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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, 05:39
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43% (01:45) correct 57% (02:21) wrong based on 183 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?
Re: When integer a is divided by 4 the remainder is 2, and when integer b [#permalink]
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01 Mar 2017, 06:33
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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.
Re: When integer a is divided by 4 the remainder is 2, and when integer b [#permalink]
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17 Aug 2017, 02:52
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.
Answer: A
Is there no other way except putting in number values to solve this problem?
Re: When integer a is divided by 4 the remainder is 2, and when integer b [#permalink]
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17 Aug 2017, 19:30
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.
Answer: A
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.
Add both of them. 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.
Re: When integer a is divided by 4 the remainder is 2, and when integer b [#permalink]
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24 Aug 2017, 13:43
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
Answer: A
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