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jlgdr
Is the sum of the integers from 54 to 153 inclusive, divisible by 100?

Hint: Can be solved fast with a property.

# of integers from 54 to 153 inclusive is 153-54+1=100.

The sum = (average)*(# of integers) = (54+153)/2*100=103.5*100=10350 --> not a multiple of 100.
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jlgdr
Is the sum of the integers from 54 to 153 inclusive, divisible by 100?

Hint: Can be solved fast with a property.

My other approach :)

Sum of 1 to 153, inclusive = [(1+153)/2] x 153 = 77 x 153
Sum of 1 to 53, inclusive = [(1+53)/2] x 53 = 27 x 153

Sum of 54 to 153, inclusive = 77*153 - 27*53
= 77*100 + 77*53 - 27*53
= 77*100 + 53*(77-27)
= 77*100 + 53*50

Only 77*100 is divisible by 100

==> The ans is: NOT divisible by 100

Hope it helps.
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Great Question.
I actually calculated the sum without looking for the property.

Also to add to the properties =>
Mean of n consecutives can be of the form x or x.5 (for any integer x)
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jlgdr
Is the sum of the integers from 54 to 153 inclusive, divisible by 100?

Hint: Can be solved fast with a property.

jlgdr Nice question

Last- First/ rate of increase +1 = # items
153-54/ 1 +1 -
99/1 +1 = 100

54 +153 = 207

According to the property 207 cannot be a multiple of 100; but here's another example 8 9 10 11- there are 4 terms and the sum is 38...38 cannot be a multiple of 4.
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jlgdr
Is the sum of the integers from 54 to 153 inclusive, divisible by 100?

Hint: Can be solved fast with a property.


I solved it using Arithmetic Progression.
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