Answer:
234 times
Step-by-step explanation:
<u>Number of times the number 7 appears in a hundred</u>
7 as units digit (07-17-27 ..... 97): 10 times
7 as tens digit (70-71-72..... 79): 10 times
20 times the digit 7 appears in first one hundred (0-100)
Let's calculate how many times 7 would be as units or tens in 7 hundreds
20X7 = 140 times digit 7 appears until number 699
<u>Now, from 700 to 777</u>
7 as hundreds digit (700-701-702 .... 777): 78 times
7 as tens digit (770-771-772 .... 777): 8 times
7 as units digit (707-717-727....777): 8 times
78 + 8 + 8 = 94 times the digit 7 appears in the range 700 - 777. Plus 140 times
140 + 94 = 234 times
There may be more brilliant solution than the following, but here are my thoughts.
We make use of Euclid's algorithm to help us out.
Consider finding the hcf of A=2^(n+x)-1 and B=2^(n)-1.
If we repeated subtract B from A until the difference C is less than B (smaller number), the hcf between A and B is the same as the hcf between B and C.
For example, we would subtract 2^x times B from A, or
C=A-2^xB=2^(n+x)-2^x(2^n-1)=2^(n+x)-2^(n+x)+2^n-1=2^n-1
By the Euclidean algorithm,
hcf(A,B)=hcf(B,C)=hcf(2^n-1,2^x-1)
If n is a multiple of x, then by repetition, we will end up with
hcf(A,B)=hcf(2^x-1,2^x-1)=2^x-1
For the given example, n=100, x=20, so
HCF(2^120-1, 2^100-1)=2^(120-100)-1=2^20-1=1048575
(since n=6x, a multiple of x).
Answer:
(0,0)
Step-by-step explanation:
x y
-2 -8
-1 0
0 0
1 -2
2 0
3 12
