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Re: If a, b, c are consecutive positive integers and a < b < c, which of [#permalink]

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07 Feb 2011, 22:58

1

This post received KUDOS

If a, b, c are consequitive positive integers and a<b<c, which of the following must be true? 1. c-a=2 2. abc is an even integer 3. (a+b+c) /3 is an integer

1. a=a b=a+1 c=a+2

c-a=a+2-a=2. Correct

2. abc a can be either even or odd if a=even; abc=even; if a=odd,b=even; abc=even

Rule: multiple of any number of integers if atleast once multiplied by an even number will result in even.

even*even=even even*odd=even

Correct

3. (a+b+c) /3 is an integer Again;

a=a b=a+1 c=a+2

a+b+c=(a+a+1+a+2)=3a+3

3a is always divisible by 3 3 is divisible by 3. Thus "3a+3" is also divisible by 3.

Rule: if integer x is divisible by n and integer y is divisible by n then; x+y must be divisible by n

Re: If a, b, c are consecutive positive integers and a < b < c, which of [#permalink]

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08 Feb 2011, 00:22

The sum of three consecutive numbers will always be divisible by 3 <<< It's a helpful result. You might want to commit it to your memory
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"Wherever you go, go with all your heart" - Confucius

If a, b, and c are consecutive positive integers and a<b<c, which of the following must be true? 1. c-a=2 2. abc is an even integer 3. (a+b+c) /3 is an integer

A. 1 B. 2 c. 1 and 2 d. 2 and 3 e. 1, 2 and 3

1. c-a=2 --> a, b, c are consequitive positive integers and a<b<c then c=a+2 --> c-a=2. So this statement is always true;

2. abc is an even integer --> out of any 3 consecutive integers at least one must be even thus abc=even. So this statement is also always true;

3. (a+b+c)/3 is an integer --> the sum of odd number of consecutive integers is ALWAYS divisible by that odd number. So this statement is also always true. Or: (a+b+c)/3=(a+a+1+a+2)/3=(3a+3)/3=a+1=integer.

Answer: E.

AmrithS wrote:

The sum of three consecutive numbers will always be divisible by 3 <<< It's a helpful result. You might want to commit it to your memory

Not only that:

• If \(k\) is odd, the sum of \(k\) consecutive integers is always divisible by \(k\). Given \(\{9,10,11\}\), we have \(k=3\) consecutive integers. The sum of 9+10+11=30, therefore, is divisible by 3.

• If \(k\) is even, the sum of \(k\) consecutive integers is never divisible by \(k\). Given \(\{9,10,11,12\}\), we have \(k=4\) consecutive integers. The sum of 9+10+11+12=42, therefore, is not divisible by 4.

• The product of \(k\) consecutive integers is always divisible by \(k!\), so by \(k\) too. Given \(k=4\) consecutive integers: \(\{3,4,5,6\}\). The product of 3*4*5*6 is 360, which is divisible by 4!=24.
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