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

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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

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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

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.