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Solving Complex Problems Using Number Line

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Solving Complex Problems Using Number Line [#permalink]

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New post 28 Apr 2015, 06:56
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We are starting this thread to show how we can solve complex divisibility problems and inequalities using the basic properties of a number line.

1: Consecutive numbers and divisibility by 2, 4 and 8:

Attachment:
no. line 1.png
no. line 1.png [ 27.38 KiB | Viewed 3336 times ]


Few facts to note:
    1. On a number line, odd and even numbers alternate. For example: 1 is odd, 2 is even. So, for any two consecutive number one will be odd and the other even.
    2. Between two consecutive even numbers, one will be divisible by 4 and the other will only be divisible by 2 and not by 4.
    For example:
    2 and 4 -> 2 is not divisible by 4 but 4 is.
    4 and 6 -> 4 is divisible by 4 but 6 is not.

Let's consider a product of any three consecutive numbers n(n+1)(n+2)

Case 1: n is odd:

n(n+1)(n+2) -> odd * even * odd

Takeaway: The product of three consecutive numbers n(n+1)(n+2) is definitely divisible by 2 if n is odd.

Case 2: n is even:

n(n+1)(n+2) -> even * odd * even
n and n+2 are two consecutive even numbers. As we know between two consecutive even numbers, one is divisible by 4 and other is only divisible by 2 and not 4. Hence, n(n+2) will be divisible by 2, 4 and 2*4 =8

Takeaway: The product of three consecutive numbers n(n+1)(n+2) is definitely divisible by 2, 4 and 8 if n is even.

In this case we can also say the last two digits of the product are divisible by 4 (a number is only divisible by 4 if its last two digits are) and the last three digits of the product are divisible by 8 (a number is only divisible by 8 if its last three digits are)

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Last edited by PrepTap on 06 May 2015, 22:56, edited 2 times in total.

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Solving Complex Problems Using Number Line [#permalink]

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New post 28 Apr 2015, 23:58
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2: Consecutive numbers and divisibility by 3 and 6:

On a number line every nth number is divisible by n (n, 2n, 3n….)

Attachment:
No. line muliples of 3.png
No. line muliples of 3.png [ 36.12 KiB | Viewed 3265 times ]


E.g.: Every 3rd number is divisible by 3 (3, 6, 9…)
Every 4th number is divisible by 4 (4, 8, 12..)

As every third number is divisible by 3, any number n on the number line:
-> Is either, divisible by 3 (it is of the form 3k)
-> Or, leaves a remainder 1 when divided by 3 (it is of the form 3k + 1) - in this case n+2 will be divisible by 3
-> Or, leaves a remainder 2 when divided by 3 (it is of the form 3k+2) - in this case n+1 will be divisible by 3

Hence, the product of three consecutive numbers n(n+1)(n+2) will always be divisible by 3.
As discussed in the previous post, the product of three consecutive numbers is always divisible by 2. So, the product of three consecutive numbers will always be divisible by 2*3 = 6 as well.

Takeaway: a product of three consecutive numbers is always divisible by 3 and 6.

We can also say that the sum of the digits of the product is divisible by 3 (a number is only divisible by 3 if the sum of its digits is divisible by 3)

Now, let’s consider the product of three numbers of the form:
(n +/- 3k)(n +1 +/- 3l)(n + 2 +/- 3m), where k, l and m are integers.

As you can see, a multiple of three is added to the individual numbers in the product of consecutive integers n(n+1)(n+2).

For example: 4 * 8 * 6
It is of the form (4) * (5 + 3) * (6) - > 3 is added to the middle number in the product of consecutive numbers 4*5*6
(4) * (5 + 3) * (6) = 4 * 5 * 6 + 4 * 3 * 6
This is clearly divisible by 3

(n +/- 3k)(n +1 +/- 3l)(n + 2 +/- 3m) = n(n+1)(n+2) + terms divisible by 3

As the product of three consecutive numbers is always divisible by 3, the above product is divisible by 3.

Takeaway: the product of three numbers of the form (n +/- 3k)(n +1 +/- 3l)(n + 2 +/- 3m) is always divisible by 3.

You can use this property for solving OG 13 PS Question 87

If n is an integer greater than 6, which of the following must be divisible by 3 ?
    (A) n(n+1)(n-4)
    (B) n(n+2)(n-1)
    (C) n(n+3)(n-5)
    (D) n(n+4)(n-2)
    (E) n(n+5)(n-6)

Option (A) - n(n+1)(n-4) = n(n+1) (n+2 - 6) (as n>6, this product will be positive)
As you can see this product is of the form n(n+1)(n+2 - 3*2)
It is divisible by 3.

Answer: Option (A)

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Last edited by PrepTap on 06 May 2015, 22:57, edited 1 time in total.

