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(1) when x is divided by 5, the result an even integer.

(2) x + 20 is a multiple of 10.

Statement 2 : x + 20 can be 10, 20, 30, 40...... So x can be -10, 0, 10, 20, 30....

Can we say that -10 and 0 are also multiples of 10? or just my reasoning above is incorrect ?

Is the integer x a multiple of 10?

(1) when x is divided by 5, the result is an even integer --> \(\frac{x}{5}=2k\) --> \(x=10k\) --> \(x\) is a multiple of 10. Sufficient.

(2) x + 20 is a multiple of 10 --> \(x+20=10n\) --> \(x=10(n-2)\) --> \(x\) is a multiple of 10. Sufficient.

Answer: D.

As for your questions: 1. Zero is divisible by EVERY integer except zero itself, since 0/integer=integer (or, which is the same, zero is a multiple of every integer except zero itself). 2. -10 is a multiple of 10 since -10/10=-1=integer.

I'm a little confused on this. Is 0 a multiple of 10? i.e., 10n, where n=0?

Integer \(a\) is a multiple of integer \(b\) means that \(a\) is "evenly divisible" by \(b\), i.e., divisible by \(b\) without a remainder. Now, since zero/integer=integer then zero is a multiple of every integer (except zero itself).

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Is the integer x a multiple of 10?

(1) when x is divided by 5, the result an even integer.

(2) x + 20 is a multiple of 10.

There is one variable (x) and 2 equations are given by the conditions, so there is high chance (D) will be the answer. For condition 1, x=5p+even (p:positive integer), and the answer to the question becomes 'yes' for x=10, but 'no' for x=7, so this is insufficient. For condition 2, from x+20=10q(q:positive integer), x=10q-20=10(q-2)= a multiple of 10. This answers the question 'yes' and is sufficient. Therefore, the answer becomes (B).

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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(1) when x is divided by 5, the result an even integer.

(2) x + 20 is a multiple of 10.

Statement 2 : x + 20 can be 10, 20, 30, 40...... So x can be -10, 0, 10, 20, 30....

Can we say that -10 and 0 are also multiples of 10? or just my reasoning above is incorrect ?

Could someone please help me? I chose answer E. I now understand why answer B is sufficient instead of insufficient (I thought that 10 itself could not be a multiple of 10. I only thought that integers greater than 10 could only be multiples of 10.

However, for answer A. I do not understand why this is sufficient. It says that when x is divided by 5, the result an even integer. X could be 25 right? 25/5 is 4 which is an even integer. X is therefore not necessarily a multiple of 10? Could someone tell me where I am wrong?

Re: Is the integer x a multiple of 10? [#permalink]

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23 Feb 2016, 12:25

saiesta wrote:

eybrj2 wrote:

Is the integer x a multiple of 10?

(1) when x is divided by 5, the result an even integer.

(2) x + 20 is a multiple of 10.

Statement 2 : x + 20 can be 10, 20, 30, 40...... So x can be -10, 0, 10, 20, 30....

Can we say that -10 and 0 are also multiples of 10? or just my reasoning above is incorrect ?

Could someone please help me? I chose answer E. I now understand why answer B is sufficient instead of insufficient (I thought that 10 itself could not be a multiple of 10. I only thought that integers greater than 10 could only be multiples of 10.

However, for answer A. I do not understand why this is sufficient. It says that when x is divided by 5, the result an even integer. X could be 25 right? 25/5 is 4 which is an even integer. X is therefore not necessarily a multiple of 10? Could someone tell me where I am wrong?

How is 25/5 = 4? It should be 25/5 =5 as 5*5=25.

What statement 1 means is that when x is divided by 5 you get a quotient = even number. So we can write it as

x= 5*(2p)+0 (from the division rule, dividend= dividend*quotient+remainder). Thus x= 10p or in other words, x is a multiple of 10.

Re: Is the integer x a multiple of 10? [#permalink]

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23 Feb 2016, 12:28

Engr2012 wrote:

saiesta wrote:

eybrj2 wrote:

Is the integer x a multiple of 10?

(1) when x is divided by 5, the result an even integer.

(2) x + 20 is a multiple of 10.

Statement 2 : x + 20 can be 10, 20, 30, 40...... So x can be -10, 0, 10, 20, 30....

Can we say that -10 and 0 are also multiples of 10? or just my reasoning above is incorrect ?

Could someone please help me? I chose answer E. I now understand why answer B is sufficient instead of insufficient (I thought that 10 itself could not be a multiple of 10. I only thought that integers greater than 10 could only be multiples of 10.

However, for answer A. I do not understand why this is sufficient. It says that when x is divided by 5, the result an even integer. X could be 25 right? 25/5 is 4 which is an even integer. X is therefore not necessarily a multiple of 10? Could someone tell me where I am wrong?

How is 25/5 = 4? It should be 25/5 =5 as 5*5=25.

What statement 1 means is that when x is divided by 5 you get a quotient = even number. So we can write it as

x= 5*(2p)+0 (from the division rule, dividend= dividend*quotient+remainder). Thus x= 10p or in other words, x is a multiple of 10.

Hope this helps.

My apologies. I guess I have been so tired from doing so many questions. I need some rest. Thank you!

Re: Is the integer x a multiple of 10? [#permalink]

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09 Mar 2016, 20:58

Here clearly both statements are sufficient we just need to realize that any number that yields a even quotient with 5 must have 0 as its Units digit
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Re: Is the integer x a multiple of 10? [#permalink]

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15 Dec 2017, 06:08

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