The digits from 1 to 9 are written in order so that the digit n is written n times. This
forms the block of digits 1223334444... 999999999. The block is written 100 times.
What is the 1953rd digit written?
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8
Answers
Answered by
2
Answer:
(c)6 is the answer.
Step-by-step explanation:
1+2+3+4+5+6+7+8+9=45
1953÷45,remainder is 18
by counting the number itself we get in 18th position 6, so 6 is the answer.
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Answered by
0
Answer:
C) 6
Step-by-step explanation:
1+2+3+4+5+6+7+8+9=45
1953 ÷ 45. The remainder is 18. If you see what the 18th position is in the block, it will be the digit 6. So the answer is C) 6.
Hope you understand! :)
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