Combinatorics
Probability
Level
4
Find the number of all 5 digit numbers each of which contains the block 15 and is divisible by 15 (eg: 31545, 41315 are two such numbers)?
The answer is 479.
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1 solution
Answer is 479.
A number is divisible by 15 iff it is divisible by both 3 & 5. There are 4 cases (wrt the position of the block 15)
CASE 1: The number is of the form _ _ _ 15
Any such number is divisible by 5. To make it divisible by 3, the first 3 digits must form a number divisible by 3. The numbers are 102, 105,...,999. There are 300 such numbers.
CASE 2: The number is of the form 15 _ _ _
To make the numbers of this case divisible by 15, it is necessary and sufficient for the numbers formed by the last 3 digits to be divisible by 15. There are 67 such numbers (000, 015,..., 990)
CASE 3: The numbers are of the form 15 _
To make such numbers divisible by 5 there are 2 sub-cases
A) the ones digit is 0
To make such numbers divisible by 3 the 2 digit numbers formed by the remaining digits must be divisible by 3. There are 30 such numbers. (12, 15,...,99)
B) the ones digit is 5
To make such numbers divisible by 3, the 2 digit numbers formed by remaining digits must leave a remainder 1 when divided by 3. There are 30 such numbers (10, 13,...,97)
Thus, total numbers in this case are 30+30=60
CASE 4: The numbers are of the form _ _ 15_
This case is similar to the previous one and has 60 numbers.
So the total numbers are 300+67+60+60= 487
But there is some double counting going on because we may have 2 blocks of 15.
1) 1515_
15150 is a number which is counted in cases 2 & 4
2) 15_15
There are 4 numbers (put 0, 3, 6,9 in the blank) which are counted in cases 1 & 2
3) _1515
There are 3 numbers (put 3,6,9 in the blank) which are counted in cases 1 & 3
Thus, 8 numbers are counted twice. So the total valid numbers are 487-8=479.