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How many numbers can we form using 0,1,2,3,4,5,6,7,8,9 only
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06 Oct 2019, 16:26
06 Oct 2019, 16:26
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Question Stats:
14% (03:13) correct
85% (04:34) wrong based on 7 sessions
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How many numbers can we form using 0,1,2,3,4,5,6,7,8,9 only once below 1000 and is divided by 5?
A) 154
B) 155
C) 156
D) 157
E) 158
A) 154
B) 155
C) 156
D) 157
E) 158
Re: How many numbers can we form using 0,1,2,3,4,5,6,7,8,9 only
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Updated on: 07 Oct 2019, 10:28
Updated on: 07 Oct 2019, 10:28
1
We are supposed to make digits divisible by 5. The numbers are less than 1000
So we can form 1 digit, 2 digit and 3 digit numbers
Starting with 1 digit numbers: there is only one 1 digit number divisible by 5 and that is the number 5. Therefore number of such digits = 1
Now 2 digit numbers can have either 0 or 5 in their units place, so lets make two cases
case1: 0 in units place x 0 where x can be any of 1,2,3,4,5,6,7,8,9. So the number of ways of filling the tens digit = 9 and number of ways of filling the units digit = 1. That makes 9*1 such combinations
case2: 5 in units place x 5 where x can be any of 1,2,3,4,6,7,8,9. So the number of ways of filling the tens digit = 8 and number of ways of filling the units digit = 1. That makes 8*1 such combinations
Thus total 9 + 8 = 17 two digit numbers divisible by 5 and less than 1000
3 digit numbers can again have either 0 or 5 in their units place, making two cases again
case1: 0 in units place x y 0 where x can be any of 1,2,3,4,5,6,7,8,9. So the number of ways of filling the tens digit = 9, y will be any number of the list except for number x, thus number of ways of filling the tens digit = 8 and number of ways of filling the units digit = 1. That makes 9*8*1 = 72 such combinations
case2: 5 in units place x y 5 where x can be any of 1,2,3,4,6,7,8,9. So the number of ways of filling the tens digit = 8, y will be any number of the list except for number x but it could also be 0, thus number of ways of filling the tens digit = 8 and number of ways of filling the units digit = 1. That makes 8*8*1 = 64 such combinations
Thus total 72 + 64 = 136 three digit numbers divisible by 5 and less than 1000
So the total such numbers are 1+17+136 = 154
So we can form 1 digit, 2 digit and 3 digit numbers
Starting with 1 digit numbers: there is only one 1 digit number divisible by 5 and that is the number 5. Therefore number of such digits = 1
Now 2 digit numbers can have either 0 or 5 in their units place, so lets make two cases
case1: 0 in units place x 0 where x can be any of 1,2,3,4,5,6,7,8,9. So the number of ways of filling the tens digit = 9 and number of ways of filling the units digit = 1. That makes 9*1 such combinations
case2: 5 in units place x 5 where x can be any of 1,2,3,4,6,7,8,9. So the number of ways of filling the tens digit = 8 and number of ways of filling the units digit = 1. That makes 8*1 such combinations
Thus total 9 + 8 = 17 two digit numbers divisible by 5 and less than 1000
3 digit numbers can again have either 0 or 5 in their units place, making two cases again
case1: 0 in units place x y 0 where x can be any of 1,2,3,4,5,6,7,8,9. So the number of ways of filling the tens digit = 9, y will be any number of the list except for number x, thus number of ways of filling the tens digit = 8 and number of ways of filling the units digit = 1. That makes 9*8*1 = 72 such combinations
case2: 5 in units place x y 5 where x can be any of 1,2,3,4,6,7,8,9. So the number of ways of filling the tens digit = 8, y will be any number of the list except for number x but it could also be 0, thus number of ways of filling the tens digit = 8 and number of ways of filling the units digit = 1. That makes 8*8*1 = 64 such combinations
Thus total 72 + 64 = 136 three digit numbers divisible by 5 and less than 1000
So the total such numbers are 1+17+136 = 154
Originally posted by goodgirlnotbad on 07 Oct 2019, 10:00.
