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0 can't be the leftmost digit, otherwise it will be a four digit number
So total numbers = 5! = 120
Numbers with 0 as leftmost digit = 4! =24
So total 5 digit numbers = 120 - 24 =96
Total 5 digit numbers = Total numbers (non 0) + Total numbers (including 0 but not at first place) = 120 + 96 =216
we can have two scenarios with a sum that is divisible by 13.
sum of 15
sum of 12
We can permute the first scenario in 5! ways.
5! = 120
We can permute the second scenario in 4*4! ways.
The first slot cannot be zero. Remove zero from the options. 4 ways to fill first slot. Replace zero in the second slot and account for the number we placed in the first slot. We have 5-1+1= 4 or 4! to permute the rest of the slots.
RULE: When sum of digits of a number is divisible by 3, that number is divisible by 3. In 0,1,2,3,4,5, we can form 5 digits out of 1,2,3,4,5 and 0,1,2,4,5 that are divisible by 3.
1. 1,2,3,4,5 -> 5! = 120 2. 0,1,2,4,5 -> 5! = 120. However, can't have leftmost digit = 0. So, numbers to be removed = 4! = 24 -> 120 - 24 = 96
Total = 120 + 96 = 216 -> D. _________________
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DS - If negative answer only, still sufficient. No need to find exact solution. PS - Always look at the answers first CR - Read the question stem first, hunt for conclusion SC - Meaning first, Grammar second RC - Mentally connect paragraphs as you proceed. Short = 2min, Long = 3-4 min