Question

let n - 6+66+666+...+666...66,where there are hundred 6'sin the last term in the sum.how many times does the digit 4 occur in the number n​

Answer 1

Answer:

Step-by-step explanation:

6+66+666+ (100 terms )

=6(1+11+111+ (100terms)

=(6/9)(9+99+999+(100 terms)

=(2/3){(10–1) + (100–1) + (1000–1) (100 terms}

=(2/3){(10+100+1000+….100terms)-(1+1+(100 terms)

=(2/3){10(10^100 - 1)/9 - 100} (since 10+100 + it is a G.P. So we can use sum to n terms formula)

=(20/27)(10^100 - 1) -200/3

                           I HOPE IT WILL HRELPS YOU

Answer 2

Answer:

n= 6+66+666+.....666....66/100 terms

⇒ 6/9 [(10-1)+(10^2-1)+........(10^100-1)]

⇒2/3[999...90/9 -100]

⇒2/3 × 111....1/9898 times 010

⇒1/3×222......2/96 times 020

⇒740740.......740/96 digits 7340

= 4 occurs 33 times in no. N

Step-by-step explanation:

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