+0  
 
0
77
5
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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?

 Oct 21, 2020
 #1
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+1

IDK but the answer is 10

 

1+3+5+7+9 is 25 which makes it a lot easier.

 Oct 21, 2020
 #2
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0

it says sum of units digit

Guest Oct 21, 2020
 #3
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0

I know. 1, 3, 5, 7, and 9 are the repeating units digits.

Guest Oct 21, 2020
 #4
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+1

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

 Oct 21, 2020
 #5
avatar+111602 
+1

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

 Oct 21, 2020

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