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If the number 892,132,24x is divisible by 11, what must . .

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by VJesus12 » Thu Jan 11, 2018 11:22 am
If the number 892,132,24x is divisible by 11, what must be the value of x?

A) 1
B) 2
C) 3
D) 4
E) 5

The OA is the option A.

How can I prove that option A is the only one possible? Can any expert give me some help? Please.
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Answer

by EconomistGMATTutor » Sat Jan 13, 2018 8:41 am
Hello Vjesus12.

Let's take a look at your question.

First, let's remember the multiplication rule of 11:

(Sum of digits at odd places - Sum of digits at even places) should be divisible by 11.

We have the number 892,132,24x, then: $$\left(8+2+3+2+x\right)-\left(9+1+2+4\right)=\left(15+x\right)-\left(16\right)=x-1.$$ The only option that makes 1-x a multiple of 11 is x=1.

This is why the correct option is [spoiler]A=1[/spoiler].

I hope this answer can help you.

Feel free to ask me again if you have any question.

Regards.
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TTT

by mariliatgd » Sun Jan 14, 2018 3:07 pm
One of the number properties is that to see if a number is multiple of 11 you must:
If the number is "abc", then c-b+a = 0 or 11.

In this case 892 132 24x needs to bee multiple of 11, so:

x - 4 + 2 - 2 + 3 - 1 + 2 - 9 - 8 must equal 0 or 11
x - 1 must equal 0 or 11

x must be 1, because is a number with one digit.
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by DrMaths » Mon Jan 15, 2018 4:14 am
8 - 9 + 2 - 1 + 3 - 2 + 2 - 4 + x = 11n (because 11n always = sum of digits alternating + - + -, etc.)
so x = 11n + 1. When n = 0, then x = 1.
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