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

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Question Stats: 17% (02:28) correct 83% (02:43) wrong based on 738 sessions

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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 $$b$$

Kudos for a correct solution.

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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.

Sufficient.

Try NEW remainders PS question.
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Re: If a and b are two-digit positive integers greater than 10  [#permalink]

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Bunuel wrote:

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]

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Bunuel wrote:

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]

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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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1
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.

Try NEW remainders PS question.
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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]

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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.

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

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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.

Try NEW remainders PS question.

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

am I missing something?

Ravi
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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.

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]

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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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Sorry about the last reply, i had a mix up.

Posted from my mobile device
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Re: If a and b are two-digit positive integers greater than 10  [#permalink]

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[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.

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

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

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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

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

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]

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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 Re: If a and b are two-digit positive integers greater than 10   [#permalink] 17 Dec 2018, 17:20

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