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

Question

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

TeamGMATIFY · Accepted Answer

Multiplication rule of 11: (Sum of digits at odd places - Sum of digits at even places) should be divisible by 11

Given number: 892,132,24x
Sum of digits at odd places = 8 + 2 + 3 + 2 + x = 15 + x (i)
Sum of digits at even places = 9 + 1 + 2 + 4 = 16 (ii)

(i) - (ii) = 15 + x - 16 = x - 1
Hence x should be = 1 to make this a multiple of 11 (0) Option A

TeamGMATIFY · Answer

The divisibility rule is not tested much on the GMAT much, but you never know

So it is better to have the rules of 2, 3, 4, 5, 8 and 9 handy. Here are the rules for your reference: 2 - The last digit should be even 3 - Sum of the digits should be multiple of 3 4 - Last two digits should be divisible by 4 5 - Last digit should be 5 or 0 6 - Number is even and if the sum of its digits is divisible by 3 7 - Quiet complex and should not be used 8 - Last three digits should be divisible by 8 9 - Sum should be divisible by 9 10 - Last digit should be 0 11 - (Sum of digits at odd places - Sum of digits at even places) should be divisible by 11

timothyhenman1 · Answer

Ah! I never learned the rule for divisibility of 11... I've done about 200 or so Quant problems and have never encountered a problem with this rule. In your experience has this rule been a common thing or more of a rarity?

Bunuel · Answer

Similar questions to practice: if-the-number-x3458623y-is-divisible-by-88-what-is-the-134183.html the-number-23a4534a01-is-divisible-by-11-how-many-values-134180.html Hope it helps.

umasarath52 · Answer

Take the alternating sum of the digits in the number, read from left to right. If that is divisible by 11, so is the original number.

In our problem, 892,132,24x is divisible by 11

alternate sum from left to right 
=> 8+2+3+4+x
=> 17+x
=> x=5 which is 22 divisible by 11

Answer E

TeamGMATIFY · Answer

Hi umasarath52,

You are missing one thing in the divisibility rule.
The rule says: (Sum of digits at odd places - Sum of digits at even places) should be divisible by 11

You missed subtracting the sum of other numbers.

To check this, try dividing 892,132,24 5  by 11. You will not have an integer.

Hop this helps!

Senthil7 · Answer

i just did long division withion 30-40 seconds was left with a remainder 1 with which you have to bring down the units digit x to divide the number by 11 and find out if it finally leaves a reminder, just clear enough that only a 1 brought down with the previous remainder 1 will make the last leg of division 11 / 11 hence 1 must be the last digit.

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

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

Prep Toolkit

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