PRMO 2018 question 19

19. Let $N = 6 + 66 + 666 + \dots + 666\ldots66$, where there are hundred 6’s in the last term in the sum. How many times does the digit 7 occur in the number $N$?

[PRE-RMO – 2018]

Solution

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$N = 6 + 66 + 666 + \dots + \underbrace{666\ldots6}_{100\text{ times}}$

$$\begin{aligned} N &= \frac{6}{9}\Bigl[(10-1) + (10^2 - 1) + \dots + (10^{100} - 1)\Bigr] \\[6pt] &= \frac{6}{9}\Bigl[\frac{10}{9}\bigl(10^{100} - 1\bigr) - 100\Bigr] \\[6pt] &= \frac{2}{3}\Bigl[\frac{\underbrace{999\ldots90}_{\text{(100 digits of 9, ending in 0)}}}{9} - 100\Bigr] \\[6pt] &= \frac{2}{3}\Bigl[\underbrace{111\ldots110}_{\text{(100 digits of 1, ending in 0)}} - 100\Bigr] \\[6pt] &= \frac{2}{3} \times \underbrace{111\ldots1010}_{\text{(some rearrangement, 98 digits total)}} \\[6pt] &= \frac{1}{3} \times \underbrace{222\ldots2020}_{\text{(98 digits total)}} \\[6pt] &= \underbrace{740740\ldots7407340}_{\text{(96 digits total)}}. \end{aligned}$$

∴ 33 sevens.

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In other words, when you sum $6 + 66 + 666 + \dots$ up to the 100‐digit string of 6’s, the resulting 96‐digit number contains 33 instances of the digit 7.