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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 \(b\)

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.

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

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

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20 Jun 2014, 08:14

1

This post received KUDOS

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

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.

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

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20 Jun 2014, 08:57

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.

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.

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

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27 Mar 2015, 20:23

1

This post received KUDOS

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.

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

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07 Apr 2016, 21:03

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.

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.

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

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13 Jun 2016, 08:42

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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