Consider the 2-digit numbers that end with 6 first.


16 26 36 46 56 66 76 86 96

There are nine 6's in the ones place.

Sum of the value of the digits in the ones place = 6 × 9 = 54

Sum of the value of the digits in the tens place
= (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9) × 10
= 45 × 10
= 450


Next, consider the 2-digit numbers that start with 6 (we do not include 66 as it was already counted above)

60 61 62 63 64 65 67 68 69

Sum of the value of the digits in the ones place

= 0 + 1 + 2 + 3 + 4 + 5 + 7 + 8 + 9
= 45 - 6
(using the result above. It is just taking away a 6 from the sum of 1 to 10, which was calculated to be 45

= 39

There are nine 6's in the tens place.

Their value = 9 × 6 × 10
= 9 × 6
= 540

Total :

54 + 450 + 39 + 540

= 1083

Another way to count :

①There are ten 60s' among these :

(60 61 62 63 64 65 66 67 68 69)

Their value = 10 × 60 = 600

Sum of the value of the digits in the ones place (cannot include 6 as it will be counted below)

= 1 + 2 + 3 + 4 + 5 + 7 + 8 + 9
= 39



②There are nine 6's among these :

(16 26 36 46 56 66 76 86 96)

9 × 6 = 54

Sum of the value of the digits in the tens place (cannot include 6 as it has been counted already)

= (1 + 2 + 3 + 4 + 5 + 7 + 8 + 9) × 10
= 39 × 10
= 390

That's ten 39s.


Total we have 11 times of 39 , 10 times of 60 and 9 times of 6


390 + 600 + 39 + 54

= 990 + 93
= 1083

A third way : pairing method


16 26 36 46 56 66 76 86 96

16 + 96 = 112
26 + 86 = 112
36 + 76 = 112
46 + 66 = 112

56 is basically half a pair.

It's as good as having nine 56's

Once we know this pairing method,
Just do

112 × 4½
= 56 × 9
= 504


60 61 62 63 64 65 66 67 68 69

60 + 69 = 129
61 + 68 = 129
..
..
64 + 65 = 129

5 pairs so 5 × 129 = 645


But we double counted 66 so we must subtract one 66 away.

645 + 504 - 66 = 1083

The third one had a miscalculation I think

46 + 56 and the last sum as well, but the student should realise that the method is logical.

Yea. Still editing.