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01 Mar 2017, 05:39
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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
A. Zero
B. One
C. Two
D. Three
E. Four
Updated on: 03 Jan 2021, 07:41
Originally posted by BrentGMATPrepNow on 01 Mar 2017, 06:33.
Last edited by BrentGMATPrepNow on 03 Jan 2021, 07:41, edited 1 time in total.
Last edited by BrentGMATPrepNow on 03 Jan 2021, 07:41, edited 1 time in total.
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Bunuel
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
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17 Aug 2017, 02:52
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GMATPrepNow
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?
