MathRevolution wrote:

What is the sum of all multiples of 3 between 1 and 200?

A. 3300

B. 6600

C. 6633

D. 10100

E. 20100

? = 3+6+9+ ... + 195 + 198

In any finite arithmetic progression, the (global) average is equal to the median and to the average of symmetric elements (to the median).

The first and last elements are always symmetrical, hence their average (3+198)/2 is equal to the (global) average.

How many terms are involved?

\(3 \leqslant 3M \leqslant 198\,\,\left( { = 180 + 18} \right)\,\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{:\,\,\,3} \,\,\,\,\,\,1 \leqslant M\,\,\, \leqslant \,\,\,66\,\,\,\,\, \Rightarrow \,\,\,\,66\,\)

Using

the homogeneity nature of the average we have:

\(? = 66 \cdot \frac{{\left( {3 + 198} \right)}}{2} = 33 \cdot 201\,\,\,\,\,\mathop \Rightarrow \limits^{choices} \,\,\,\,\,\left\langle ? \right\rangle = \left\langle {33 \cdot 201} \right\rangle = \left\langle 3 \right\rangle\)

\(\left\langle N \right\rangle \,\,\, = \,\,\,\,{\text{units}}\,\,{\text{digit}}\,\,{\text{of}}\,\,N\)

The above follows the notations and rationale taught in the GMATH method.

_________________

Fabio Skilnik :: http://www.GMATH.net (Math for the GMAT)

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