An integer consists of 110 digits such that any 10 consecutive digits is different. What is the sum of all the digits!?
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Answers
EXPLANATION.
An integer consists of 110 digits.
Such that any 10 consecutive digits is different.
As we know that,
Total digits = 110 - 10 = 100.
10 consecutive digits is different.
Let, we take.
⇒ 0, 1, 2, 3, . . . . . 9.
First term = a = 0.
Common difference = d = b - a = c - b.
Common difference = d = 1 - 0 = 1.
As we know that,
Formula of sum of nth term of an A.P.
⇒ Sₙ = n/2[2a + (n - 1)d].
⇒ S₁₀ = 10/2[2(0) + (10 - 1)1].
⇒ S₁₀ = 5[0 + 9].
⇒ S₁₀ = 45.
Since, 110 digits makes 11 combinations.
Sum of all digits = 11 x 45 = 495.
MORE INFORMATION.
Supposition of terms on A.P.
(1) = Three terms as : a - d, a, a + d.
(2) = Four terms as : a - 3d, a - d, a + d, a + 3d.
(3) = Five terms as : a - 2d, a - d, a, a + d, a + 2d.
Given :-
An integer consists of 110 digits such that any 10 consecutive digits is different.
To Find :-
Sum
Solution :-
Total digit = 110 - 10
Total digit = 100
Sum of the 10 digit = 9 × 10
Sum of the 10 digit = 90
Now
Sum = 90/2
Sum = 45
Given,
Total digit is 110
110/10
11/1
11
Now
Sum = 45 × 11
Sum = 495