Tuesday, September 20, 2011

Math Beast Challenge problem

From Manhattan GRE's Math Beast Challenge page:

This Week's Problem: "Awesome Odds Sum"

Which of the following numbers cannot be written as the sum of a sequence of numbers, a1 + a2 + a3 + … + an, where a1 = 1 and an = an–1 + 2 for all n > 1?

(A) 121
(B) 144
(C) 169
(D) 201
(E) 225

My answer will be in the comments section, as usual. MGRE's official answer will be out next week.



Charles said...

Isn't that just the sequence of all odd numbers?

The answer is D, but I figured that out using the brute force method (actually running through the sequence).

After looking at the list of numbers again, though, I noticed that every number but 201 is the square of some other number: 11 squared is 121, 12 squared is 144, 13 squared is 169, and 15 squared is 225. D is the odd man out, and would have to be 196 to fit in this sequence.

So I guess the answer has something to do with squares, but I can't quite make the connection. That is, I now know that any sum of the sequence of odd numbers will be the square of an integer (4, 9, 16, 25, 36, etc.), but I'm not sure what the mathematical reasoning behind that is. Help me, Math Hominid!

Kevin Kim said...


Well, there goes my need to explain the problem! Heh. Yes, indeed: one quickly comes to realize that the sum of a set of consecutive odd integers, where the first term is 1, is always a perfect square.

1 + 3 = 4

1 + 3 + 5 = 9

1 + 3 + 5 + 7 = 16


So by that logic, 201 stands out as not being a perfect square, and voilĂ .

As to why it works out this way... beats me. Just one of those funky number properties, I guess.

NB: The rule is that the sequence of consecutive odd integers must always begin with 1 as the first term. If any cluster of consecutive odd integers were being considered, you wouldn't necessarily end up with a perfect square. Examples:

3 + 5 + 7 = 15

9 + 11 + 13 = 33

5 + 7 + 9 + 11 = 32


To be honest, you figured this out faster than I did. As with many GRE Quant problems, the quick solution to this problem involves a trick-- in this case, knowledge of a weird property of numbers.

Horace Jeffery Hodges said...

I refuse to answer any more math problems that are clearly designed to make me look stupid!

Everyone else, of course, is free to continue this mathematical bestiality . . .

Jeffery Hodges

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