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LM
What is the tens' digit of the sum of the first 40 terms of 1, 11, 111, 1111, 11111, 111111, ...?

A. 2
B. 3
C. 4
D. 8
E. 9

40*1 (units digits) = 40, carry over 4
39*1 + 4 = 43, carry over 4

Answer: B
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LM
What is the tens' digit of the sum of the first 40 terms of 1, 11, 111, 1111, 11111, 111111, ...?

A. 2
B. 3
C. 4
D. 8
E. 9

:( I misunderstood when reading the problem. Tens's digit = tenth digit :( OMG
Anyways, the answer should be B
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1 + 11 + 111 + .........

Unit digit: Sum of [1's] till 40: 40 * 1 = 40 [carried over '4' to ten's digit]

Ten's digit: We have 39 [1's]. Therefore, sum: 39 * 1 = 39 + 4[carry over] = 43.

Ten's digit is 3.

Answer B
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What is the tens' digit of the sum of the first 40 terms of 1, 11, 111, 1111, 11111, 111111, ...?

A. 2
B. 3
C. 4
D. 8
E. 9
______________________________

1
+ 11
+111
.
.
40 terms
That means for unit digit there are 40 1s hence unit digit will be 0 and carry over is 4

Now while adding the second term there are only 39 terms ( as we have used up 1st term which was "1"
Now 39 + 4 = 43

Hence tens' digit is 3

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