The number 75 can be written as the sum of the squares

[Hei]

The number 75 can be written as the sum of the squares of 3 different positive integers. What is the sum of these 3 integers?
A. 17
B. 16
C. 15
D. 14
E. 13

What's the efficient way to locate the answer?
Thanks.

[shaji]

The number 75 can be written as the sum of the squares of 3 different positive integers. What is the sum of these 3 integers?
A. 17
B. 16
C. 15
D. 14
E. 13

What's the efficient way to locate the answer?
Thanks.

Source not specified!!!. E. is however the correct answer. Can't ellaborate without knowing the source as Forum descipline.

[Hei]

Oh, I thought that this sub-forum was for asking questions from GMATPrep software ONLY.
No?
Anyway, it is from GMATPrep software at MBA.com.
Yes, the answer is E. I know why but I don't know how to come up with 1, 5 and 7 as the 3 numbers in an efficient way.
Thanks in advance.

[RonPurewal]

hei is right: there is no need for formal attributions in this folder (everything is assumed to come from the gmatprep software). if you are posting problems here that are NOT from the gmatprep software, you will soon be afflicted with terrible diseases.

--

ok, there might be some crazy elegant way to solve this problem in ten seconds flat. but if it isn't rather obvious in retrospect, who cares? here's a better method:

consider odd / even numbers
note that squares of odds remain odd, and squares of evens remain even. therefore, the 3 numbers are either 3 odds or 1 odd and 2 evens.
notice that this observation kills choices b and d right away (nice if you're pressed for time and have to guess).

next observation: none of the individual numbers can be greater than 8 (because 9^2 = 81 by itself is already too big)

at this point, you have a very good idea of the sizes of numbers you're looking for, so just start trying numbers in an organized way.
3 odds:
sum = 13: (1, 5, 7)
sum = 15: (3, 5, 7)
sum = 17: impossible unless at least one number > 8

2 evens + odd:
sum = 13: (1, 4, 8) (3, 2, 8) (3, 4, 6) (5, 2, 6) (7, 2, 4)
sum = 15: (1, 6, 8) (3, 4, 8) (5, 2, 8) (5, 4, 6) (7, 2, 6) all of which are waywaywaywayway too big
sum = 17: (3, 6, 8) (5, 4, 8) (7, 2, 8) (7, 4, 6) all of which are waywaywaywayway too big

that isn't so bad.

...especially because you can stop as soon as you find (1, 5, 7); i.e., you're not going to have to go through anywhere close to everything in this list, unless you are extraordinarily unlucky.

--

if you're interested (and hardcore):
all squares of odds give remainder = 1 upon division by 4. (try it if you don't believe me)
all squares of evens give remainder = 0 upon division by 4.
(if you reallyreallyreally want to know, i can show you why these rules work, although the proofs will be 100% irrelevant to the gmat)
since 75 has remainder = 3 upon division by 4, it follows that all three numbers are odd. as evidenced in the lists above, that reduces your shortlist of candidates to two possibilities.

[Hei]

Thanks a lot Ron!

[shaji]

hei is right: there is no need for formal attributions in this folder (everything is assumed to come from the gmatprep software). if you are posting problems here that are NOT from the gmatprep software, you will soon be afflicted with terrible diseases.

--

ok, there might be some crazy elegant way to solve this problem in ten seconds flat. but if it isn't rather obvious in retrospect, who cares? here's a better method:

consider odd / even numbers
note that squares of odds remain odd, and squares of evens remain even. therefore, the 3 numbers are either 3 odds or 1 odd and 2 evens.
notice that this observation kills choices b and d right away (nice if you're pressed for time and have to guess).

next observation: none of the individual numbers can be greater than 8 (because 9^2 = 81 by itself is already too big)

at this point, you have a very good idea of the sizes of numbers you're looking for, so just start trying numbers in an organized way.
3 odds:
sum = 13: (1, 5, 7)
sum = 15: (3, 5, 7)
sum = 17: impossible unless at least one number > 8

2 evens + odd:
sum = 13: (1, 4, 8) (3, 2, 8) (3, 4, 6) (5, 2, 6) (7, 2, 4)
sum = 15: (1, 6, 8) (3, 4, 8) (5, 2, 8) (5, 4, 6) (7, 2, 6) all of which are waywaywaywayway too big
sum = 17: (3, 6, 8) (5, 4, 8) (7, 2, 8) (7, 4, 6) all of which are waywaywaywayway too big

that isn't so bad.

...especially because you can stop as soon as you find (1, 5, 7); i.e., you're not going to have to go through anywhere close to everything in this list, unless you are extraordinarily unlucky.

--

if you're interested (and hardcore):
all squares of odds give remainder = 1 upon division by 4. (try it if you don't believe me)
all squares of evens give remainder = 0 upon division by 4.
(if you reallyreallyreally want to know, i can show you why these rules work, although the proofs will be 100% irrelevant to the gmat)
since 75 has remainder = 3 upon division by 4, it follows that all three numbers are odd. as evidenced in the lists above, that reduces your shortlist of candidates to two possibilities.

