The sum of the first 'n' natural numbers is a 3 digit number in which all the digits are the same. How many numbers have been summed? Some people just added up 1+2+3+ etc. until they found the first total in which all the digits are the same. The sum of the first 36 numbers from 1 to 36 add up to 666 so the answer must be that 36 numbers have been summed. There are better methods.

Soh Yong Sheng, 13, Raffles Institution, Singapore recognsied
that if the sum has all the digits the same it must be a multiple
of 111 and used the fact that the sum of the first n whole number
is *n(n+1)/2*

Since the sum has to be a three digit number in which all the digits are equal, the sum has to be a multiple of 111 = 37 x 3.

Hence *n* or *n+1* is a multiple of 37. But the
product is a three digit number and hence higher multiples can be
ignored. Therefore *n* or *n+1* is 37.

If *n* = 37 then *n* +1 = 38 and the product is
not a multiple of 3. So *n*+1 = 37 and *n* = 36 .

Soh Yong Sheng also used this method:

The question gives an equation 111a = *n(n+1)/2* . Thus
we need to find *n* where 222a = *n* ^{2} +
*n*, 1

Listing the possibilities:

a=1: 14 ^{2} + 14 < 222 and 15 ^{2} + 15 >
222

a=2: 20 ^{2} + 20 < 444 and 21 ^{2} + 22 >
444

a=3: 25 ^{2} + 25 < 666 and 26 ^{2} + 26 >
666

a=4: 29 ^{2} + 29 < 888 and 30 ^{2} + 30 >
888

a=5: 32 ^{2} + 32 < 1110 and 33 ^{2} + 33 >
1110

a=6: 36 ^{2} + 36 = 1332

Hence *n* = 36