Question

I How many times does the digit 3 occur in the hundred place
' in the numbers between loo and 10,000?
​

Answer 1

Answer:

You can solve this question using AP

$a_{n} = a+(n-1)d$

where,

$a_{n}$ = 10000

a = 1000

d = 3

n = ?

putting them in place,

10000 = 1000 + (n-1)3

9000 = (n-1)3

$\frac{9000}{3}$ = (n-1)

3000 = n-1

3001 = n

Pls mark this as brainliest!

Answer 2

Answer:

1000 times

Step-by-step explanation:

first no= 300

last number= 9399

no are

300, 301, 302....... ...399 =100

1300, 1301, 1302.... ..1399 = 100

2300, 2301.................2399 = 100

3300 ,3301............. ..3399 = 100

.....

....

9300, 9301 ,9301.....9399 = 100

Total= 10×100=1000