In Mathematics what's elegant is never 'crazy' and what's 'crazy is never elegant.
The 'better' method can be surmised as :
Having elliminated the impossible, that whatever is left, however improbable has to be the correct answer!!!
Having said so, the 'better' method has its share of elegance , no doubt.

[StaceyKoprince]

By "crazy elegant" I think Ron meant "incredibly elegant." That's a common slang meaning for "crazy." :)

[navdeep_bajwa]

Hi Ron i appreciate your explanation especially the remainder one

[editor: thanks, but the remainder one is WAY less important and WAY less useful. if you appreciate it more than you appreciate the trial-and-error / number plugging solution, then you don't have the right study priorities.
see my comments below. --ron]

[mikrodj]

hei is right: there is no need for formal attributions in this folder (everything is assumed to come from the gmatprep software). if you are posting problems here that are NOT from the gmatprep software, you will soon be afflicted with terrible diseases.

--

ok, there might be some crazy elegant way to solve this problem in ten seconds flat. but if it isn't rather obvious in retrospect, who cares? here's a better method:

consider odd / even numbers
note that squares of odds remain odd, and squares of evens remain even. therefore, the 3 numbers are either 3 odds or 1 odd and 2 evens.
notice that this observation kills choices b and d right away (nice if you're pressed for time and have to guess).

next observation: none of the individual numbers can be greater than 8 (because 9^2 = 81 by itself is already too big)

at this point, you have a very good idea of the sizes of numbers you're looking for, so just start trying numbers in an organized way.
3 odds:
sum = 13: (1, 5, 7)
sum = 15: (3, 5, 7)
sum = 17: impossible unless at least one number > 8

2 evens + odd:
sum = 13: (1, 4, 8) (3, 2, 8) (3, 4, 6) (5, 2, 6) (7, 2, 4)
sum = 15: (1, 6, 8) (3, 4, 8) (5, 2, 8) (5, 4, 6) (7, 2, 6) all of which are waywaywaywayway too big
sum = 17: (3, 6, 8) (5, 4, 8) (7, 2, 8) (7, 4, 6) all of which are waywaywaywayway too big

that isn't so bad.

--

if you're interested (and hardcore):
all squares of odds give remainder = 1 upon division by 4. (try it if you don't believe me)
all squares of evens give remainder = 0 upon division by 4.
(if you reallyreallyreally want to know, i can show you why these rules work, although the proofs will be 100% irrelevant to the gmat)
since 75 has remainder = 3 upon division by 4, it follows that all three numbers are odd. as evidenced in the lists above, that reduces your shortlist of candidates to two possibilities.

Hi Ron, could you recommend any resources in which I can find these advanced number properties?
Thank you in advanced.

[pritesh.suvarna]

In the last section -
Any specific reason you have "divided by 4" to come to a conclusion that all numbers should be odd?


Thanks
Pritesh

[kevinmarmstrong]

Another way is approach this is to list the perfect squares less than 75:

1 4 9 16 25 36 49 64

Could the sum involve 64? The sum of the other two squares would have to be 11: impossible.

How about 49? The sum of the other two squares would have to be 26= 25 + 1

Thus 75= 7^2 + 5^2 +1^2

[Ben Ku]

In the last section -
Any specific reason you have "divided by 4" to come to a conclusion that all numbers should be odd?

When you add numbers together, you can also add their remainders when dividing them by a common number. For example, 14 divided by 6 is 2 R 2, and 25 divided by 6 is 4 R 1. So when we take 39 divided by 6, the remainder will be 3.

Because Ron stated that perfect squares of odd numbers have remainder of 1 when divided by 4, and perfect squares of even numbers divide evenly into 4, we want to find out the remainder when 75 is divided by 4 to determine if our three numbers are all odd, or if we have one odd and two evens.

Hope that makes sense.

[sudaif]

elegant indeed.

[RonPurewal]

In the last section -
Any specific reason you have "divided by 4" to come to a conclusion that all numbers should be odd?


Thanks
Pritesh

well, notice that i prefaced that part with "if you're interested (and hardcore)". i would not expect very many people to come up with a solution like that (or even to understand it, really), unless they had previously undertaken an impressively extensive study of number theory.

those are inherent properties of remainders upon division by 4 -- when you take the square of an odd integer and divide it by 4, you get a remainder of 1, and when you take the square of an even integer and divide it by 4, you get a remainder of 0.
there aren't any direct hints about this in the problem; hence, the "hardcore" introduction.

in other words, it's really not much worth worrying about (unless you have an unusually deep knowledge of number theory); this problem is perfectly accessible if you just throw numbers into the equation until you get it to work.

[RonPurewal]

by the way, people, the "remainders upon division by 4" solution is MUCH MUCH MUCH less important than the solution by pure brute force (i.e., just throwing possibilities into the equation until it works).
not even close.

DO NOT IGNORE THE "TRIAL & ERROR" SOLUTION, AND DO NOT BECOME UNDULY OBSESSED WITH THE ADVANCED NUMBER THEORY.
the chance that you will see another problem on which you'll be able to apply the "remainders upon division by 4" method is pretty much zero.
the chance that you will see another problem on which you'll be able to use trial and error is very, very, very high.

priorities, people; priorities.