Nice mental math problem - if you're bold enough - with the sum from 1 to 9 is 45: 4500/9 + 450/10 + 45/9 = 500 + 45 + 5 = 550. The divisors are 9 in hundreds and ones place to avoid 0 as a leading digit.
My solution was quicker: you can pick the 1st digit from 1-9, the second digit independently from 0-9 and the third digit is the same as the first one. So the average of the first and third digits is 5, the average of the 2nd digit is 4½, and the average of all of all the 3 digit palindromes is 5 hundred 4½-ty 5: 5×100 + 4½×10 + 5 = 550.
I think you get 101*a for first digit average of first digit is 5 as stated then the average of the middle digit is 4.5 and multiplies by 10 ... Yeah 550
I got no where with this question, I attempted solution like below but got stuck. One of the solutions given included: the arithmetic mean of all n digit palindromes is just the average of the smallest and largest n digit palindromes, in this case 101 and 999, hence answer is 550. Very clever I thought!
Nice mental math problem - if you're bold enough - with the sum from 1 to 9 is 45:
4500/9 + 450/10 + 45/9
= 500 + 45 + 5
= 550.
The divisors are 9 in hundreds and ones place to avoid 0 as a leading digit.
My solution was quicker: you can pick the 1st digit from 1-9, the second digit independently from 0-9 and the third digit is the same as the first one. So the average of the first and third digits is 5, the average of the 2nd digit is 4½, and the average of all of all the 3 digit palindromes is 5 hundred 4½-ty 5: 5×100 + 4½×10 + 5 = 550.
Nice, much more straightforward than my way.
I think you get 101*a for first digit average of first digit is 5 as stated then the average of the middle digit is 4.5 and multiplies by 10 ... Yeah 550
I got no where with this question, I attempted solution like below but got stuck. One of the solutions given included: the arithmetic mean of all n digit palindromes is just the average of the smallest and largest n digit palindromes, in this case 101 and 999, hence answer is 550. Very clever I thought!
Sounds similar to the Gauss technique of summing an arithmetic series!
@@mathoutloud Yep, so it matters not how many of them there are, which seems like the obvious first step to the question.