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Remainder question

Expert replies
by Osirus@VeritasPrep » Sun Feb 07, 2010 5:21 pm
If the integer P leaves a remainder of 4 when divided by 9, all of the following must be treu EXCEPT:

A) The number that is 4 less than P is a multiple of 9

B) The number that is 5 more than P is a multiple of 9

C) The number that is 2 more than P is a multiple of 3.

D) When divided by 3, P will leave a remainder of 1.

E) When divided by 2, P will leave a remainder of 1.

I know the answer and can solve the question by plugging in, can anyone explain to me how to solve this problem WITHOUT plugging in?
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Source: — Problem Solving |

by shashank.ism » Sun Feb 07, 2010 7:39 pm
osirus0830 wrote:If the integer P leaves a remainder of 4 when divided by 9, all of the following must be treu EXCEPT:

A) The number that is 4 less than P is a multiple of 9

B) The number that is 5 more than P is a multiple of 9

C) The number that is 2 more than P is a multiple of 3.

D) When divided by 3, P will leave a remainder of 1.

E) When divided by 2, P will leave a remainder of 1.

I know the answer and can solve the question by plugging in, can anyone explain to me how to solve this problem WITHOUT plugging in?

Since integer P leaves a remainder of 4 when divided by 9

so P-4 is divisible by 9 and P+5 is divisible by 9 so A) and b) are correct.

4+2=6 6 is divisible by 3 and P is divisible by 9 and hence divisible by 3 so C) is correct.

when 4 is divided by 3 it leaves remainder 1 and P is divisble by 9 so overall a remainder of 1 will be left. so D) is correct.

when 4 is divided by 2 it will leave no remainder but P divided by 2 will remainder of 1 if P is odd and 0 if P is even.
so for P=even E) is not correct

Hence Correct answer is E
Last edited by shashank.ism on Sun Feb 07, 2010 8:04 pm, edited 2 times in total.
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by ajith » Sun Feb 07, 2010 7:51 pm
osirus0830 wrote:If the integer P leaves a remainder of 4 when divided by 9, all of the following must be treu EXCEPT:

A) The number that is 4 less than P is a multiple of 9

B) The number that is 5 more than P is a multiple of 9

C) The number that is 2 more than P is a multiple of 3.

D) When divided by 3, P will leave a remainder of 1.

E) When divided by 2, P will leave a remainder of 1.

I know the answer and can solve the question by plugging in, can anyone explain to me how to solve this problem WITHOUT plugging in?
Say the number is P; P=9k+4

A) says 9k+4-4 = 9k is divisible by 9
B) says 9k+4+5 = 9k+9 = 9(k+1) is divisible by 9
C)9k+4+2 = 9k+6 = 3(3k+2) is a multiple of 3
D) 9k+4 = 9k+3+1 = 3(3k+1) +1 leaves a remainder of 1
E) 9k+4= 2(4k+2) + k ; it depends on the value of k what remainder it leaves when divided by 2

Hence E
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by Osirus@VeritasPrep » Sun Feb 07, 2010 7:53 pm
shashank.ism wrote:
osirus0830 wrote:If the integer P leaves a remainder of 4 when divided by 9, all of the following must be treu EXCEPT:

A) The number that is 4 less than P is a multiple of 9

B) The number that is 5 more than P is a multiple of 9

C) The number that is 2 more than P is a multiple of 3.

D) When divided by 3, P will leave a remainder of 1.

E) When divided by 2, P will leave a remainder of 1.

I know the answer and can solve the question by plugging in, can anyone explain to me how to solve this problem WITHOUT plugging in?

Ans is D
SMH...Did you even read my question? Not only that, but you even got the answer wrong.
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