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01 Jun 2019, 11:23
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cf92
Maybe it's a typo because the OA on GMATPrep only talks about the value of k but the answer is the same because it says k is the same digit in both spaces (so only 4 choices).
If k could be a different digit then n would have 4*4 = 16 different values divisible by 3.
29 Sep 2019, 16:12
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NoHalfMeasures
n is of the form :- 1k2k24 ....
Some of the digits except k :- 1 + 2 + 2 + 4 = 9
So for n to be divisible by 3 , k can take :- 0 or 3 or 6 or 9 . (4 values)
So one 'k' can be of 4 kinds and for each kind , another 'k' can be of that kind only ( similar to it ) i.e 1 kind.
So the number 'n' can be of :- 4 *1 = 4
Option C is the answer.
Please give me kudo s if you liked my explanation.
24 Apr 2020, 01:43
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SOLUTION:
A positive integer is divisible by 3 if and only if the sum of its digits is divisible by 3.
1 K 2 K 2 4
9+ K + K
K= 0: 9+0+0 = 9 divisible by 3.
K= 1: 9+1+1 = 11 it is not divisible by 3.
K= 2: 9+2+2 = 13 it is not divisible by 3
K= 3: 9+3+3 = 15 divisible by 3
K= 4: 9+4+4 = 17 it is not divisible by 3.
K= 5: 9+5+5 = 19 it is not divisible by 3
K= 6: 9+6+6 = 21 divisible by 3
K= 7: 9+7+7 = 23 it is not divisible by 3.
K= 8: 9+8+8 = 25 it is not divisible by 3.
K= 9: 9+9+9 = 27 divisible by 3.
K could be: {0,3,6,9} 4 values.
ANSWER C
A positive integer is divisible by 3 if and only if the sum of its digits is divisible by 3.
1 K 2 K 2 4
9+ K + K
K= 0: 9+0+0 = 9 divisible by 3.
K= 1: 9+1+1 = 11 it is not divisible by 3.
K= 2: 9+2+2 = 13 it is not divisible by 3
K= 3: 9+3+3 = 15 divisible by 3
K= 4: 9+4+4 = 17 it is not divisible by 3.
K= 5: 9+5+5 = 19 it is not divisible by 3
K= 6: 9+6+6 = 21 divisible by 3
K= 7: 9+7+7 = 23 it is not divisible by 3.
K= 8: 9+8+8 = 25 it is not divisible by 3.
K= 9: 9+9+9 = 27 divisible by 3.
K could be: {0,3,6,9} 4 values.
ANSWER C
