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Re: If a and b are two-digit positive integers greater than 10 [#permalink]
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Bunuel wrote:

New project from GMAT Club: Topic-wise questions with tips and hints!



If \(a\) and \(b\) are two-digit positive integers greater than 10, is the remainder when \(a\) is divided by 11 less than the remainder when \(b\) is divided by 11?

(1) The remainder when \(a\) is divided by 69 is the fifth power of a prime number

(2) The remainder when \(b\) is divided by 12 is \(b\)

Kudos for a correct solution.



a=11x+r1
b=11y+r2

now question is is r1>r2

st.1 : a=69k+p^5

p is a prime number and its fifth value is the remainder.

if p=2 then P^5 =32
if p=3 then P^5 = 243 which is not possible (remainder cannot be greater than the divisor)
hence p=2
a=69k+32
also, as per the question 'a' is two digit number therefore, k=0 hence a=32 ( when k=1 value of 'a' becomes a=69+32=101)

but we need to know the value of b to compare the remainders hence statement 1 is not sufficient

st.2 : remainder of b, when b is divided by 12 is only possible if b is less than 12. also, since b is a two digit number therefore possible values of b are 10 and 11. but since b is greater than 10 hence value of b=11

combining 1 and 2
remainder when a=32 is divided by 11 is 10
remainder when b=11 is divided by 11 is 0
clearly r1 is greater than r2

hence answer should be C.
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Re: If a and b are two-digit positive integers greater than 10 [#permalink]
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1) The remainder when a is divided by 69 is the fifth power of a prime number

Let us consider prime numbers - 2, 3 ... . Note that remainder will have a value less than 69
2^5 = 32 -> Possible
3^5 = 243 -> This cannot the remainder

Therefore from 1) , the only possible 2-digit integer that can represent a is => 69*(0)+32 = 32

Note that other numbers are 3 digit numbers eg : 69*1 + 32 = 101

Since, we don't have any information about b , this is insufficient

2) The remainder when b is divided by 12 is b

This implies that the number b itself is less than 12, ranging from 1 - 11 . As B) is a 2-digit +ve integer > 10, the number is 11

Since, we don't have any information about a , this is insufficient


Combining 1) and 2) , we know that a is 32 and b is 11

Therefore, the remainder when a (32) is divided by 11 is 10 => 11*(2) + 10
the remainder when b(11) is divided by 11 is 0


Hence, both the statements together are sufficient to answer this question. C) is the answer
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Re: If a and b are two-digit positive integers greater than 10 [#permalink]
Bunuel wrote:

New project from GMAT Club: Topic-wise questions with tips and hints!



If \(a\) and \(b\) are two-digit positive integers greater than 10, is the remainder when \(a\) is divided by 11 less than the remainder when \(b\) is divided by 11?

(1) The remainder when \(a\) is divided by 69 is the fifth power of a prime number

(2) The remainder when \(b\) is divided by 12 is \(b\)

Kudos for a correct solution.



As per statement 2, B can only be 11. Since 11 divided 11 has a remainder of 0, let A be any value but its remainder when A is divided by 11 can never be less than 0. Hence statement 2 is sufficient.
Statement one is clearly insufficient to ans the question.

Ans=B
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Re: If a and b are two-digit positive integers greater than 10 [#permalink]
I'll bite.

1)rem A/69 = fifth power of a prime number. . . only 2 has a 5th power below 69 (32), so the remainder is 32. 69 + 32 = 101, so A must then be 32, since 32 can be divided by 69 zero times with a remainder of 32. 32/11 = 2 with a remainder of 10. Rem. 10 is the highest possible remainder when dividing by 11, so remainder of B/11 cannot be larger. No is sufficient.

A-D

2)Rem when b/12 = b. B could be 11, causing a remainder of 11, or could be 10 causing a remainder of 10. We know nothing about A from this statement, so it's insufficient.

Therefore, A?

Nvmd - I see my error. . .
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Re: If a and b are two-digit positive integers greater than 10 [#permalink]
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Wow! No correct solution for this question. Guess it's 700+ then.
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Re: If a and b are two-digit positive integers greater than 10 [#permalink]
Hi Bunuel - can you explain why the answer isn't C? From Statement 1, we can't tell if B's remainder is also 10...
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Re: If a and b are two-digit positive integers greater than 10 [#permalink]
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cg0588 wrote:
Hi Bunuel - can you explain why the answer isn't C? From Statement 1, we can't tell if B's remainder is also 10...


I assume that you understood how we calculated the remainder for a/11 to be 10.
Since the question states that "is the remainder when a is divided by 11 less than the remainder when b is divided by 11?"
Its a "Yes" or "No" DS question.
So if the statement gives a Yes for all possible scenarios, then it is sufficient.
Similarly, if the statement gives a No for all possible scenarios, then it is sufficient.

