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If x and y are positive integers, what is the remainder when

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

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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

Originally posted by dzodzo85 on 19 Mar 2012, 23:40.
Last edited by Bunuel on 20 Jun 2014, 01:14, edited 1 time in total.
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Re: If x and y are positive integers, what is the remainder when  [#permalink]

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New post 19 Mar 2012, 23:51
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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,

(1) x=5. Not sufficient.
(2) y=2. Sufficient.

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

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New post 09 May 2012, 13:07
So the question should be ((10^x)+y)/3?

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

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

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New post 05 Sep 2012, 02:17
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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.
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Re: If x and y are positive integers, what is the remainder when  [#permalink]

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New post 20 Jun 2014, 01:03
The question is not really understandable. Please post it either in the "formula form" or like this (10^x)+y.
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Re: If x and y are positive integers, what is the remainder when  [#permalink]

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New post 20 Jun 2014, 01:14
unceldolan wrote:
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.
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Re: If x and y are positive integers, what is the remainder when  [#permalink]

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

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New post 13 Dec 2015, 01:34
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.
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Re: If x and y are positive integers, what is the remainder when  [#permalink]

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

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New post 13 Dec 2015, 09:13
1
redfield wrote:
Bunuel wrote:
redfield wrote:
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.
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Re: If x and y are positive integers, what is the remainder when  [#permalink]

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

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New post 13 Dec 2015, 20:40
1
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.
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if integers x,y > 0, what is the remainder when (10^x + y)/3? Logic?  [#permalink]

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New post Updated on: 20 Feb 2016, 08:09
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
The correct answer is B;

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)

Originally posted by nilem94 on 20 Feb 2016, 07:50.
Last edited by nilem94 on 20 Feb 2016, 08:09, edited 1 time in total.
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Re: if integers x,y > 0, what is the remainder when (10^x + y)/3? Logic?  [#permalink]

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New post 20 Feb 2016, 08:08
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nilem94 wrote:
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
The correct answer is B;

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

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Re: if integers x,y > 0, what is the remainder when (10^x + y)/3? Logic?  [#permalink]

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New post 20 Feb 2016, 08:13
Ah now I see, the remainder is always 0, no matter what power!
Thank you so much!
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Re: If x and y are positive integers, what is the remainder when  [#permalink]

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New post 20 Feb 2016, 13:33
nilem94 wrote:
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
The correct answer is B;

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.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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

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New post 14 Aug 2018, 19:55
dzodzo85 wrote:
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


Answer B.

\(\frac{10^x +y}{3}\) = \(\frac{10^x}{3}+ \frac{y}{3}\) = \(\frac{(9+1)^x}{3}\) + \(\frac{y}{3}\)

For \((9+1)^x\), all the terms will have a factor of 9 (divisible by 3) except the last term \(1^x\) [According to binomial theorem], so the remainder is always 1.

Now we just need to know the value of y for the 2nd term \(\frac{y}{3}\)

Hence the answer is B.
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Re: If x and y are positive integers, what is the remainder when &nbs [#permalink] 14 Aug 2018, 19:55
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