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Yeetyeti38

sujoykrdatta
Question:
What is the summation of all the 2-digit numbers that are divisible by 3?

Solution:
The first 2 digit number is 12 = 3 * 4
The first 2 digit number is 99 = 3 * 33

The number of such terms = 33-4+1 = 30

Thus, required sum

= 3*4 + 3*5 + ... 3*33
= 3(4 + 5 + ... 33)

Since the numbers 4,5,6... Have a constant difference between consecutive terms (i.e. are in Arithmetic Progression), the sum can be calculated as:

Sum of 4,5,6,...33
= Mean * Number of terms
= [(1st term + Last term)/2] * (Number of terms)

= [(4+33)/2] * 30
= 37*15 = (37*3)*5
= 111*5 = 555

Thus, the required sum

= 3(4 + 5 + ... 33)
= 3 * 555
= 1665

Answer B


Note: The answer options aren't so well thought of. First, they should be in ascending order.
Next, it's too easy to decide the answer by looking at the options - only 111 and 1665 are multiples of 3; of which 111 is just too small.
Third, what is 0 doing as an option here!!!


Posted from my mobile device
­I don't understand why 0 is wrong...the question asks for all 2-digit numbers divisible by three, not all positive 2-digit numbers divisible by three. If we include the negative numbers into the summation, it would be 0.

Why shouldn't we include negative numbers?
­
Yes, the question should ask about positive integers. Not a good wording. So, ingonre and move on.
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