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# More counting.....

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Find the no. of natural nos. less than 10,000 and divisible by 5 that can be formed with the ten digits 0 to 9, each digit not occurring more than once in each number. Thanks for help.

Jun 14, 2018

#1
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This is a tough one, because there are so many exceptions:

Here is my attempt:

There are 18 numbers between 5 and 99
There are 16 numbers each, between 100 and 999, with the exception of 500, in which there are only 8 such numbers. So:
16 x 8 + 8 + 18 =154 numbers.
1025, 1035, 1045, 1065, 1075, 1085, 1095.................7 such numbers.
1230, 1240, 1250, 1260, 1270, 1280, 1290.................7........................
1205, 1235, 1245, 1265, 1275, 1285, 1295.................7........................
5120, 5130, 5140, 5160, 5170, 5180, 5190.................7 such numbers x 7
So, the total should be=154 + (112 x 8) + (7 x 7) =1,099 such numbers.

Jun 14, 2018
edited by Guest  Jun 14, 2018
#2
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I agree with the "Guest" of having 154 numbers between 1 and 999
Between 1,000 and 1999: There are 8 options x 7 options for each number ending in "0" and ending in "5": 8 x 7 x 2 = 112
This applies to 2000, 3000, 4000, 6000, 7000, 8000 and 9000
112 x 8 =896
Now 5000 to 5999 will be different because of numerous repeats:
Only numbers ending in "0" will be allowed. There are the 8 options x 7 options =8 x 7 =56
So, the total should be:
154 + 112 x 8 + 8 x 7 =1,106 Numbers divisible by 5 with no repeated digits.

Jun 15, 2018