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Question: The digit in the tens place in the number \(\angle 1 + \angle 2 + \angle 3 + ... + \angle 99\) is...

The digit in the tens place in the number

∠1+∠2+∠3+...+∠99\angle 1 + \angle 2 + \angle 3 + ... + \angle 99 is

A

1

B

2

C

3

D

4

Answer

1

Explanation

Solution

Now

∠1+∠2+∠3+∠4+∠5+∠6\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6

= 1+ 2 + 6 + 24 + 120 + 720 = 873 and ∠7+∠8+∠9\angle 7 + \angle 8 + \angle 9

= 5040 + 40320+ 362880 = 408240

∴ ∠1 6mu+ 6mu∠2 6mu+ 6mu∠3 6mu+ 6mu...+ 6mu∠9 6mu= 6mu409113\angle 1\mspace{6mu} + \mspace{6mu}\angle 2\mspace{6mu} + \mspace{6mu}\angle 3\mspace{6mu} + \mspace{6mu}... + \mspace{6mu}\angle 9\mspace{6mu} = \mspace{6mu} 409113

All the numbers from ∠10\angle 10 to ∠99\angle 99 are divisible by 100, therefore, ∠10\angle 10 + ∠11+...+∠99\angle 11 + ... + \angle 99 will have 0 in the units place and 0 in the hundreds place.

Hence ∠1+∠2+∠3+...+∠99\angle 1 + \angle 2 + \angle 3 + ... + \angle 99 will have the digit ‘1’ in the tens place (and the digit 3 in the units place).