The following list of rows of odd numbers continues in the same pattern:
1
3, 5
7, 9, 11
13, 15, 17, 19.
Each row starts where the previous left off, and has one more term than the previous row. There is one row in which the sum of the units digits of the numbers in that row equals 50. How many terms are in that row?
1
3, 5
7, 9, 11
13, 15, 17, 19.
21, 23. 25, 27, 29
31, 33, 35, 37, 39, 41
43, 45, 47, 49, 51, 53, 55
57, 59, 61, 63, 65, 67. 69, 71
73, 75, 77, 79, 81, 83, 85, 87, 89,
91, 93, 95, 97, 99, 101, 103, 105, 107, 109
so 10
+118667 1
1+3=4
1+3+5=9
1+3+5+7=16
1+3+5+7+9=25
1+3+5+7+9+1=26
1+3+5+7+9+1+3=29
1+3+5+7+9+1+3+5=34
1+3+5+7+9+1+3+5+7=41
1+3+5+7+9+1+3+5+7+9=50
First number of row, 1,3,7,13, 21
| number of terms | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| first number in row | 1 | 3 | 7 | 13 | 21 | 31 | 43 | 57 | 73 | 91 | 111 |
| *** |
So it is the 10th row