Last edited by goodgirlnotbad on 07 Oct 2019, 10:28, edited 1 time in total.
Last edited by goodgirlnotbad on 07 Oct 2019, 10:28, edited 1 time in total.
Re: How many numbers can we form using 0,1,2,3,4,5,6,7,8,9 only
[#permalink]
07 Oct 2019, 10:02
07 Oct 2019, 10:02
1
goodgirlnotbad wrote:
We are supposed to make digits divisible by 5. The numbers are less than 1000
So we can form 1 digit, 2 digit and 3 digit numbers
Starting with 1 digit numbers: there is only one 1 digit number divisible by 5 and that is the number 5. Therefore number of such digits = 1
Now 2 digit numbers can have either 0 or 5 in their units place, so lets make two cases
case1: 0 in units place x 0 where x can be any of 1,2,3,4,5,6,7,8,9. So the number of ways of filling the tens digit = 9 and number of ways of filling the units digit = 1. That makes 9*1 such combinations
case2: 5 in units place x 5 where x can be any of 1,2,3,4,6,7,8,9. So the number of ways of filling the tens digit = 8 and number of ways of filling the units digit = 1. That makes 8*1 such combinations
Thus total 9 + 8 = 17 two digit numbers divisible by 5 and less than 1000
3 digit numbers can again have either 0 or 5 in their units place, making two cases again
case1: 0 in units place x y 0 where x can be any of 1,2,3,4,5,6,7,8,9. So the number of ways of filling the tens digit = 9, y will be any number of the list except for number x, thus number of ways of filling the tens digit = 8 and number of ways of filling the units digit = 1. That makes 9*8*1 = 72 such combinations
case1: 5 in units place x y 0 where x can be any of 1,2,3,4,6,7,8,9. So the number of ways of filling the tens digit = 8, y will be any number of the list except for number x but it could also be 0, thus number of ways of filling the tens digit = 8 and number of ways of filling the units digit = 1. That makes 8*8*1 = 64 such combinations
Thus total 72 + 64 = 136 three digit numbers divisible by 5 and less than 1000
So the total such numbers are 1+17+136 = 154
So we can form 1 digit, 2 digit and 3 digit numbers
Starting with 1 digit numbers: there is only one 1 digit number divisible by 5 and that is the number 5. Therefore number of such digits = 1
Now 2 digit numbers can have either 0 or 5 in their units place, so lets make two cases
case1: 0 in units place x 0 where x can be any of 1,2,3,4,5,6,7,8,9. So the number of ways of filling the tens digit = 9 and number of ways of filling the units digit = 1. That makes 9*1 such combinations
case2: 5 in units place x 5 where x can be any of 1,2,3,4,6,7,8,9. So the number of ways of filling the tens digit = 8 and number of ways of filling the units digit = 1. That makes 8*1 such combinations
Thus total 9 + 8 = 17 two digit numbers divisible by 5 and less than 1000
3 digit numbers can again have either 0 or 5 in their units place, making two cases again
case1: 0 in units place x y 0 where x can be any of 1,2,3,4,5,6,7,8,9. So the number of ways of filling the tens digit = 9, y will be any number of the list except for number x, thus number of ways of filling the tens digit = 8 and number of ways of filling the units digit = 1. That makes 9*8*1 = 72 such combinations
case1: 5 in units place x y 0 where x can be any of 1,2,3,4,6,7,8,9. So the number of ways of filling the tens digit = 8, y will be any number of the list except for number x but it could also be 0, thus number of ways of filling the tens digit = 8 and number of ways of filling the units digit = 1. That makes 8*8*1 = 64 such combinations
Thus total 72 + 64 = 136 three digit numbers divisible by 5 and less than 1000
So the total such numbers are 1+17+136 = 154
Starting with 1 digit numbers: there is only one 1 digit number divisible by 5 and that is the number 5. Therefore number of such digits = 1
but not only one digit but also 0 and 5(2 digit) both are also can be divide by 5.
