Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

If x and y are positive integers, what is the remainder when 10^x +y is divided by 3?

Since, the sum of the digits of 10^x is always 1 then the remainders when 10^x+y is divided by 3 is only dependant on the value of the number added to 10^x, so on y. If y is a multiple of 3 then 10^x+y will yield the remainder of 1 (since the sum of the digits of 10^x+y will be one more than a multiple of 3), if y is one more than a multiple of 3 then 10^x+y will yield the remainder of 2 and finally if y is two more than a multiple of 3 then 10^x+y will yield the remainder of 0,

Re: If x and y are positive integers, what is the remainder when [#permalink]

Show Tags

05 Sep 2012, 02:17

2

This post received KUDOS

The OG explanations sometimes really baffle me; I reached the solution simply by realizing that it did not matter what power the 10 was to be elevated to, and to know the value of y was sufficient; just like Bunuel explained above. The OG explanation should not be the primary route, in my opinion.
_________________

The question is not really understandable. Please post it either in the "formula form" or like this (10^x)+y.

10^x +y means \(10^x +y\), so no ambiguity there. If it were 10^(x +y) it would be written that way. Still edited the original post to avoid further confusions.
_________________

Re: If x and y are positive integers, what is the remainder when [#permalink]

Show Tags

12 Dec 2015, 10:58

Bunuel wrote:

If x and y are positive integers, what is the remainder when 10^x +y is divided by 3?

Since, the sum of the digits of 10^x is always 1

I'm sorry if this is something obvious i'm missing but I'm not quite following what you mean by that; sum of digits 10^x is always 1 as in...10 and x?
_________________

If x and y are positive integers, what is the remainder when 10^x +y is divided by 3?

Since, the sum of the digits of 10^x is always 1

I'm sorry if this is something obvious i'm missing but I'm not quite following what you mean by that; sum of digits 10^x is always 1 as in...10 and x?

If x = 1, then 10^1 = 10 --> the sum of the digits = 1 + 0 = 1; If x = 2, then 10^3 = 100 --> the sum of the digits = 1 + 0 + 0 = 1; If x = 3, then 10^3 = 1000 --> the sum of the digits = 1 + 0 + 0 + 0= 1; ...

Re: If x and y are positive integers, what is the remainder when [#permalink]

Show Tags

13 Dec 2015, 09:07

Bunuel wrote:

redfield wrote:

Bunuel wrote:

If x and y are positive integers, what is the remainder when 10^x +y is divided by 3?

Since, the sum of the digits of 10^x is always 1

I'm sorry if this is something obvious i'm missing but I'm not quite following what you mean by that; sum of digits 10^x is always 1 as in...10 and x?

If x = 1, then 10^1 = 10 --> the sum of the digits = 1 + 0 = 1; If x = 2, then 10^3 = 100 --> the sum of the digits = 1 + 0 + 0 = 1; If x = 3, then 10^3 = 1000 --> the sum of the digits = 1 + 0 + 0 + 0= 1; ...

Hope it's clear.

Sorry I'm still struggling with this; what are you equating the 1 and the 0's to, the 1 being the 10 and the 0's being the power it is risen to?
_________________

I'm sorry if this is something obvious i'm missing but I'm not quite following what you mean by that; sum of digits 10^x is always 1 as in...10 and x?

If x = 1, then 10^1 = 10 --> the sum of the digits = 1 + 0 = 1; If x = 2, then 10^3 = 100 --> the sum of the digits = 1 + 0 + 0 = 1; If x = 3, then 10^3 = 1000 --> the sum of the digits = 1 + 0 + 0 + 0= 1; ...

Hope it's clear.

Sorry I'm still struggling with this; what are you equating the 1 and the 0's to, the 1 being the 10 and the 0's being the power it is risen to?

We are talking about the sum of the digits of a number. For example, the sum of the digits of 17 is 8 because 1 + 7 = 8. Or, the sum of the digits of 2^5 is 5 because 2^5 = 32 and 3 + 2 = 5.
_________________

Re: If x and y are positive integers, what is the remainder when [#permalink]

Show Tags

13 Dec 2015, 09:19

Bunuel wrote:

We are talking about the sum of the digits of a number. For example, the sum of the digits of 17 is 8 because 1 + 7 = 8. Or, the sum of the digits of 2^5 is 5 because 2^5 = 32 and 3 + 2 = 5.

Ohhhhh; my apologies I was trying to make it some abstract/esoteric concept rather than literally just the sum of the numbers that make up the number!
_________________

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.

If x and y are positive integers, what is the remainder when 10 x +y is divided by 3?

(1) x = 5 (2) y = 2

In the original condition, there are 2 variables(x,y), which should match with the number of equations. So, you need 2 equations. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer. In 1) & 2), it becomes 10^5+2=100,002. Also, the remainder after divided by 3 is same as the remainder after the sum of all digits divided by 3. That is, 1+0+0+0+0+2=3 can be divided by 3 and therefore it is yes, which is sufficient. So the answer is C. This is an integer question which is one of the key questions. When applying 4(A) of the mistake type, 1) x=5 but y=2- > yes, y=3 -> no, which is not sufficient. 2) y=2 -> 10^x+2=12,102,1002,10002......... When it comes to number like these, the sum of all digits are 3 and also can be divided by 3 and it is yes, which is sufficient. Therefore, both B and C can be the answer. When both B and C can be the answer, the answer is B. This type of question is given in Math Revolution lectures and you need to get this type of question right to reach the score range 50-51.

-> For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
_________________

I understood the explanation, and indeed B was my first choice, too. However, I'm struggling with its logic. Because the question asks for a value, so I thought it depends on the power of 10th? I see the similar pattern it follows (33,334.. etc), yet different powers would yield in different results, therefore I won't be able to find out the exact value.

i.e. (10^1 + 2)/ 3 = 4 or (10^2 + 2 /3) = 34)

Could you explain, why in I say B is sufficient, even though there are many values? (Find the question in the 2nd Quant Review)

Last edited by nilem94 on 20 Feb 2016, 08:09, edited 1 time in total.

I understood the explanation, and indeed B was my first choice, too. However, I'm struggling with its logic. Because the questions ask for a value, so I thought it depends on the power of 10th? I see the similar pattern it follows (33,334.. etc), yet different powers would yield in different results, therefore I won't be able to find out the exact value.

i.e. (10^1 + 2)/ 3 = 4 or (10^2 + 2 /3) = 34)

Could you explain, why in I say B is sufficient, even though there are many values? (Find the question in the 2nd Quant Review)

Hi,

INFO:-

since the div rule by 3 states that the sum of digits should be div by 3 and the remainder of any integer is same as remainder of the sum of integer.. lets see what is 10^x + y... here irrespective of value of x, the sum of integers will be 1, as it will be 1,10,100,1000... so we require to know y to find the remainder..

lets see the choices.. 1) X=5 nothing about y insuff

(2) y=2 now we know sum = 1+2.. remainder is 0.. Suff B Hope it helped _________________

I understood the explanation, and indeed B was my first choice, too. However, I'm struggling with its logic. Because the question asks for a value, so I thought it depends on the power of 10th? I see the similar pattern it follows (33,334.. etc), yet different powers would yield in different results, therefore I won't be able to find out the exact value.

i.e. (10^1 + 2)/ 3 = 4 or (10^2 + 2 /3) = 34)

Could you explain, why in I say B is sufficient, even though there are many values? (Find the question in the 2nd Quant Review)

Merging topics. Please search before posting.
_________________

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Happy 2017! Here is another update, 7 months later. With this pace I might add only one more post before the end of the GSB! However, I promised that...