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3 years ago

How many times does the digit 7 appear among the terms of the sequence of consecutive integer numbers 7, 8, 9, ...., 777?

Mathematics

2 answers:

harina [27]3 years ago

7 0

Answer:

It appears 234 times

Step-by-step explanation:

For first 6 hundreds i.e, 17-107, 117-207, 217-307, 317-407, 417-507, 517-607, there are 20 sevens for each. Which gives a total of 20*6 = 120 sevens.

The number '7' along with this makes a total of 121 sevens.

From 617-707, there are 28 sevens. TOTAL = 121 + 28 = 149

From 708-716, there are 9 sevens.

Now, from 717-777, we have a total of 76 sevens

Adding these all makes total= 149+9+76 = 234

Dahasolnce [82]3 years ago

4 0

Answer:

234 times

Step-by-step explanation:

Number of times the number 7 appears in a hundred

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

Now, from 700 to 777

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

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