Q-2: How many two digit numbers of distinct digits can be formed by using digits 1, 2, 3, 4, 5, 6 and 7 such that the numbers are divisible by 3?
A) 9
B) 10
C) 11
D) 12
E) 14
oa is e.
please show your solution to the correct answer. my answer here in my calculation is A
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Number properties
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Hi Roland2rule,
For this question, you would likely find it helpful to use the 'division rule of 3': for an integer to be divisible by 3, the sum of the DIGITS of the integer will be divisible by 3.
For example:
24 is divisible by 3 because 2+4 = 6 and 6 is divisible by 3
25 is NOT divisible by 3 because 2+5 = 7 and 7 is NOT divisible by 3
Using that pattern, you can quickly list out the 'pairs' of numbers that would form a 2-digit number that is divisible by 3:
1 and 2 (forming 12 and 21)
1 and 5 (forming 15 and 51)
2 and 4 (forming 24 and 42)
2 and 7 (forming 27 and 72)
3 and 6 (forming 36 and 63)
4 and 5 (forming 45 and 54)
5 and 7 (forming 57 and 75)
Total numbers = 14
Final Answer: E
GMAT assassins aren't born, they're made,
Rich
For this question, you would likely find it helpful to use the 'division rule of 3': for an integer to be divisible by 3, the sum of the DIGITS of the integer will be divisible by 3.
For example:
24 is divisible by 3 because 2+4 = 6 and 6 is divisible by 3
25 is NOT divisible by 3 because 2+5 = 7 and 7 is NOT divisible by 3
Using that pattern, you can quickly list out the 'pairs' of numbers that would form a 2-digit number that is divisible by 3:
1 and 2 (forming 12 and 21)
1 and 5 (forming 15 and 51)
2 and 4 (forming 24 and 42)
2 and 7 (forming 27 and 72)
3 and 6 (forming 36 and 63)
4 and 5 (forming 45 and 54)
5 and 7 (forming 57 and 75)
Total numbers = 14
Final Answer: E
GMAT assassins aren't born, they're made,
Rich
