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Question

Mathematics Question on Arithmetic Progression

The sum of 55 digit numbers in which only odd digits occur without any repetition is

A

277775277775

B

555550555550

C

11111001111100

D

66666006666600

Answer

66666006666600

Explanation

Solution

The digits that make the numbers are 1,3,5,71, 3, 5, 7 and 99. The number of numbers with one of these in the first place =4!= 4!. \therefore The required sum of all the numbers =25(104+103+102+10+1)×4!= 25\left(10^{4} + 10^{3} +10^{2} + 10 + 1\right)\times 4! =600×1051101= 600 \times\frac{ 10^{5} -1}{10 -1 } =600×11111= 600 \times 11111 =6666600= 6666600.