Correct Relation

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GMATSUCKER: Senior | Next Rank: 100 Posts; Posts: 46; Joined: Sat Feb 27, 2010 2:58 am; Location: GMAT

by GMATSUCKER » Wed Mar 03, 2010 11:55 am

Q: Which one of the following is not the correct relation ;

a)25^3 + 38^3 + 87^3 =90^3

b)3^3 + 4^3 + 5^3 = 6^3

c)17^3 + 29^3 = 57^3

d)1^3 + 6^3 + 8^3 = 9^3

e)All of these

OAC

Now, is there a short cut formula to find the relation ?

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Source: Beat The GMAT — Problem Solving | Wed Mar 03, 2010 11:55 am

ajith: Legendary Member; Posts: 1275; Joined: Thu Sep 21, 2006 11:13 pmFollowed by: 2 members; Location: Arabian Sea; Thanked: 125 times

by ajith » Wed Mar 03, 2010 12:22 pm

GMATSUCKER wrote:Q: Which one of the following is not the correct relation ;

a)25^3 + 38^3 + 87^3 =90^3

b)3^3 + 4^3 + 5^3 = 6^3

c)17^3 + 29^3 = 57^3

d)1^3 + 6^3 + 8^3 = 9^3

e)All of these

OAC

Now, is there a short cut formula to find the relation ?

I cannot think about a formula but, can certainly think about a trick

a)25^3 + 38^3 + 87^3 =90^3

Last digit of 25^3 is 5; 38^3 is 2 and 87^3 is 3 => the sum ends in zero so does 90^3; so cannot rule out this option

b)3^3 + 4^3 + 5^3 = 6^3

This we can calculate 125+64 +27 = 216

c)17^3 + 29^3 = 57^3

Last Digit of 17^3 is 3; last digit of 29^3 is 9; sum ends in 2 but 57^3 must end in 3 => this equation cannot be true

d)1^3 + 6^3 + 8^3 = 9^3

We can calculate 512+216+1 = 729

Hence C

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GMATSUCKER: Senior | Next Rank: 100 Posts; Posts: 46; Joined: Sat Feb 27, 2010 2:58 am; Location: GMAT

by GMATSUCKER » Wed Mar 03, 2010 12:39 pm

ajith wrote:
GMATSUCKER wrote:Q: Which one of the following is not the correct relation ;

a)25^3 + 38^3 + 87^3 =90^3

b)3^3 + 4^3 + 5^3 = 6^3

c)17^3 + 29^3 = 57^3

d)1^3 + 6^3 + 8^3 = 9^3

e)All of these

OAC

Now, is there a short cut formula to find the relation ?
I cannot think about a formula but, can certainly think about a trick

a)25^3 + 38^3 + 87^3 =90^3

Last digit of 25^3 is 5; 38^3 is 2 and 87^3 is 3 => the sum ends in zero so does 90^3; so cannot rule out this option

b)3^3 + 4^3 + 5^3 = 6^3

This we can calculate 125+64 +27 = 216

c)17^3 + 29^3 = 57^3

Last Digit of 17^3 is 3; last digit of 29^3 is 9; sum ends in 2 but 57^3 must end in 3 => this equation cannot be true

d)1^3 + 6^3 + 8^3 = 9^3

We can calculate 512+216+1 = 729

Hence C

First of all many congratulations to Ajith on becoming the moderator !!

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kstv: Legendary Member; Posts: 610; Joined: Fri Jan 15, 2010 12:33 amFollowed by: 2 members; Thanked: 47 times

by kstv » Thu Mar 04, 2010 7:48 am

a)25^3 + 38^3 + 87^3 =90^3

b)3^3 + 4^3 + 5^3 = 6^3

c)17^3 + 29^3 = 57^3

d)1^3 + 6^3 + 8^3 = 9^3

aÂ³+bÂ³+cÂ³ < (a+b+c)Â³
This rule is not violated except in C where 17+29 < 57
57 = 17 + 29 + 11 so 57Â³ has to be > just 17Â³+29Â³

If the above is confusing consider aÂ²+bÂ² = cÂ²
always a+b > c
eg. 3Â²+4Â²=5Â², 8Â² has to be > 3Â²+4Â²

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Ian Stewart: GMAT Instructor; Posts: 2623; Joined: Mon Jun 02, 2008 3:17 amFollowed by: 355 members; Location: Montreal; Thanked: 1090 times; GMAT Score:780

by Ian Stewart » Thu Mar 04, 2010 10:23 am

GMATSUCKER wrote:
c)17^3 + 29^3 = 57^3

You do not need any inequality or units digit relationships to see that this cannot be true: on the left side of the equation, we're adding two odd numbers. That must give us an even number, and 57^3 is not even.

