Question

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?

Answer 1

10 terms

Explanation:

Given : A list of rows of odd numbers continues in the same pattern: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19 such that each row starts where the previous left off, and has one more term than the previous row.

To find : Number of terms in the row in which the sum of the units digits of the numbers equal to 50.

List of rows of odd numbers :

First Row: 1

Second Row: 1 3

Third Row: 1 3 5

Fourth Row : 1 3 5 7

and so on.

Now if we write 1 3 5 ... 19, sum of unit digits = 1+3+5+7+9+1+3+5+7+9 = 50

Therefore, there are 10 terms in this row .

Answer 2

10 entries

Explanation:

Given that odd numbers starting from 1 are written as

1

1 3

1 3 5

and so on.

We have to find the number of digits such that the unit digits total 50

We find that when we write 1 3 5 ...upto 19, unit digits add upto

1+3+5+7+9+1+3+5+7+9 = 50

Thus there are 10 entries.