Question

When the natural numbers 1, 2, 3, .... 500 are written, then the digit 3 is used n times in this way. the value of n is?

Answer 1

we have to use progressions for this
that is
a.p=1,2,3..........500
from this 3,13,13........493
the common difference of the a.p=a2-a1
we get common difference=10
$an = a + (n - 1) \times d$
subistuting the values in this equation
493=3+(n-1)×10
493-3=10n-10
490+10=10n
500/10=n
therefore the value of n=50