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DS: Remainder problem
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Anurag@Gurome
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The remainder of n/3 is 2 -> n = (3 * k1) + 2, where k1 =0,1,2....
when 10^z is divided by 3, remainder is 1.
10^z = 9 * k2 + 1, where k2 =0,1,2...
so, 10^m + n = (9 * k2 + 1) + 3*k1 + 2 = (9 * k2) + (3 * k1) + 3.
which is divisible by 3. hence, remainder is 0.
for a number is divided by 3. its remainder can be 0, 1, or 2.
But as the remainder of 10^m + n is 0. It will never be greater than remainder of (10^n + 3), which can be 0,1 or 2. it can be at most be equal to remainder of (10^n + 3).
is it larger can be answered by a no.
hence, it is B
when 10^z is divided by 3, remainder is 1.
10^z = 9 * k2 + 1, where k2 =0,1,2...
so, 10^m + n = (9 * k2 + 1) + 3*k1 + 2 = (9 * k2) + (3 * k1) + 3.
which is divisible by 3. hence, remainder is 0.
for a number is divided by 3. its remainder can be 0, 1, or 2.
But as the remainder of 10^m + n is 0. It will never be greater than remainder of (10^n + 3), which can be 0,1 or 2. it can be at most be equal to remainder of (10^n + 3).
is it larger can be answered by a no.
hence, it is B
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GMATGuruNY
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If the sum of the digits of integer x is a multiple of 3, then x itself is a multiple of 3.nisagl750 wrote:I didn't understand how statement 2 can help us?
Can anybody please explain?
10^m + N implies the following:
If m=1, then 10^m + N = 10 + N = 1N.
If m=2, then 10^m + N = 100 + N = 10N.
If m=3, then 10^m + N = 1000 + N = 100N.
And so on.
In each case:
The first digit is 1, the units digit is N, and the intervening digits are all 0.
Thus, the sum of the digits = 1+N.
Statement 2: The remainder of N/3 is 2.
In other words, N is 2 more than a multiple of 3:
N = 3k + 2 = 2, 5, 8, 11, 14...
Thus, the sum of the digits of 10^m + N = 1+N = 3, 6, 9, 12, 15...
Since in each case the sum of the digits is a multiple of 3, 10^m + N is a multiple of 3.
Thus, when 10^m + N is divided by 3, the remainder is 0.
Thus, the remainder of (10^m + N)/3 cannot be greater than the remainder of (10^n + m)/3.
SUFFICIENT.
The correct answer is B.
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nisagl750
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Do we need to spend more time in putting different values for m & n to show that statement 1 is not sufficient Or there is some logic to rule out Statement 1?
P.S: Thanks for the explanations
P.S: Thanks for the explanations
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Anurag@Gurome
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we can have all the values of remainder of m or n when divided by 3, satisfying m > n.nisagl750 wrote:Do we need to spend more time in putting different values for m & n to show that statement 1 is not sufficient Or there is some logic to rule out Statement 1?
P.S: Thanks for the explanations
so, it won`t help.
substituting simple values would be easy to rule out.
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