Here, the remainder when a is divided by 11 is 10 (10 is the maximum possible remainder when the divisor is 11 right? )

Now the remainder when b is divided by 11 can be from 0,1,2,3,4,5,6,7,8,9 or 10.
Even if you assume the highest possible remainder when b is divided by 11 i.e. 10.

There is no way 10<10? Its always a NO for every possible scenario.
Hence it's sufficient.

use the same approach for Statement 2 as well. Hence the answer is D.

Hope it's clear :)
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Re: If a and b are two-digit positive integers greater than 10 [#permalink]
Bunuel wrote:
SOLUTION

If \(a\) and \(b\) are two-digit positive integers greater than 10, is the remainder when \(a\) is divided by 11 less than the remainder when \(b\) is divided by 11?

First of all, note that the remainder when a positive integer is divided by 11 could be 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 or 10.

(1) The remainder when \(a\) is divided by 69 is the fifth power of a prime number:

\(a=69q+prime^5\) and \(prime^5<69\) (the remainder must be less than the divisor). The only prime number whose fifth power is less than 69 is 2: \((2^5=32) < 69\). So, we have that \(a=69q+32\): 32, 101, 170, ... Since \(a\) is a two-digit integer, then \(a=32\).

Now, \(a=32\) divided by 11 gives the remainder of 10. Since 10 is the maximum remainder possible when divided by 11, then this remainder cannot be less then the remainder when \(b\) is divided by 11 (0, 1, 2, 3, 4, 5, 6, 7, 8, 9 or 10). So, the answer to the question is NO.

Sufficient.

(2) The remainder when \(b\) is divided by 12 is \(b\):

The remainder must be less than the divisor, hence \(b\) must be less than 12 and since we are told that \(b\) is a two-digit integers greater than 10, then \(b\) must be 11.

Now, \(b=11\) divided by 11 gives the remainder of 0. Since 0 is the minimum remainder possible, than the remainder when \(a\) is divided by 11 cannot possible be less than 0. So, the answer to the question is NO.

Sufficient.

Answer: D.

Try NEW remainders PS question.



Is the assumption a=b not correct in this questions?

am I missing something?

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Re: If a and b are two-digit positive integers greater than 10 [#permalink]
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email2vm wrote:
Bunuel wrote:
SOLUTION

If \(a\) and \(b\) are two-digit positive integers greater than 10, is the remainder when \(a\) is divided by 11 less than the remainder when \(b\) is divided by 11?

First of all, note that the remainder when a positive integer is divided by 11 could be 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 or 10.

(1) The remainder when \(a\) is divided by 69 is the fifth power of a prime number:

\(a=69q+prime^5\) and \(prime^5<69\) (the remainder must be less than the divisor). The only prime number whose fifth power is less than 69 is 2: \((2^5=32) < 69\). So, we have that \(a=69q+32\): 32, 101, 170, ... Since \(a\) is a two-digit integer, then \(a=32\).

Now, \(a=32\) divided by 11 gives the remainder of 10. Since 10 is the maximum remainder possible when divided by 11, then this remainder cannot be less then the remainder when \(b\) is divided by 11 (0, 1, 2, 3, 4, 5, 6, 7, 8, 9 or 10). So, the answer to the question is NO.

Sufficient.

(2) The remainder when \(b\) is divided by 12 is \(b\):

The remainder must be less than the divisor, hence \(b\) must be less than 12 and since we are told that \(b\) is a two-digit integers greater than 10, then \(b\) must be 11.

Now, \(b=11\) divided by 11 gives the remainder of 0. Since 0 is the minimum remainder possible, than the remainder when \(a\) is divided by 11 cannot possible be less than 0. So, the answer to the question is NO.

Sufficient.

Answer: D.

Try NEW remainders PS question.



Is the assumption a=b not correct in this questions?

am I missing something?

Ravi


Hi,
you have to cater for a=b in this case and all other Qs, if not specified other way..

why a=b does not make a difference ?
Because the Q asks us
Quote:
is the remainder when \(a\) is divided by 11 less than the remainder when \(b\) is divided by 11?
..
the KEY word is LESS..
even if a=b, the remainder will be same BUT not LESS..


Hope it's clear
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Re: If a and b are two-digit positive integers greater than 10 [#permalink]
It is amazing question. I builds up the reminder concepts in mind.
thank you Bunuel.I got the answer E. but I reached the figures 32 , 10 , and 0 reminders from statements but couldnot decide on the right answer.
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Re: If a and b are two-digit positive integers greater than 10 [#permalink]
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Hi. A cannot be 32 as the remainder is 32. A should be 37. As 37/69 is 1 remainder 32.

