Problem
Let \( 7 \overset{r}{\overbrace{5…5}}7 \) denote the (r + 2) digit number where the first and the last digits are 7 and the remaining r digits are 5. Consider the sum S = 77 + 757 + 7557 + … + \(7 \overset{98}{\overbrace{5…5}}7\). If S = \( \frac{7 \overset{99}{\overbrace{5…5}}7+m}{n} \), where m and n are natural numbers less than 3000, then find the value of m + n.
Solution
To begin, let’s find an expression Tr for the number \( 7 \overset{r}{\overbrace{5…5}}7 \).
For that, let’s expand Tr into a sum, i.e. 7 x 10r+1 + 5 x 10r + 5 x 10r-1 + … + 5 x 102 + 5 x 101 + 7.
Now, we can express both the 7s as 5 + 2 to get Tr as 5 x (10r+1 + 10r + … + 102 + 101 + 1) + 2(10r+1 + 1).
Now, the sum in the left bracket is a GP, whose sum is (10r+2 – 1)/9.
So, Tr simplifies to 5(10r+2 – 1)/9 + 2(10r+1 + 1), which equals (68.10r+1 + 13)/9.
Let’s do a sanity check! If r = 0, we get T0 = (680 + 13)/9 or 77, which is what we expect (i.e. a number with zero 5s and two 7s at the end)
If r = 1, T1 = (6800 + 13)/9 or 757. Two 7s again, and one 5 in between. Nice!
Let’s do one more. If r = 2, T2 = (68000 + 13)/9 or 7557. Okay, we’re good!
Now, if we take a look at S, it’s nothing but T0 + T1 + T2 + … T98.
That is, S = (68.100+1 + 13)/9 + (68.101+1 + 13)/9 + (68.102+1 + 13)/9 + … + (68.1098+1 + 13)/9
This simplifies to 68(101 + 102 + 103 + … + 1099)/9 + 13.99/9.
The sum in the bracket is a GP again, whose sum is 10.(1099 – 1)/9.
So, S further simplifies to 680(1099 – 1)/81 + 13.11.
Now, our job is to express S in terms of T99 and try to figure out m and n.
Since T99 equals (68.10100 + 13)/9, we can write 1099 as (9.T99 – 13)/680. Let’s substitute this back in S.
We’ll get S = 680((9.T99 – 13)/680 – 1)/81 + 13.11. Looks like a mess!
But, on simplification, we’ll get S = (T99 + 1210)/9.
Comparing this with the given expression, we get m = 1210 and n = 9, implying m + n = 1219.
Comments
Kind of a tedious problem, but not too difficult. If you don’t get scared, and don’t give up on the way, you’ll surely emerge victorious.
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