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When integer a is divided by 4 the remainder is 2, and when integer b

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When integer a is divided by 4 the remainder is 2, and when integer b  [#permalink]

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01 Mar 2017, 04:39
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95% (hard)

Question Stats:

44% (02:41) correct 56% (03:15) wrong based on 111 sessions

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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

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Re: When integer a is divided by 4 the remainder is 2, and when integer b  [#permalink]

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01 Mar 2017, 05:33
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Top Contributor
Bunuel wrote:
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

When it comes to remainders, we have a nice rule that says:
If N divided by D leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc.
For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc.
------NOW ONTO THE QUESTION----------------------------

When integer a is divided by 4 the remainder is 2
So, the possible values of a are: 2, 6, 10, 14, 18, 22, 26,....

When integer b is divided by 5 the remainder is 1
So, the possible values of b are: 1, 6, 11, 16, 21, 26, 31,....

How many integers between 20 and 29, inclusive, CANNOT be the sum a + b?
Let's see which sums we CAN get:
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
We can get ALL of the sums.

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Re: When integer a is divided by 4 the remainder is 2, and when integer b  [#permalink]

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17 Aug 2017, 01:52
1
GMATPrepNow wrote:
Bunuel wrote:
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

When it comes to remainders, we have a nice rule that says:
If N divided by D leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc.
For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc.
------NOW ONTO THE QUESTION----------------------------

When integer a is divided by 4 the remainder is 2
So, the possible values of a are: 2, 6, 10, 14, 18, 22, 26,....

When integer b is divided by 5 the remainder is 1
So, the possible values of b are: 1, 6, 11, 16, 21, 26, 31,....

How many integers between 20 and 29, inclusive, CANNOT be the sum a + b?
Let's see which sums we CAN get:
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
We can get ALL of the sums.

Is there no other way except putting in number values to solve this problem?
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Joined: 01 Jun 2016
Posts: 27
Re: When integer a is divided by 4 the remainder is 2, and when integer b  [#permalink]

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17 Aug 2017, 18:30
1
1
saswatdodo wrote:
GMATPrepNow wrote:
Bunuel wrote:
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

When it comes to remainders, we have a nice rule that says:
If N divided by D leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc.
For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc.
------NOW ONTO THE QUESTION----------------------------

When integer a is divided by 4 the remainder is 2
So, the possible values of a are: 2, 6, 10, 14, 18, 22, 26,....

When integer b is divided by 5 the remainder is 1
So, the possible values of b are: 1, 6, 11, 16, 21, 26, 31,....

How many integers between 20 and 29, inclusive, CANNOT be the sum a + b?
Let's see which sums we CAN get:
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
We can get ALL of the sums.

Is there no other way except putting in number values to solve this problem?

saswatdodo This is how I solved it:
A = 4q + 2.
B = 5p + 1.

A + B = 4q + 5p + 3.

now, 20 <= 4q + 5p + 3 <= 29
17<= 4q + 5p <=26

plugin in a couple of numbers and you will find that it will satisfy the whole range.
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Re: When integer a is divided by 4 the remainder is 2, and when integer b  [#permalink]

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24 Aug 2017, 12:43
Bunuel wrote:
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

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Re: When integer a is divided by 4 the remainder is 2, and when integer b  [#permalink]

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11 Feb 2019, 19:19
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Re: When integer a is divided by 4 the remainder is 2, and when integer b   [#permalink] 11 Feb 2019, 19:19
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