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Re: If a and b are two-digit positive integers greater than 10 [#permalink]
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Sorry about the last reply, i had a mix up.

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Re: If a and b are two-digit positive integers greater than 10 [#permalink]
[quote="Bunuel"]SOLUTION

If \(a\) and \(b\) are two-digit positive integers greater than 10, is the remainder when \(a\) is divided by 11 less than the remainder when \(b\) is divided by 11?

First of all, note that the remainder when a positive integer is divided by 11 could be 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 or 10.

(1) The remainder when \(a\) is divided by 69 is the fifth power of a prime number:

\(a=69q+prime^5\) and \(prime^5<69\) (the remainder must be less than the divisor). The only prime number whose fifth power is less than 69 is 2: \((2^5=32) < 69\). So, we have that \(a=69q+32\): 32, 101, 170, ... Since \(a\) is a two-digit integer, then \(a=32\).

Now, \(a=32\) divided by 11 gives the remainder of 10. Since 10 is the maximum remainder possible when divided by 11, then this remainder cannot be less then the remainder when \(b\) is divided by 11 (0, 1, 2, 3, 4, 5, 6, 7, 8, 9 or 10). So, the answer to the question is NO.

Sufficient.

(2) The remainder when \(b\) is divided by 12 is \(b\):

The remainder must be less than the divisor, hence \(b\) must be less than 12 and since we are told that \(b\) is a two-digit integers greater than 10, then \(b\) must be 11.

Now, \(b=11\) divided by 11 gives the remainder of 0. Since 0 is the minimum remainder possible, than the remainder when \(a\) is divided by 11 cannot possible be less than 0. So, the answer to the question is NO.

Sufficient.

Answer: D.

I got C.
there are two issues, one in each statement

1) we got remainder for a i.e. 10. The remainder value of "b" could be either 10 or less. Since we dont know b's value, this choice is insufficient.

2) we got b=11 and hence remainder 0. but we dont know "a" value. which could be either 11 or any other number, making it insufficient.

Hence C
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Re: If a and b are two-digit positive integers greater than 10 [#permalink]
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hanyhamdani wrote:
I got C.
there are two issues, one in each statement

1) we got remainder for a i.e. 10. The remainder value of "b" could be either 10 or less. Since we dont know b's value, this choice is insufficient.

2) we got b=11 and hence remainder 0. but we dont know "a" value. which could be either 11 or any other number, making it insufficient.

Hence C


Please re-read the solution carefully.

If \(a\) and \(b\) are two-digit positive integers greater than 10, is the remainder when \(a\) is divided by 11 less than the remainder when \(b\) is divided by 11?

First of all, note that the remainder when a positive integer is divided by 11 could be 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 or 10.

(1) The remainder when \(a\) is divided by 69 is the fifth power of a prime number:

\(a=69q+prime^5\) and \(prime^5<69\) (the remainder must be less than the divisor). The only prime number whose fifth power is less than 69 is 2: \((2^5=32) < 69\). So, we have that \(a=69q+32\): 32, 101, 170, ... Since \(a\) is a two-digit integer, then \(a=32\).

Now, \(a=32\) divided by 11 gives the remainder of 10.Since 10 is the maximum remainder possible when divided by 11, then this remainder cannot be less then the remainder when \(b\) is divided by 11 (0, 1, 2, 3, 4, 5, 6, 7, 8, 9 or 10). So, the answer to the question is NO.

Sufficient.

(2) The remainder when \(b\) is divided by 12 is \(b\):

The remainder must be less than the divisor, hence \(b\) must be less than 12 and since we are told that \(b\) is a two-digit integers greater than 10, then \(b\) must be 11.

Now, \(b=11\) divided by 11 gives the remainder of 0. Since 0 is the minimum remainder possible, than the remainder when \(a\) is divided by 11 cannot possible be less than 0. So, the answer to the question is NO.

Sufficient.
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Re: If a and b are two-digit positive integers greater than 10 [#permalink]
Bunuel wrote:
hanyhamdani wrote:
I got C.
there are two issues, one in each statement

1) we got remainder for a i.e. 10. The remainder value of "b" could be either 10 or less. Since we dont know b's value, this choice is insufficient.

2) we got b=11 and hence remainder 0. but we dont know "a" value. which could be either 11 or any other number, making it insufficient.

Hence C


Please re-read the solution carefully.

If \(a\) and \(b\) are two-digit positive integers greater than 10, is the remainder when \(a\) is divided by 11 less than the remainder when \(b\) is divided by 11?

First of all, note that the remainder when a positive integer is divided by 11 could be 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 or 10.

