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Find the sum of all three digit numbers that leave a remainder of '1' when divided by 3?
897
1,64,850
1,64,749
1,49,700
The smallest three digit number that will leave a remainder of two when divided by three is 100. The next no. that will leave a remainder of two when divided by three is 103, 106, The largest three digit number that will leave a remainder of two when divided by three is 997. So, it is an Arithmetic Progression with the first term being 100 and the last term being 997 and common difference being three. We know that in an Arithmetic Progression, the nth term an = a1 + (n - 1) d In this particular case, therefore, 997 = 100 + (n - 1)* 3 i.e., 897 = (n - 1) * 3 Therefore, n - 1 = 299 Or n = 300. So, the sum of the Arithmetic Progression will be = n/2[2a+ (n−1) d] =164,850
By: Amit Kumar ProfileResourcesReport error
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