How many times does the digit one appear from 101 to 200

Step-by-step explanation:

Let's get the answer first and then we can discuss how it is done. The answer is 119.

There is a problem. The question does not state whether or not 101 is included. If it is 119 is correct. If the question is between 101 and 200 then the answer will be 117.

How was it done? A programming language was used. So it counted by brute force.

From 101 to 120 we get

101  102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120

32  Certainly there must be a quicker way

101 = 2

102 to 109 = 1 * 8 = 8

110 to 119 = 10*2 = 20  

111 has an extra 1  = 1

120 has                    1

Total                       32  

From here it gets a bit easier.

121                      2

122 - 129            8

total                  10

130                     1

131                      2

132 - 139           8  

total                  11

140 to 149        11

150 to 159        11

160 to 169        11

170 to 179         11    

180 to 189         11

190 to 199         11

total 7 * 11 + 32 + 10 = 119

the answer is 99.

answer: c, the green line

step-by-step explanation:

there is no black line, eliminate b.

simple interest implies periodic payments, but the red line shows no payments, eliminate d.

simple interest has equal payments. if paid say every day, the graph is about a straight line (actually a stairstep).

the yellow line is not straight, eliminate a.

check: the green line is straight, answer is c.

simple interest with positive interest rate gives a straight line sloping up to the right. green line:

g(t) = $100 + ($200-$100) ÷ (20 years) × (t years)

= $100 + (4%/year) × ($100)(t years)

incidentally, the yellow line curving upward represents continuously compounded interest

y = $100 × 2^(t years / (14 assuming the yellow line hits $200 at 14 years. 2^0 is 1, so y(0) = $100; 2^(14/14) is 2, so y(14) = $200.