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.
Hei wrote: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.
RPurewal wrote: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.
RonPurewal wrote: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.
pritesh.suvarna wrote:In the last section -
Any specific reason you have "divided by 4" to come to a conclusion that all numbers should be odd?
pritesh.suvarna wrote: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