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Solving Complex Problems Using Number Line [#permalink]

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New post 30 Apr 2015, 04:11
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3. Solving Inequalities using number line:

What does a>b mean:
It means "a" is always on the right side of "b" on the number line.
Attachment:
Screen Shot 2015-04-30 at 5.12.43 pm.png
Screen Shot 2015-04-30 at 5.12.43 pm.png [ 29.78 KiB | Viewed 3204 times ]

Also, "b" is always on the left side of "a" on the number line.
Attachment:
Screen Shot 2015-04-30 at 5.15.47 pm.png
Screen Shot 2015-04-30 at 5.15.47 pm.png [ 31.18 KiB | Viewed 3209 times ]


Now, lets consider this inequality:
4a > 5b
⇨ a > 5b/4
⇨ a > b + b/4
We can say that a will always be on right side of (b + b/4)
Hence, when "b" is positive:
-> "a" will also be positive and a > b
Attachment:
Screen Shot 2015-04-30 at 5.26.31 pm.png
Screen Shot 2015-04-30 at 5.26.31 pm.png [ 17.04 KiB | Viewed 3201 times ]

But, when "b" is negative:
-> a will always be on the right side of (b + b/4) and so it can fall between (b + b/4) and b.
Attachment:
Screen Shot 2015-04-30 at 4.56.56 pm.png
Screen Shot 2015-04-30 at 4.56.56 pm.png [ 18.78 KiB | Viewed 3205 times ]

Thus, it is not necessary that a > b.
For values of "a" satisfying the inequality 5b/4 < a < b (where b is negative), 4a > 5b but a is not greater than b.

Takeaway: For an inequality ka > lb, where k and l are positive integers and k<l, we can infer that a > b only when b is positive.

With this concept in mind take a look at the question:
is-n-m-where-n-and-m-are-real-numbers-196938.html

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Last edited by PrepTap on 06 May 2015, 22:58, edited 1 time in total.

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Re: Solving Complex Problems Using Number Line [#permalink]

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4. Absolute Values and the Number Line

On a number line, \(|x-a|\) denotes the distance between x and a irrespective of the signs of x and a.

Case 1: x and a are positive

Let's say x = 1 and a = 3
Attachment:
x and a positive.png
x and a positive.png [ 15.09 KiB | Viewed 3114 times ]

|x-a| = |1 - 3| = |-2| = 2

The distance remains the same when the values of x and a are swapped.
Attachment:
x and a positive 2.png
x and a positive 2.png [ 14.91 KiB | Viewed 3115 times ]


Case 2: x and a are negative

Let's say x = -1 and a = -3
Attachment:
x and a negative.png
x and a negative.png [ 13.67 KiB | Viewed 3111 times ]

|x-a| = |-1 + 3| = |2| = 2

The distance remains the same when the values of x and a are swapped.


Case 3: x and a have opposite signs

|x - a| won't change as long as the distance between x and a is the same.

Attachment:
x negative a positive.png
x negative a positive.png [ 10.85 KiB | Viewed 3118 times ]



Takeaway: To calculate |x - a|, find out the distance between x and a on the number line, irrespective of the signs of x and a.

Now let's take a look at an inequality involving absolute values using the number line.

|x - a| > |x - b| > |x - c|

All this inequality is saying that the distance between x and a is greater than the distance between x and b, which in turn is greater than the distance between x and c.
This is true whether x, a, b and c are positive, negative or zero.

Therefore, all the arrangements illustrated below are possible where the distances satisfy the inequality above.

x c b a
Attachment:
x c b a.png
x c b a.png [ 25.34 KiB | Viewed 3115 times ]


a b c x
Attachment:
a b c x.png
a b c x.png [ 26.28 KiB | Viewed 3117 times ]


c x b a
Attachment:
c x b a.png
c x b a.png [ 25.62 KiB | Viewed 3117 times ]


a b x c
Attachment:
a b x c.png
a b x c.png [ 23.94 KiB | Viewed 3117 times ]


b x c a
Attachment:
b x c a.png
b x c a.png [ 25.16 KiB | Viewed 3114 times ]


a c x b
Attachment:
a c x b.png
a c x b.png [ 26.47 KiB | Viewed 3120 times ]


It doesn't matter whether a, b, and c are to the left or to the right of x as long as the inequality regarding distances is maintained.
In the arrangement of a, b, c and x, the farthest point from x (namely a) will always be at one of the edges and never in the middle. Think about it and you'll see why this is so.
Also, if b and c are on the same side of x, c will always lie between x and b, i.e. these two arrangements -> c b x a and a x b c are not valid.

Takeaway: An inequality like |x - a| > |x - b| > |x - c| doesn't mean a > b > c or a < b < c. It only means either a < b, c or a > b, c. The relationship between b and c can't be established unless more information is provided.

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Re: Solving Complex Problems Using Number Line [#permalink]

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Re: Solving Complex Problems Using Number Line   [#permalink] 05 Aug 2017, 20:10
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