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.

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 a and b are two-digit positive integers greater than 10 [#permalink]
20 Jun 2014, 04:03

2

This post received KUDOS

Expert's post

4

This post was BOOKMARKED

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?

(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]
20 Jun 2014, 07:14

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]
20 Jun 2014, 07:57

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 [#permalink]
22 Jun 2014, 03:34

1

This post received KUDOS

Expert's post

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.