 23

18 + 6 = 24

14 + 11 = 25

10 + 16 = 26

6 + 21 = 27

22 + 6 = 28

18 + 11 = 29

Answer: A