(1) The remainder when \(a\) is divided by 69 is the fifth power of a prime number:

\(a=69q+prime^5\) and \(prime^5<69\) (the remainder must be less than the divisor). The only prime number whose fifth power is less than 69 is 2: \((2^5=32) < 69\). So, we have that \(a=69q+32\): 32, 101, 170, ... Since \(a\) is a two-digit integer, then \(a=32\).

Now, \(a=32\) divided by 11 gives the remainder of 10.Since 10 is the maximum remainder possible when divided by 11, then this remainder cannot be less then the remainder when \(b\) is divided by 11 (0, 1, 2, 3, 4, 5, 6, 7, 8, 9 or 10). So, the answer to the question is NO.

Sufficient.

(2) The remainder when \(b\) is divided by 12 is \(b\):

The remainder must be less than the divisor, hence \(b\) must be less than 12 and since we are told that \(b\) is a two-digit integers greater than 10, then \(b\) must be 11.

Now, \(b=11\) divided by 11 gives the remainder of 0. Since 0 is the minimum remainder possible, than the remainder when \(a\) is divided by 11 cannot possible be less than 0. So, the answer to the question is NO.

Sufficient.


Bunuel

I understand B is sufficient.

But I don't grasp why A is sufficient. From A, remainder could be lots of different values. For an example when a=170, remainder is 5. Clearly, the remainder of b/11 could be easily bigger or smaller than that. We have no info about it.

Consider the values, a=170, b=15 and a=170, b=19. Both are possible since we don't have any info about b.
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Re: If a and b are two-digit positive integers greater than 10 [#permalink]
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josico111 wrote:
Bunuel wrote:
hanyhamdani wrote:
I got C.
there are two issues, one in each statement

1) we got remainder for a i.e. 10. The remainder value of "b" could be either 10 or less. Since we dont know b's value, this choice is insufficient.

2) we got b=11 and hence remainder 0. but we dont know "a" value. which could be either 11 or any other number, making it insufficient.

Hence C


Please re-read the solution carefully.

If \(a\) and \(b\) are two-digit positive integers greater than 10, is the remainder when \(a\) is divided by 11 less than the remainder when \(b\) is divided by 11?

First of all, note that the remainder when a positive integer is divided by 11 could be 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 or 10.

(1) The remainder when \(a\) is divided by 69 is the fifth power of a prime number:

\(a=69q+prime^5\) and \(prime^5<69\) (the remainder must be less than the divisor). The only prime number whose fifth power is less than 69 is 2: \((2^5=32) < 69\). So, we have that \(a=69q+32\): 32, 101, 170, ... Since \(a\) is a two-digit integer, then \(a=32\).

Now, \(a=32\) divided by 11 gives the remainder of 10.Since 10 is the maximum remainder possible when divided by 11, then this remainder cannot be less then the remainder when \(b\) is divided by 11 (0, 1, 2, 3, 4, 5, 6, 7, 8, 9 or 10). So, the answer to the question is NO.

Sufficient.

(2) The remainder when \(b\) is divided by 12 is \(b\):

The remainder must be less than the divisor, hence \(b\) must be less than 12 and since we are told that \(b\) is a two-digit integers greater than 10, then \(b\) must be 11.

Now, \(b=11\) divided by 11 gives the remainder of 0. Since 0 is the minimum remainder possible, than the remainder when \(a\) is divided by 11 cannot possible be less than 0. So, the answer to the question is NO.

Sufficient.


Bunuel

I understand B is sufficient.

But I don't grasp why A is sufficient. From A, remainder could be lots of different values. For an example when a=170, remainder is 5. Clearly, the remainder of b/11 could be easily bigger or smaller than that. We have no info about it.

Consider the values, a=170, b=15 and a=170, b=19. Both are possible since we don't have any info about b.


Not to go any deeper, notice that we are given that \(a\) and \(b\) are two-digit positive integers.
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Re: If a and b are two-digit positive integers greater than 10 [#permalink]
Hi

Great question. I do see lot of people are confused why B is 11 and not 10? It's because the question states a and b are greater than 10. Hence B has to be 11.
Second why D is the answer instead of C - it's because Stm 1 proves A=32 and when 32/11 gives remainder 10 which is the highest remainder for 11 hence even if we say a=b remainder of b will also be 10 which clearly ans the question "the remainder of a is less than the remainder of b" as NO else all the other values of b will provide the remainder to be less than the remainder of a. That's why Stm 1 is sufficient.
Same goes for Stm 2, even if we consider a=b the question is answered as NO again since the remainder of a will either be equal to or greater than the remainder of b but never be smaller.

Hope it helps!!

Thanks
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