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properties of numbers

Expert replies
by Lattefah84 » Sun Jan 31, 2010 1:55 pm
If M and Nare integers, is {10^M + N}/3 an integer?

1. N = 5
2. MN is even



* Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
* Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
* BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
* EACH statement ALONE is sufficient
* Statements (1) and (2) TOGETHER are NOT sufficient

How to solve this ?
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Source: — Data Sufficiency |

by ajith » Sun Jan 31, 2010 2:07 pm
Lattefah84 wrote:If M and Nare integers, is {10^M + N}/3 an integer?

1. N = 5
2. MN is even



* Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
* Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
* BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
* EACH statement ALONE is sufficient
* Statements (1) and (2) TOGETHER are NOT sufficient

How to solve this ?
A number is divisible by 3 if sum of digits is divisible by 3

The sum of digits for 10^M+N = N+1 (if M>0 and N>0)

1. Not sufficient ( had the question been that M, N are positive integers then this would have been sufficient)
2. Not Sufficient (M or N can be negative)

1&2 combined also is not good enough

IMO E

(A if M,N are strictly positive)
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by money9111 » Sun Jan 31, 2010 4:20 pm
i didn't know where to start with this one
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by papgust » Sun Jan 31, 2010 6:27 pm
ajith wrote:
A number is divisible by 3 if sum of digits is divisible by 3

The sum of digits for 10^M+N = N+1 (if M>0 and N>0)

1. Not sufficient ( had the question been that M, N are positive integers then this would have been sufficient)
2. Not Sufficient (M or N can be negative)

1&2 combined also is not good enough

IMO E

(A if M,N are strictly positive)
Ajith,

How do you calculate sum of digits in a decimal?

You say that statement 1 is insufficient. Whether M is -ve or +ve, you always have the sum of digits as 1 in 10^M. Am i wrong?
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by sars72 » Sun Jan 31, 2010 7:10 pm
papgust wrote:
ajith wrote:
A number is divisible by 3 if sum of digits is divisible by 3

The sum of digits for 10^M+N = N+1 (if M>0 and N>0)

1. Not sufficient ( had the question been that M, N are positive integers then this would have been sufficient)
2. Not Sufficient (M or N can be negative)

1&2 combined also is not good enough

IMO E

(A if M,N are strictly positive)
Ajith,

How do you calculate sum of digits in a decimal?

You say that statement 1 is insufficient. Whether M is -ve or +ve, you always have the sum of digits as 1 in 10^M. Am i wrong?
You can only use the "sum of digits" rule to check for divisibilty by 3 if the value is an integer. Not a decimal e.g. 2.01 is not divisible by 3 even though the sum of its digits =3.

So, since M can be negative, it is insufficient.
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by Lattefah84 » Mon Feb 01, 2010 12:59 am
ajith wrote:
Lattefah84 wrote:If M and Nare integers, is {10^M + N}/3 an integer?

1. N = 5
2. MN is even



* Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
* Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
* BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
* EACH statement ALONE is sufficient
* Statements (1) and (2) TOGETHER are NOT sufficient

How to solve this ?
A number is divisible by 3 if sum of digits is divisible by 3

The sum of digits for 10^M+N = N+1 (if M>0 and N>0)

1. Not sufficient ( had the question been that M, N are positive integers then this would have been sufficient)
2. Not Sufficient (M or N can be negative)

1&2 combined also is not good enough

IMO E

(A if M,N are strictly positive)
And how can I know if the 10^M is equal to N+1? Shouldn't it be 10^M= - 5 :(
Join the discussion