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The set X consists of the following terms: {4, 44, 444, 4444, ...), where the nth term has n 4's in it for example 10th term is 4444444444. What is the hundreds digit of the sum of the first 45 terms of set S?

A. 5
B. 3
C. 1
D. 6
E. 9

Explanation:
unit place digit = sum of all 4s at unit place (45 times) i.e 180

so, 0 will take the units place and 18 will carry forward.

similarly for tens place digit =sum of all 4s at tens place (44 times)+ carry forward value from unit place(18)
i.e 44*4+ (18)
=194
so, 4 will take the tens place and 19 will carry forward.

similarly for hundreds place=sum of all 4s at hundreds place (43 times)+ carry forward value from tens place(19)
i.e 43*4+ (19)
=191
so, 1 will be the hundreds digit in given sum.

Hence C is the correct answer.
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S consists of the numbers 4, 44, 444, 4444,..... Now summing the 45 digits means that the units digit of each number has all 45 places. So 45*4=180(18 in hand). Next for tens digit, we now have 44 places. So 44*4=176. Also, we had 18 in hand previously, so 194(19 in hand). Finally, for hundreds digit, we have 43 places. So 43*4=172. Adding the carry over of 19, we get 191. Hence 1 is the hundreds digit as 19 is the carry over for thousands digit.

Hence, C is the answer.

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Since we only need the 100s Digit, the question can be simplified to 45(4) + 44(40) + 43(400)

I personally solved it as 43(444) + 40(1) + 4(2) which gives ....092 + 40 + 8 = ...140

So the answer is C. 1
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