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# Answer to Question #28108 in Discrete Mathematics for Kyle

Question #28108
Let S ={100, 101, 102,....999} so that |S| = 900.
a) How many numbers in S have at least one digit that is a 3 or 7? Examples... 300, 707, 103.
b) How many numbers in S have at least one digit that is a 3 and a 7? Examples.... 736 and 377
a) 7*(2*(10-2)+2*(10-1))+2*(100-9*1-10*1)=428
The amount of our numbers is the same on these diapasons (100 - 199), (200 -299), (400 - 499), (500 - 599), (600 - 699), (800 - 899), (900 - 999).
Let&#039;s look at (100 - 199). There we have: 103,107,113,117,...,130-136,138-139,143,147,...,170-172,174-179,183,187,193,197 =
2*(10-2)+2*(10-1). So we have 7*(2*(10-2)+2*(10-1)). And the amount of our numbers is the same on these diapasons:
(300 - 399), (700 - 799).
Let&#039;s look at (300 - 399). There we have all numbers, except: 307,317,...,370-379,387,397 = 100-9*1-10*1.
So we have 2*(100-9*1-10*1).

b) 7*2*1+2*((10-1)*1+10*1)=52
The amount of our numbers is the same on these diapasons (100 - 199), (200 - 299), (400 - 499), (500 - 599), (600 - 699), (800 - 899), (900 - 999). Let&#039;s look at (100 - 199). There we have: 137,173 = 2*1=2.
So we have 7*2=14. And the amount of our numbers is the same on these diapasons: (300 - 399), (700 - 799).
Let&#039;s look at (300 - 399). There we have: 307,317,...,397,370-379 = (10-1)*1+10*1=19 So we have 2*19=38.

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