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When the positive integer N is divided by 6
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12 Nov 2020, 19:21
12 Nov 2020, 19:21
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Question Stats:
53% (01:58) correct
46% (01:32) wrong based on 43 sessions
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When the positive integer N is divided by 6, the remainder is 3. When the positive integer M is divided by 4, the remainder is 1. Which of the following integers could be the remainder when the product of the two numbers, NM, is divided by 12?
Indicate all such integers.
A. 1
B. 3
C. 5
D. 7
E. 9
Indicate all such integers.
A. 1
B. 3
C. 5
D. 7
E. 9
Re: Divisibility
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12 Nov 2020, 20:56
12 Nov 2020, 20:56
3
N=6a+3
M=4b+1
N*M= 24ab+12b+6a+3 (after simplification)
1st two-term of the R.H.S is always divisible by 12.
However, the remainder depends on the last two-terms
case-1: when a' is odd then the remainder is 9 (E)
case-2: when 'a' is even then the remainder is 3 (B)
answers: B, E
M=4b+1
N*M= 24ab+12b+6a+3 (after simplification)
1st two-term of the R.H.S is always divisible by 12.
However, the remainder depends on the last two-terms
case-1: when a' is odd then the remainder is 9 (E)
case-2: when 'a' is even then the remainder is 3 (B)
answers: B, E
Re: When the positive integer N is divided by 6
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11 Jan 2022, 10:31
11 Jan 2022, 10:31
endurance wrote:
N=6a+3
M=4b+1
N*M= 24ab+12b+6a+3 (after simplification)
1st two-term of the R.H.S is always divisible by 12.
However, the remainder depends on the last two-terms
case-1: when a' is odd then the remainder is 9 (E)
case-2: when 'a' is even then the remainder is 3 (B)
answers: B, E
M=4b+1
N*M= 24ab+12b+6a+3 (after simplification)
1st two-term of the R.H.S is always divisible by 12.
However, the remainder depends on the last two-terms
case-1: when a' is odd then the remainder is 9 (E)
case-2: when 'a' is even then the remainder is 3 (B)
answers: B, E
i reached half correctly but the 2 cases thats where i stuck. Please explan
using numbers.