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melvinsgmat: Newbie | Next Rank: 10 Posts; Posts: 5; Joined: Sun Jan 17, 2010 5:46 pm

by melvinsgmat » Thu Mar 04, 2010 11:29 am

Ian Stewart wrote:
GMATSUCKER wrote:
c)17^3 + 29^3 = 57^3
You do not need any inequality or units digit relationships to see that this cannot be true: on the left side of the equation, we're adding two odd numbers. That must give us an even number, and 57^3 is not even.

Hi Ian,

Can you explain that point in detail. Do you mean that I need to add the the numbers on the left side to get a even number and thus it satisfies the relation.

taking the case of a

25+38+87=150 = LHS and it is a even number . RHS=90 that is also even and thus relation fine.

3+4+5=12= LHS and it is even also and RHS=6 is also even

17+29=46=LHS and even and RHS=57 and it is not even

1+6+8=15=LHS and odd and RHS=9 and is odd

Please clarify. thanks in advance.

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Ian Stewart: GMAT Instructor; Posts: 2623; Joined: Mon Jun 02, 2008 3:17 amFollowed by: 355 members; Location: Montreal; Thanked: 1090 times; GMAT Score:780

by Ian Stewart » Thu Mar 04, 2010 6:42 pm

melvinsgmat wrote:
Ian Stewart wrote:
GMATSUCKER wrote:
c)17^3 + 29^3 = 57^3
You do not need any inequality or units digit relationships to see that this cannot be true: on the left side of the equation, we're adding two odd numbers. That must give us an even number, and 57^3 is not even.
Hi Ian,

Can you explain that point in detail. Do you mean that I need to add the the numbers on the left side to get a even number and thus it satisfies the relation.

taking the case of a

25+38+87=150 = LHS and it is a even number . RHS=90 that is also even and thus relation fine.

3+4+5=12= LHS and it is even also and RHS=6 is also even

17+29=46=LHS and even and RHS=57 and it is not even

1+6+8=15=LHS and odd and RHS=9 and is odd

Please clarify. thanks in advance.

I'm not sure I understand your question, but I will try to explain: I was only pointing out that 17^3 is odd, and that 29^3 is odd. So when we add them, we must get an even number; 17^3 + 29^3 could not possibly be equal to 57^3, which is an odd number. The question asks us which relation is not correct, and this gives us a quick way to see that the relation in answer choice C is not correct. I wasn't doing any calculations using numbers at all; I was just considering whether the sum ought to be even or odd.

For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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GMATSUCKER: Senior | Next Rank: 100 Posts; Posts: 46; Joined: Sat Feb 27, 2010 2:58 am; Location: GMAT

by GMATSUCKER » Fri Mar 05, 2010 4:06 am

Thanks to everybody for providing their insights. Ajith's way was quite good : checking the unit's digit which is a surefire technique in case of doubt ; KSTV's way was also good using the formula method , but Ian's method was the fastest :- checking odd/even. With these three methods I don't think any student can ever make an error.
I thank all for their participation & look forward to their future participation.

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STEVEN SPIELBERG: Senior | Next Rank: 100 Posts; Posts: 43; Joined: Sat Mar 13, 2010 4:11 am; Thanked: 1 times

by STEVEN SPIELBERG » Sat Mar 13, 2010 11:24 am

C . Great ways to solve it !

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Aman verma: Master | Next Rank: 500 Posts; Posts: 304; Joined: Wed Jan 27, 2010 8:35 amFollowed by: 3 members; Location: International Space Station; Thanked: 11 times

by Aman verma » Sat Mar 13, 2010 11:57 am

Yes, good approach suggested by everybody . I would also like to know whether there is any formula based approach to solve these type of problems !!

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birdistheword: Newbie | Next Rank: 10 Posts; Posts: 1; Joined: Mon Apr 19, 2010 2:12 am; Location: london

by birdistheword » Mon Apr 19, 2010 2:44 am

Ian's suggested approach is best and fastest I suppose.

I used the approach to equate last digits. Though I would like to add a point that might help in saving a few precious seconds

In such questions I always tend to start from choice E. Since in this case you need to evaluate only 4 choices, if you had started from D, you would have found the answer in C.

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